Triangle Trigonometry: 16 Applications, Worked Step by Step

The Height of a Cliff Sighted at Two Angles of Elevation

methodWrite the Height Twice, Once from Each Point of Sighting, with the Exact Values of $\tan 30^\circ$ and $\tan 60^\circ$, and Set the Two Expressions Equal

A walker on level ground heads straight toward the foot of a vertical cliff. At point A the angle of elevation of the top of the cliff is 30^°. She walks 40 m straight toward the cliff to point B, where the angle of elevation is 60^°. Treat her eye as being at ground level, and give exact answers as well as answers to 1 decimal place. (a) How high is the cliff? (b) How far is B from the foot of the cliff?

Both angles of elevation are measured up from the horizontal ground, and the cliff stands at a right angle to it.

Let the foot of the cliff be F and its top C. Let the height CF be h m and the distance BF be x m. Both triangles CBF and CAF are right-angled at F, and each angle of elevation is measured from the horizontal ground.

step 1 of 5

An angle of elevation is measured up from the horizontal, so each sighting makes a right-angled triangle with the ground and the cliff. Neither triangle can be solved alone, because each has two unknown lengths. The height is the same in both, so write it from each triangle and set the two expressions equal.

Let the foot of the cliff be F and its top C. Let the height CF be h m and the distance BF be x m. Both triangles CBF and CAF are right-angled at F, and each angle of elevation is measured from the horizontal ground.
In triangle CBF the height is opposite the 60^° angle and BF is adjacent to it, so tan 60^° = hx. The exact value is tan 60^° = √3, so h = √3 x.
In triangle CAF the distance AF is x + 40, so tan 30^° = hx + 40. The exact value is tan 30^° = 1√3, so h = x + 40√3.
Set the two expressions for h equal: √3 x = x + 40√3. Multiply both sides by √3 to get 3x = x + 40, so 2x = 40 and x = 20. (b) B is 20 m from the foot of the cliff.
(a) The height is h = √3 × 20 = 20√3 m, which is 34.6 m to 1 decimal place. Check: A is 20 + 40 = 60 m from the foot, and 20√360 = √33 = 1√3, which is tan 30^°.

answer(a) 20√3 m, which is 34.6 m to 1 decimal place; (b) 20 m

techniqueElevation and Depression · Exact Sine, Cosine and Tangent at 30, 45 and 60° · Exact Trigonometric Ratios with Surds

examsO-Level · SAT · GCSE Higher

Common pitfalls

Writing tan 30^° = h40. The 40 m is only the walk from A to B; the side adjacent to the angle at A is the whole distance from A to the foot of the cliff, x + 40.
Rounding √3 to 1.7 at the start and carrying it through. The error grows at every step, so keep √3 exact until the last line and round once.

A Lighthouse Keeper Watching a Boat Come In

methodMeasure the Angle of Depression Down from the Horizontal Through the Lamp, Move It to the Boat as an Equal Alternate Angle, and Use Tangent Because the Height and the Distance Out Are the Two Sides Beside the Right Angle

The lamp L of a lighthouse is 42 m above the sea. The keeper sees a boat at P at an angle of depression of 35^°. Later the boat has sailed straight toward the foot F of the lighthouse, to Q, and the angle of depression is 50^°. Take tan 35^° = 0.700 and tan 50^° = 1.192, and give answers to 1 decimal place. (a) How far is the boat from the foot of the lighthouse at first? (b) How far does the boat sail between the two sightings?

The angle of depression is below the horizontal through the lamp; the alternate angle at the boat is also 35^°.

Draw the horizontal through the lamp. The angle of depression of 35^° lies between that horizontal and the line of sight LP. The horizontal and the sea are parallel, so the angle LPF at the boat is also 35^°: they are alternate angles.

step 1 of 5

An angle of depression is measured down from the horizontal at the observer's eye. The horizontal through the lamp is parallel to the sea, so the angle of depression from the lamp equals the angle of elevation of the lamp seen from the boat. That puts the angle inside a right-angled triangle whose two shorter sides are the height and the distance out.

Draw the horizontal through the lamp. The angle of depression of 35^° lies between that horizontal and the line of sight LP. The horizontal and the sea are parallel, so the angle LPF at the boat is also 35^°: they are alternate angles.
Triangle LFP is right-angled at F. The height LF = 42 m is opposite the 35^° angle at P and the distance FP is adjacent to it, so tan 35^° = 42FP.
(a) FP = 42tan 35^° = 420.700 = 60.0 m.
At the second sighting the angle LQF is 50^°, so FQ = 42tan 50^° = 421.192 = 35.23 m.
(b) The boat sails PQ = 60.0 − 35.23 = 24.77 m, which is 24.8 m to 1 decimal place. Check: the steeper angle belongs to the nearer boat, and 35.23 m is less than 60.0 m.

answer(a) 60.0 m; (b) 24.8 m

techniqueElevation and Depression · Solving Right Triangles · Sine, Cosine and Tangent

examsO-Level · SAT · GCSE Higher

Common pitfalls

Measuring the 35^° from the lighthouse tower instead of from the horizontal. The angle between the tower and the line of sight is 90 − 35 = 55^°; the angle of depression is the one below the horizontal.
Using sin 35^° = 42FP. Sine pairs the opposite side with the hypotenuse, which is the line of sight LP; the distance along the sea is the adjacent side, so tangent is the ratio that links it to the height.

A Wheelchair Ramp Built to a Limit on Its Slope

methodFind the Angle from the Rise and the Run with Inverse Tangent, Compare It with the Limit, Then Turn the Limit Round to Find the Run It Needs

A wheelchair ramp must rise 0.35 m from a path to a doorway. The building rules say a ramp may make an angle of at most 5^° with the horizontal. The builder plans a horizontal run of 3.8 m. Take tan⁻¹(0.0921) = 5.26^° and tan 5^° = 0.0875. (a) What angle does the planned ramp make with the horizontal, to 1 decimal place, and is it within the rules? (b) What is the shortest horizontal run the rules allow, to 2 decimal places?

The rise is opposite the angle of the ramp and the run is adjacent: tan θ = 0.353.8 = 0.0921.

Let θ be the angle between the ramp and the horizontal. The rise of 0.35 m is opposite θ and the run of 3.8 m is adjacent to it, so tan θ = 0.353.8 = 0.0921.

step 1 of 4

The ramp, the ground and the doorstep make a right-angled triangle. The rise is opposite the angle of the ramp and the run is adjacent to it, so tangent links the two lengths to the angle. Inverse tangent finds the angle from the lengths, and the same equation, with the angle fixed at the limit, finds the run.

Let θ be the angle between the ramp and the horizontal. The rise of 0.35 m is opposite θ and the run of 3.8 m is adjacent to it, so tan θ = 0.353.8 = 0.0921.
(a) θ = tan⁻¹(0.0921) = 5.26^°, which is 5.3^° to 1 decimal place. That is more than 5^°, so the planned ramp is too steep.
For the steepest ramp the rules allow, the angle is 5^° and the run r m is unknown: tan 5^° = 0.35r.
Multiply both sides by r and divide by tan 5^°: r = 0.350.0875 = 4.00. (b) The run must be at least 4.00 m. Check: a longer run gives a gentler slope, and 4.00 m is longer than the planned 3.8 m, which is why the plan was too steep.

answer(a) 5.3^°, which is more than 5^°, so the ramp is too steep; (b) 4.00 m

techniqueFinding an Angle from a Ratio · Sine, Cosine and Tangent · Solving Right Triangles

examsO-Level · SAT · GCSE Higher

Common pitfalls

Working out tan⁻¹(3.80.35), with the run on top. That gives the angle at the doorstep, 84.7^°; the angle with the ground has the rise, the opposite side, on top.
Using the length of the sloping ramp as the run. The rule and the tangent both use the HORIZONTAL run along the ground; the sloping surface is the hypotenuse, which is a little longer.

A Boat Sighted from Two Lifeguard Towers on a Straight Beach

methodFind the Third Angle, Pair the Wanted Side with the Sine of the Angle Opposite It in the Sine Rule, Then Drop a Perpendicular to the Beach for the Distance Out

Two lifeguard towers A and B stand 2 km apart on a straight beach. A fishing boat F is out at sea. The angle FAB is 62^° and the angle FBA is 48^°. Take sin 48^° = 0.743, sin 70^° = 0.940 and sin 62^° = 0.883, and give answers to 2 decimal places. (a) How far is the boat from tower A? (b) How far is the boat from the beach?

The angles of the triangle add to 180^°, so the angle at the boat is 180 − 62 − 48 = 70^°.

The angles of triangle ABF add to 180^°, so the angle at the boat is 180 − 62 − 48 = 70^°.

step 1 of 5

Triangle ABF has one known side and two known angles, and it has no right angle, so the sine rule is the tool: in any triangle each side divided by the sine of the angle opposite it gives the same number. Once one slant distance is known, a perpendicular to the beach makes a right-angled triangle for the distance out.

The angles of triangle ABF add to 180^°, so the angle at the boat is 180 − 62 − 48 = 70^°.
AF is opposite the 48^° angle at B, and AB = 2 km is opposite the 70^° angle at F. The sine rule gives AFsin 48^° = 2sin 70^°.
(a) AF = 2 × 0.7430.940 = 1.4860.940 = 1.581, so the boat is 1.58 km from tower A.
The distance from the beach is the perpendicular FN from the boat to the beach. Triangle ANF is right-angled at N, AF is its hypotenuse, and FN is opposite the 62^° angle at A, so FN = AF sin 62^°.
(b) FN = 1.581 × 0.883 = 1.396, so the boat is 1.40 km from the beach. Check: 1.40 km is less than AF, as the perpendicular to a line is the shortest distance to it.

answer(a) 1.58 km; (b) 1.40 km

techniqueThe Sine Rule · Solving Right Triangles · Sine, Cosine and Tangent

examsO-Level · GCSE Higher

Common pitfalls

Pairing AF with sin 62^°, the angle at A itself. In the sine rule a side goes with the angle OPPOSITE it, and the angle opposite AF is the one at B.
Giving AF as the distance from the beach. AF runs at a slant to the beach; the distance from the beach is measured along the perpendicular FN.

A Surveyor Measuring Across a Lake from a Point on the Shore

methodUse the Cosine Rule for the Side Opposite the Measured Angle, Then the Sine Rule for an Angle, Rejecting the Obtuse Angle Because the Side Opposite It Is Not the Longest

A surveyor cannot measure straight across a lake from a jetty J to a boathouse H, so she stands at a point P on the shore where she can see both. She measures PJ = 350 m, PH = 420 m and the angle JPH = 72^°. Take cos 72^° = 0.309, sin 72^° = 0.951 and sin⁻¹(0.8757) = 61.1^°. (a) How far is it across the lake from the jetty to the boathouse, to 1 decimal place? (b) What is the angle PJH at the jetty, to 1 decimal place?

Two sides and the angle between them are known, so the cosine rule gives the third side.

The sides PJ and PH and the angle between them are known, so the cosine rule gives the side opposite that angle: JH² = PJ² + PH² − 2 × PJ × PH × cos 72^°.

step 1 of 5

Two sides and the angle between them fix a triangle, and the cosine rule gives the third side: it is Pythagoras with a correction for an angle that is not 90^°. With all three sides known, the sine rule gives another angle, and the sizes of the sides decide which of the two angles with that sine is the right one.

The sides PJ and PH and the angle between them are known, so the cosine rule gives the side opposite that angle: JH² = PJ² + PH² − 2 × PJ × PH × cos 72^°.
JH² = 122500 + 176400 − 2 × 350 × 420 × 0.309 = 298900 − 90846 = 208054.
(a) JH = √208054 = 456.1 m, the distance across the lake.
For the angle at J, PH = 420 m is the side opposite it. The sine rule gives sin J420 = sin 72^°456.1, so sin J = 420 × 0.951456.1 = 399.42456.1 = 0.8757.
(b) J = sin⁻¹(0.8757) = 61.1^°. The other angle with the same sine, 180 − 61.1 = 118.9^°, is rejected: PH is shorter than JH, so the angle at J must be smaller than the 72^° angle at P. Check: the third angle is 180 − 72 − 61.1 = 46.9^°, the smallest, opposite the shortest side PJ.

answer(a) 456.1 m; (b) 61.1^°

techniqueThe Cosine Rule · Finding Angles with the Sine Rule · The Sine Rule

examsO-Level · GCSE Higher

Common pitfalls

Writing JH² = 350² + 420². That is Pythagoras, which holds only when the angle at P is 90^°; here it is 72^°, so the term 2 × 350 × 420 × cos 72^° must be taken off.
Working out 350² + 420² − 2 × 350 × 420 first and then multiplying by cos 72^°. The product 2 × 350 × 420 × cos 72^° is a single term, taken off the sum of the two squares.

The Pitch of a Barn Roof from the Lengths of Its Timbers

methodRearrange the Cosine Rule for an Angle, Subtracting the Square of the Side Opposite It, Recognize the Exact Value, Then Use the Exact Sine for the Height of the Ridge

The end frame of a barn roof is a triangle ABC. The horizontal tie beam AC is 8 m long, and two rafters meet at the ridge B: AB is 5 m and BC is 7 m. (a) What angle does the rafter AB make with the tie beam? (b) How high is the ridge B above the tie beam? Give the exact value and the value to 2 decimal places.

The side opposite the angle at A is BC, so its square is the one subtracted: cos A = AB² + AC² − BC²2 × AB × AC.

All three sides are known and the angle at A is wanted. The side opposite A is BC, so it is the one subtracted: cos A = AB² + AC² − BC²2 × AB × AC.

step 1 of 5

When all three sides of a triangle are known, the cosine rule rearranged gives any angle: cos A = b² + c² − a²2bc, where a is the side opposite A. A cosine of exactly 12 is an exact angle, and its exact sine then gives the height without a calculator.

All three sides are known and the angle at A is wanted. The side opposite A is BC, so it is the one subtracted: cos A = AB² + AC² − BC²2 × AB × AC.
cos A = 25 + 64 − 492 × 5 × 8 = 4080 = 12.
(a) cos 60^° = 12 exactly, so the rafter AB makes an angle of 60^° with the tie beam.
Drop the perpendicular BN from the ridge to the tie beam. In the right-angled triangle ABN the rafter AB is the hypotenuse and the height BN is opposite the 60^° angle, so BN = 5 sin 60^° = 5 × √32.
(b) BN = 5√32 m, which is 4.33 m to 2 decimal places. Check: AN = 5 cos 60^° = 2.5 m, so NC = 8 − 2.5 = 5.5 m, and 5.5² + 4.33² = 30.25 + 18.75 = 49 = 7², the length of the other rafter.

answer(a) 60^°; (b) 5√32 m, which is 4.33 m to 2 decimal places

techniqueFinding Angles with the Cosine Rule · Exact Sine, Cosine and Tangent at 30, 45 and 60° · Exact Trigonometric Ratios with Surds

examsO-Level · GCSE Higher

Common pitfalls

Subtracting the wrong square: 5² + 7² − 8²2 × 5 × 7 is the cosine of the angle at the ridge B, not at A. The square subtracted is always that of the side opposite the angle wanted.
Taking the height as 5 cos 60^° = 2.5 m. Cosine gives the side ADJACENT to the angle, which is AN along the tie beam; the height is opposite the angle, so it needs sine.

Fertilizing a Triangular Field with an Obtuse Corner

methodTake Half the Product of Two Sides and the Sine of the Angle Between Them, Using the Sine of the Supplement for the Obtuse Angle, Then Count the Hundreds of Square Meters

A farmer's field is a triangle PQR. Two fences meet at the gate P: PQ = 80 m and PR = 110 m, and the angle QPR between them is 150^°. (a) What is the area of the field? (b) Fertilizer for the field costs $3.50 for every 100 m². How much does it cost to fertilize the whole field?

Two sides from the gate and the angle between them: area = 12ab sin C.

The two sides from the gate and the angle between them are known, so the area is 12ab sin C with a = 80, b = 110 and C = 150^°.

step 1 of 4

The area of a triangle is half the base times the height, and when two sides and the angle between them are known the height is one side times the sine of that angle. That gives area = 12ab sin C, which works for an obtuse angle too, because an angle and its supplement have the same sine.

The two sides from the gate and the angle between them are known, so the area is 12ab sin C with a = 80, b = 110 and C = 150^°.
The angle is obtuse. An angle and its supplement have the same sine, so sin 150^° = sin 30^° = 12.
(a) The area is 12 × 80 × 110 × 12 = 2200 m².
The field holds 2200 ÷ 100 = 22 lots of 100 m². (b) The fertilizer costs 22 × 3.50 = $77. Check: the height from R to the line QP, extended past P, is 110 sin 30^° = 55 m, and 12 × 80 × 55 = 2200 m².

answer(a) 2200 m²; (b) $77

techniqueArea with Sine · Sine and Cosine of Angles Past 90 Degrees

examsO-Level · GCSE Higher

Common pitfalls

Taking sin 150^° as negative, or using cos 150^°. The sine of an obtuse angle is positive, equal to the sine of its supplement, so the area comes out positive, as an area must.
Working out 80 × 110 × sin 150^° and forgetting the 12. That is the area of a parallelogram with these two sides; the triangle is half of it.

Where a Driver on a Straight Road Picks Up and Loses a Radio Signal

methodUse the Sine Rule for the Angle at the Driver and Keep Both Angles with That Sine, Because the Signal Circle Cuts the Road Twice, Then Drop a Perpendicular from the Mast for the Two Distances

A straight road runs from a junction A. A radio mast M stands 8 km from the junction, and the angle between the road and the line AM is 30^°. The mast's signal reaches 5 km in every direction. A driver on the road picks up the signal at C and loses it at D, further along, so MC = MD = 5 km. Take sin⁻¹(0.8) = 53.1^°. (a) Find the angle ACM and the angle ADM. (b) How far along the road from the junction are C and D? Give exact values and values to 2 decimal places.

The sine rule in triangle AMC: sin C8 = sin 30^°5, so sin C = 0.8.

In triangle AMC, MC = 5 km is opposite the 30^° angle at A and AM = 8 km is opposite the angle at C. The sine rule gives sin C8 = sin 30^°5, so sin C = 8 × 0.55 = 0.8.

step 1 of 5

The side 5 km opposite the known 30^° angle is shorter than the other known side, 8 km, so the sine rule gives a sine that belongs to two angles, one acute and one obtuse. That is the ambiguous case, and here both triangles exist: they are the triangles at the two points where the road enters and leaves the circle of the signal.

In triangle AMC, MC = 5 km is opposite the 30^° angle at A and AM = 8 km is opposite the angle at C. The sine rule gives sin C8 = sin 30^°5, so sin C = 8 × 0.55 = 0.8.
Two angles between 0^° and 180^° have a sine of 0.8: sin⁻¹(0.8) = 53.1^° and 180 − 53.1 = 126.9^°. Both are possible, because 5 km is less than 8 km but more than the distance from the mast to the road, so the circle of radius 5 km about M cuts the road twice.
(a) At the nearer point C the triangle AMC has its obtuse angle at C, so the angle ACM is 126.9^°. At the farther point D the angle ADM is 53.1^°. Triangle MCD is isosceles, so its base angle at C is also 53.1^°, and 126.9 + 53.1 = 180 along the straight road.
Drop the perpendicular MN to the road. MN = 8 sin 30^° = 4 km and AN = 8 cos 30^° = 4√3 km. In the right-angled triangle MNC, CN = √5² − 4² = √9 = 3 km, and ND = 3 km as well.
(b) AC = 4√3 − 3, which is 3.93 km, and AD = 4√3 + 3, which is 9.93 km, to 2 decimal places. Check: the driver has the signal for 9.93 − 3.93 = 6 km, which is CN + ND = 3 + 3.

answer(a) angle ACM = 126.9^° and angle ADM = 53.1^°; (b) AC = 4√3 − 3, which is 3.93 km, and AD = 4√3 + 3, which is 9.93 km

techniqueThe Ambiguous Case of the Sine Rule · Finding Angles with the Sine Rule · The Sine Rule

examsO-Level

Common pitfalls

Stopping at 53.1^°, the only angle a calculator gives for sin⁻¹(0.8). The sine of the obtuse angle 126.9^° is also 0.8, and here that second triangle is real: it is the one at the point where the signal is picked up.
Expecting two triangles every time the sine rule gives an angle. The second triangle exists only when the side opposite the known angle, here 5 km, is longer than the perpendicular distance, 4 km, and shorter than the other known side, 8 km.

A Ship's Two Legs and the Course Straight Back to Port

methodTurn the Two Bearings into the Angle Between the Legs, Use the Cosine Rule for the Distance Home and Again for the Angle at the Ship, Then Measure That Angle from the North Line to Give the Bearing

A ship leaves port P and sails on a bearing of 060^° for 8 km to a buoy A. It then sails due south for 3 km to a point B. Take cos⁻¹(−17) = 98.2^°. (a) How far is the ship from the port? (b) On what bearing, to the nearest degree, must it sail to return straight to the port?

The bearing of P from A is 060 + 180 = 240^° and the ship leaves on 180^°, so the angle at A is 60^°.

Draw a north line at A. The ship arrived on 060^°, so the bearing of P from A is 060 + 180 = 240^°. It leaves A due south, on 180^°. The angle PAB between the two legs is 240 − 180 = 60^°.

step 1 of 5

A bearing is measured clockwise from north at the point where it is taken. To use the cosine rule, the two bearings at the turning point are first turned into the angle between the two legs. The cosine rule then gives the distance home, and, rearranged, the angle at the ship, whose sign shows whether it is obtuse.

Draw a north line at A. The ship arrived on 060^°, so the bearing of P from A is 060 + 180 = 240^°. It leaves A due south, on 180^°. The angle PAB between the two legs is 240 − 180 = 60^°.
Two sides and the angle between them are known, so the cosine rule gives the third side: PB² = 8² + 3² − 2 × 8 × 3 × cos 60^° = 64 + 9 − 24 = 49.
(a) PB = √49 = 7 km.
For the angle at B the side opposite it is PA = 8 km: cos B = 3² + 7² − 8²2 × 3 × 7 = −642 = −17, so the angle ABP is 98.2^°. The cosine is negative, so the angle is obtuse.
A is due north of B, so the angle ABP is measured from the north line at B, turning toward the west. (b) The bearing of the port from B is 360 − 98.2 = 261.8^°, which is 262^° to the nearest degree. Check: the port lies a little south of due west of B, and 262^° is between 180^° and 270^°.

answer(a) 7 km; (b) 262^°

techniqueBearings with the Sine and Cosine Rules · The Cosine Rule · Finding Angles with the Cosine Rule

examsO-Level · GCSE Higher

Common pitfalls

Finding the angle at B with the sine rule: sin B = 8 sin 60^°7 = 0.990 gives 81.8^°. The sine rule cannot tell 81.8^° from 98.2^°, and B is obtuse here; the negative cosine is what shows it.
Giving the bearing as 098^°, the angle inside the triangle. A bearing is measured clockwise from north, and the port lies to the west of B, so the bearing is 360^° minus that angle.

A Camera Cable Across a Sports Hall from Corner to Corner

methodFind the Diagonal of the Floor First, Then Stand a Right-Angled Triangle on It, and Measure the Angle with the Floor Between the Cable and Its Projection

A camera hangs from a cable stretched straight from a top corner G of a sports hall to the bottom corner A diagonally opposite. The hall is a cuboid 12 m long, 9 m wide and 6 m high. Take tan⁻¹(0.4) = 21.8^°. (a) How long is the cable, to 1 decimal place? (b) What angle does the cable make with the floor?

The cable AG runs from the top corner G down to A. Straight below it on the floor is the diagonal AC.

Label the floor ABCD with AB = 12 m and BC = 9 m, and let G be the top corner above C, so CG = 6 m. The cable is AG. Straight below it on the floor is the diagonal AC, the projection of the cable.

step 1 of 5

A problem in three dimensions is solved in flat right-angled triangles picked out of the solid. The angle between a line and a plane is the angle between the line and its projection on the plane, the line straight below it, so the floor diagonal under the cable is the side to work with.

Label the floor ABCD with AB = 12 m and BC = 9 m, and let G be the top corner above C, so CG = 6 m. The cable is AG. Straight below it on the floor is the diagonal AC, the projection of the cable.
Triangle ABC on the floor is right-angled at B, so AC² = 12² + 9² = 144 + 81 = 225 and AC = 15 m.
Triangle ACG stands upright and is right-angled at C. (a) AG² = 15² + 6² = 225 + 36 = 261, so AG = √261 = 16.2 m to 1 decimal place.
The angle between the cable and the floor is the angle GAC between the cable and its projection. The height 6 m is opposite it and AC = 15 m is adjacent, so tan θ = 615 = 0.4.
(b) θ = tan⁻¹(0.4) = 21.8^°. Check: sin 21.8^° = 0.371, and 16.2 × 0.371 = 6.0 m, the height of the hall.

answer(a) 16.2 m; (b) 21.8^°

techniqueTrigonometry in Three Dimensions · Finding an Angle from a Ratio · Solving Right Triangles

examsO-Level · GCSE Higher

Common pitfalls

Using the length of the hall, 12 m, as the adjacent side: tan⁻¹(612) is the angle with the floor of a line up the end wall. The cable crosses the hall, so its projection on the floor is the diagonal AC.
Finding tan⁻¹(156) = 68.2^°. That is the angle at G, between the cable and the upright corner post; opposite and adjacent swap when the angle moves to the other end of the triangle.

The Depth of Water at a Harbor Mouth Through the Day

methodRead the Amplitude as the Rise Above the Mean Level, the Principal Axis as Mean Sea Level and $360^\circ$ Divided by the Coefficient of $x$ as the Time Between High Waters

At a harbor mouth the depth of water, d meters, x hours after midnight is modeled by d = 3 sin (30x)^° + 5 for 0 ≤ x ≤ 24. (a) What is the depth at high water, the depth at low water, and the mean depth about which the water rises and falls? (b) How long is it from one high water to the next, and at what time is the first high water after midnight?

The sine lies between −1 and 1, so the water rises and falls 3 m about its middle level: the amplitude is 3 m.

The sine of any angle lies between −1 and 1, so 3 sin (30x)^° lies between −3 and 3. The amplitude is 3 m: the water rises 3 m above its middle level and falls 3 m below it.

step 1 of 4

In a model d = a sin (bx)^° + c each number has a meaning. The principal axis d = c is the middle level, the amplitude a is how far the curve rises above it and falls below it, and the period 360b is the length of one full cycle, here the time from one high water to the next.

The sine of any angle lies between −1 and 1, so 3 sin (30x)^° lies between −3 and 3. The amplitude is 3 m: the water rises 3 m above its middle level and falls 3 m below it.
The principal axis is d = 5, from the + 5 in the model. (a) The mean depth is 5 m, the depth at high water is 5 + 3 = 8 m and the depth at low water is 5 − 3 = 2 m.
The sine repeats every 360^°, and 30x grows by 360 when x grows by 36030 = 12. The period is 12 hours, the time from one high water to the next.
High water is where sin (30x)^° = 1, first when 30x = 90, so x = 3. (b) It is 12 hours from one high water to the next, and the first high water is at 3 a.m. Check: at x = 3, d = 3 sin 90^° + 5 = 3 + 5 = 8 m, and the next high water is at x = 3 + 12 = 15, which is 3 p.m.

answer(a) 8 m at high water, 2 m at low water, and a mean depth of 5 m; (b) 12 hours, and the first high water is at 3 a.m.

techniqueModeling Tides and Daylight with a Sine Function · Amplitude and the Principal Axis · The Period and Phase Shift of a Wave

examsO-Level

Common pitfalls

Reading the amplitude 3 as the depth at high water. The amplitude is how far the water rises ABOVE the mean level, so the depth at high water is 5 + 3 = 8 m.
Taking the period as 30 hours, from the 30 in the model. The 30 is how many degrees the angle turns in one hour, and a full cycle of 360^° takes 360 ÷ 30 = 12 hours.

When a Ferris Wheel Rider Is High Above the Ground

methodSolve for the Cosine, Take the Principal Value and the Second Angle in One Turn, Add Whole Turns Across the Stretched Interval of $12x$, and Only Then Divide by $12$

A rider boards a Ferris wheel at its lowest point. Her height above the ground, h meters, x minutes after boarding is h = 20 − 18 cos (12x)^°. (a) At what times in the first hour is she exactly 29 m above the ground? (b) For how long in each turn of the wheel is she higher than 29 m?

Set the height equal to 29: −18 cos (12x)^° = 9, so cos (12x)^° = −12.

Set the height equal to 29: 20 − 18 cos (12x)^° = 29, so −18 cos (12x)^° = 9 and cos (12x)^° = −12.

step 1 of 5

Setting the height equal to 29 gives a trigonometric equation in the angle 12x. A calculator gives only the principal value, so the other angles with the same cosine come from the symmetry of the cosine curve and from adding whole turns. Because the angle is 12x, an hour of x is 720^° of angle, two full turns, and every solution in that range counts.

Set the height equal to 29: 20 − 18 cos (12x)^° = 29, so −18 cos (12x)^° = 9 and cos (12x)^° = −12.
The principal value is cos⁻¹(−12) = 120^°. The cosine curve is symmetrical about 180^°, so the other angle in one turn with the same cosine is 360 − 120 = 240^°.
In the first hour 0 ≤ x ≤ 60, so the angle 12x runs from 0^° to 720^°, two full turns. Adding 360^° to each angle gives four solutions: 12x = 120, 240, 480 and 600.
(a) Divide each angle by 12: she is exactly 29 m up at x = 10, 20, 40 and 50 minutes after boarding.
Between x = 10 and x = 20 the cosine is less than −12, so the height is more than 29 m. (b) She is higher than 29 m for 20 − 10 = 10 minutes in each 30-minute turn. Check: at x = 15, h = 20 − 18 cos 180^° = 20 + 18 = 38 m, the top of the wheel.

answer(a) 10, 20, 40 and 50 minutes after boarding; (b) 10 minutes

techniqueTrigonometric Equations with a Multiple Angle · Solving Simple Trigonometric Equations · The Principal Values of the Inverse Functions

examsO-Level

Common pitfalls

Giving only x = 10, the principal value divided by 12. The cosine has two angles in every turn, and the hour holds two turns of 12x, so there are four times.
Looking for angles between 0^° and 360^° only. That range of 12x covers just the first 30 minutes; the range for 12x is twelve times the range for x, so it runs to 720^°.

A Painter on a Ladder Who Needs the Angle at the Wall

methodFind the Angle at the Top as the Complement of the Angle at the Foot, Then Read Its Sine and Cosine from the Cosine and Sine of the Angle at the Foot

A ladder 6 m long stands on level ground and leans against a vertical wall, making an angle of 76^° with the ground. There is no calculator, only the values sin 76^° = 0.9703 and cos 76^° = 0.2419. (a) What angle does the ladder make with the wall, and what are the sine and cosine of that angle? (b) A painter stands on a rung 1.5 m from the top of the ladder, measured along the ladder. How far is that rung from the wall, and how far below the top of the ladder is it? Give both to 2 decimal places.

The wall and the ground meet at a right angle, so the angle at the top is 90^° − 76^° = 14^°.

The wall meets the ground at a right angle, so the angles of the triangle at the foot of the ladder and at the top add to 90^°. The angle between the ladder and the wall is 90 − 76 = 14^°.

step 1 of 5

The ladder, the wall and the ground make a right-angled triangle, and its two acute angles add to 90^°. Each side beside the right angle is opposite one of those angles and adjacent to the other, so the sine of one angle is the cosine of the other: sin (90^° − θ) = cos θ and cos (90^° − θ) = sin θ. That is how two given values answer questions about a second angle.

The wall meets the ground at a right angle, so the angles of the triangle at the foot of the ladder and at the top add to 90^°. The angle between the ladder and the wall is 90 − 76 = 14^°.
The distance from the foot of the ladder to the wall is opposite the 14^° angle and adjacent to the 76^° angle, so sin 14^° = cos 76^°. The height up the wall is adjacent to 14^° and opposite 76^°, so cos 14^° = sin 76^°. (a) The ladder makes 14^° with the wall, with sin 14^° = 0.2419 and cos 14^° = 0.9703.
Let R be the rung, T the top of the ladder and K the point on the wall level with R. Triangle RKT is right-angled at K, its hypotenuse RT is 1.5 m, and its angle at T is the 14^° between the ladder and the wall.
RK is opposite the 14^° angle, so RK = 1.5 sin 14^° = 1.5 × 0.2419 = 0.36285 m. KT is adjacent to it, so KT = 1.5 cos 14^° = 1.5 × 0.9703 = 1.45545 m.
(b) The rung is 0.36 m from the wall and 1.46 m below the top of the ladder. Check: 0.36285² + 1.45545² = 0.132 + 2.117 = 2.249, which is 1.5² = 2.25 to the accuracy of the values used.

answer(a) 14^°, with sin 14^° = 0.2419 and cos 14^° = 0.9703; (b) 0.36 m from the wall and 1.46 m below the top

techniqueSine and Cosine of Complementary Angles · Sine, Cosine and Tangent · Solving Right Triangles

examsSAT

Common pitfalls

Using sin 76^° for the distance from the wall. In the small triangle at the top, the side out from the wall is opposite the 14^° angle, so it is 1.5 sin 14^°; 1.5 sin 76^° = 1.45545 m is the drop down the wall.
Deciding that a new angle needs a new value from a calculator. The two acute angles of a right-angled triangle share their sides, so sin 14^° is already known: it is cos 76^°.

Hours of Daylight in a City in Northern Europe Through the Year

methodTake the Principal Axis Halfway Between the Longest and Shortest Days and the Amplitude as Half Their Difference, Find $b$ from a Period of $12$ Months, Then Use the Symmetry of the Cosine Graph About Its Peak

In a city in northern Europe the longest day, June 21, has 16.5 hours of daylight and the shortest, December 21, has 7.5 hours. The number of hours of daylight, D, x months after June 21 is modeled by D = a cos (bx)^° + c, taking each month as one twelfth of a year. (a) Find a, b and c. (b) Between which dates does the city have more than 14.25 hours of daylight?

The principal axis is halfway between the longest and shortest days: c = 16.5 + 7.52 = 12.

The principal axis is halfway between the greatest and least values: c = 16.5 + 7.52 = 12 hours.

step 1 of 5

The graph of y = a cos (bx)^° + c is the cosine wave moved and stretched. Its principal axis, the midline, is y = c; the amplitude a is how far it rises above that line; one full cycle takes 360b; and the graph of cosine starts at a peak. Once the model is built, the symmetry of the graph about each peak gives a second crossing of a level from the first.

The principal axis is halfway between the greatest and least values: c = 16.5 + 7.52 = 12 hours.
The amplitude is how far the curve rises above the principal axis: a = 16.5 − 7.52 = 4.5 hours. At x = 0, June 21, the daylight is at its greatest, and the graph of cosine starts at its peak, so the model is a cosine with a positive.
One cycle is a year of 12 months, and bx must grow by 360 in that time, so b = 36012 = 30. (a) a = 4.5, b = 30 and c = 12, so D = 4.5 cos (30x)^° + 12. Check: at x = 6, December 21, D = 4.5 cos 180^° + 12 = 12 − 4.5 = 7.5 hours.
Set D = 14.25: 4.5 cos (30x)^° = 2.25, so cos (30x)^° = 12. The principal value is 30x = 60, so x = 2, which is August 21.
The graph is symmetrical about each of its peaks. The next peak is at x = 12, June 21 of the next year, so the curve also crosses D = 14.25 two months before it, at x = 10, which is April 21, and it is above the line from x = 10 to x = 14. (b) The city has more than 14.25 hours of daylight from April 21 to August 21, four months of the year. Check: at x = 11, May 21, D = 4.5 cos 330^° + 12 = 15.9 hours.

answer(a) a = 4.5, b = 30, c = 12; (b) from April 21 to August 21, 4 months, for x from 10 to 14

techniqueThe Graphs of Sine and Cosine · Amplitude and the Principal Axis · The Period and Phase Shift of a Wave

examsO-Level · GCSE Higher

Common pitfalls

Using a sine: D = 4.5 sin (30x)^° + 12 gives 12 hours at x = 0, but June 21 is the longest day, where the graph must be at its peak. The graph of sine starts on its principal axis; the graph of cosine starts at its peak.
Giving the dates from August 21 to April 21. Between those dates the curve is below the line D = 14.25; the stretch above it lies on either side of the peak at June 21.

A Rotating Warning Light and the Spot It Sweeps Along a Wall

methodWrite the Spot's Distance Along the Wall as Opposite over Adjacent, $y = 8 \tan \theta$, and Read the Graph of Tangent Toward Its Asymptote

A rotating warning light at a building site is 8 m from a long straight wall, and F is the point of the wall nearest to it. As the light turns, its beam makes an angle θ with the line from the light to F and lights a spot on the wall y meters from F, where y = 8 tan θ for −90^° < θ < 90^°. (a) How far from F is the spot when θ = 30^° and when θ = 60^°? Give exact values and values to 2 decimal places. (b) The wall runs 40 m from F in each direction. For what values of θ is the spot on the wall? How far from F would the wall have to run to catch the beam at θ = 89^°? Take tan⁻¹(5) = 78.7^° and tan 89^° = 57.29.

The line from the light to F is at right angles to the wall, so tan θ = y8 and y = 8 tan θ.

The line from the light to F is at right angles to the wall, so the triangle from the light to F to the spot is right-angled at F. The 8 m is adjacent to θ and y is opposite it, so tan θ = y8 and y = 8 tan θ.

step 1 of 5

The light, the point F and the spot make a right-angled triangle, and the distance along the wall is opposite the angle of the beam. The graph of y = tan θ rises ever more steeply toward θ = 90^° and has an asymptote there: the beam turns parallel to the wall, and the spot runs off along it without limit.

The line from the light to F is at right angles to the wall, so the triangle from the light to F to the spot is right-angled at F. The 8 m is adjacent to θ and y is opposite it, so tan θ = y8 and y = 8 tan θ.
(a) At 30^°, y = 8 tan 30^° = 8√3 = 8√33 ≈ 4.62 m. At 60^°, y = 8 tan 60^° = 8√3 ≈ 13.86 m. The second 30^° of turn moves the spot 9.24 m, twice as far as the first, because the graph of tangent grows steeper as θ grows.
The spot reaches the end of the wall where 8 tan θ = 40, so tan θ = 5 and θ = tan⁻¹(5) = 78.7^°. The graph of y = tan θ is symmetrical about the origin, so the spot reaches the other end, 40 m on the other side of F, at θ = −78.7^°.
At 89^°, y = 8 × 57.29 = 458.3 m. As θ nears 90^° the beam turns parallel to the wall and tan θ grows without limit: the line θ = 90^° is an asymptote of the graph, and at 90^° itself the beam never meets the wall.
(b) The spot is on the wall for −78.7^° ≤ θ ≤ 78.7^°, and to catch the beam at 89^° the wall would have to run 458.3 m from F. Check: 8 tan 78.7^° = 8 × 4.99 = 39.9, which is 40 m to the accuracy of the angle.

answer(a) 8√33 ≈ 4.62 m and 8√3 ≈ 13.86 m from F; (b) −78.7^° ≤ θ ≤ 78.7^°, and 458.3 m

techniqueThe Graph of Tangent · Exact Sine, Cosine and Tangent at 30, 45 and 60° · Exact Trigonometric Ratios with Surds

examsO-Level · GCSE Higher

Common pitfalls

Treating the spot as moving the same distance for every degree: 30^° of turn moves it 4.62 m, so 60^° would move it 9.24 m. The distance is 8 tan θ, not a fixed amount per degree, and 8 tan 60^° = 13.86 m.
Adding the period of 180^° and including θ = 258.7^°, where tan θ = 5 as well. The graph of tangent repeats every 180^°, but at 258.7^° the beam points away from the wall, which is why the model holds only for −90^° < θ < 90^°.

The Angle of a Ski Jump Kicker for a Chosen Height in the Air

methodSet the Model Equal to the Height, Substitute a Letter for $\sin \theta$, Factor the Quadratic, and Keep Only a Sine Between $-1$ and $1$ with $\theta$ in the Stated Range

A freestyle skier rides up a straight kicker 7.5 m long, built at an angle θ to the flat snow, and leaves its lip at 10 m/s. Ignoring air resistance and taking g = 10 m/s², the highest point of the jump is H meters above the snow, where H = 7.5 sin θ + 5 sin² θ for 0^° < θ < 90^°. (a) At what angle must the kicker be built for the highest point to be 5 m above the snow? (b) How high is the lip above the snow, and how far does the skier rise above the lip?

Set H = 5: 7.5 sin θ + 5 sin² θ = 5, a quadratic in sin θ.

Set the height equal to 5: 7.5 sin θ + 5 sin² θ = 5. The sine appears squared and on its own, so this is a quadratic in sin θ.

step 1 of 5

An equation with a sin² θ term and a sin θ term is a quadratic in sin θ. Substitute a letter for sin θ, factor as usual, and then turn each root back into a sine. A sine always lies between −1 and 1, so a root outside that interval gives no angle, and the range in the question decides which angles remain.

Set the height equal to 5: 7.5 sin θ + 5 sin² θ = 5. The sine appears squared and on its own, so this is a quadratic in sin θ.
Let s = sin θ. Then 5s² + 7.5s − 5 = 0, and multiplying every term by 25 gives 2s² + 3s − 2 = 0.
Factor: 2s² + 3s − 2 = (2s − 1)(s + 2), so s = 12 or s = −2. Check: (2s − 1)(s + 2) = 2s² + 4s − s − 2 = 2s² + 3s − 2.
No angle has a sine less than −1, so sin θ = −2 is rejected. sin θ = 12 gives 30^° or 180 − 30 = 150^°, and only 30^° lies between 0^° and 90^°. (a) The kicker must be built at 30^°.
(b) The kicker is the hypotenuse of a right-angled triangle, so the lip is 7.5 sin 30^° = 7.5 × 12 = 3.75 m above the snow. The rest of the height is the rise above the lip: 5 sin² 30^° = 5 × 14 = 1.25 m. Check: 3.75 + 1.25 = 5 m.

answer(a) 30^°; (b) the lip is 3.75 m above the snow, and the skier rises 1.25 m above it

techniqueQuadratic Equations in Sine, Cosine or Tangent · Solving Simple Trigonometric Equations · The Principal Values of the Inverse Functions

examsO-Level

Common pitfalls

Dividing through by sin θ or taking a square root term by term. The equation has both a sin² θ term and a sin θ term, so bring every term to one side and factor it as a quadratic.
Keeping sin θ = −2 and looking for sin⁻¹(−2). No angle has a sine below −1, and a calculator gives an error; that root comes from the algebra, and no kicker can be built at it.