SOHCAHTOA (Trigonometry) - Worksheets, Questions and Revision

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GCSE · Geometry and Measures

5.19 SOHCAHTOA (Trigonometry)

EDEXCEL 1MA1 · Calculator allowed · about 90 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Revision notes: SOHCAHTOA (Trigonometry)

SOHCAHTOA is a memory aid for the three trigonometric ratios used in right-angled triangles: Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent. It is used to find a missing side or a missing angle in a right-angled triangle when one side and one other angle (or two sides) are known.

Label the triangle's sides relative to the angle you are using (or finding): the Hypotenuse (opposite the right angle), the Opposite (across from the angle), and the Adjacent (next to the angle, not the hypotenuse).
Choose the correct ratio (SOH, CAH or TOA) based on which two sides are involved: the two you know, plus the one you want to find.
To find a missing SIDE, substitute the known angle and side into the ratio, then rearrange to make the unknown side the subject.
To find a missing ANGLE, substitute the two known sides into the ratio, then use the inverse function (sin^-1, cos^-1 or tan^-1) on your calculator to find the angle.

Worked example. In a right-angled triangle, the hypotenuse is 12 cm and one angle is 35 degrees. Work out the length of the side opposite this angle, to 1 decimal place.

The known angle links the Opposite and the Hypotenuse, so use SOH: sin(angle) = Opposite / Hypotenuse.
Substitute the known values: sin(35) = Opposite / 12.
Rearrange to make Opposite the subject: Opposite = 12 x sin(35).
Opposite = 12 x 0.5736 = 6.883 cm.

Answer: 6.9 cm (to 1 decimal place).

The diagram shows a right-angled triangle ABC, with the right angle at B and angle θ marked at vertex A.
Diagram: right-angled triangle ABC, right angle marked at B, angle θ marked at A, side AB along the base, side BC vertical, side AC the sloping side joining A to C.

(a)Write down the letter of the side that is the hypotenuse.(1)

(b)Write down the letter of the side that is opposite angle θ.(1)

(c)Write down the letter of the side that is adjacent to angle θ.(1)

(Total for Question 1 is 3 marks)

For each right-angled triangle described below, write down which trigonometric ratio (sine, cosine or tangent) should be used.

(a)The hypotenuse and angle θ are known. The side opposite θ is unknown. Which ratio should be used to find it?(1)

(b)The hypotenuse and angle θ are known. The side adjacent to θ is unknown. Which ratio should be used to find it?(1)

(c)The side opposite θ and the side adjacent to θ are known. Angle θ is unknown. Which ratio should be used to find it?(1)

(Total for Question 2 is 3 marks)

In triangle ABC, angle B = 90 degrees, angle A = 34 degrees and the hypotenuse AC = 9.6 cm.
Diagram: right-angled triangle ABC, right angle at B, angle A = 34 degrees, hypotenuse AC = 9.6 cm, side BC = x cm (opposite angle A) unmarked.
Calculate the length of BC, marked x. Give your answer correct to 3 significant figures.

Diagram NOT accurately drawn

(Total for Question 3 is 3 marks)

In triangle PQR, angle Q = 90 degrees, angle P = 47 degrees and side PQ (adjacent to angle P) = 7.3 cm.
Diagram: right-angled triangle PQR, right angle at Q, angle P = 47 degrees, side PQ = 7.3 cm, hypotenuse PR = y cm unmarked.
Calculate the length of the hypotenuse, PR. Give your answer correct to 3 significant figures.

Diagram NOT accurately drawn

(Total for Question 4 is 3 marks)

In triangle XYZ, angle Y = 90 degrees, angle X = 38 degrees and side XY (adjacent to angle X) = 11.2 cm.
Diagram: right-angled triangle XYZ, right angle at Y, angle X = 38 degrees, side XY = 11.2 cm, side YZ = z cm (opposite angle X) unmarked.
Calculate the length of YZ, marked z. Give your answer correct to 3 significant figures.

Diagram NOT accurately drawn

(Total for Question 5 is 3 marks)

In triangle DEF, angle E = 90 degrees, the side opposite angle D (EF) = 6.4 cm and the hypotenuse DF = 10.5 cm.
Diagram: right-angled triangle DEF, right angle at E, side EF = 6.4 cm opposite angle D, hypotenuse DF = 10.5 cm, angle D marked but unknown.
Calculate the size of angle D. Give your answer correct to 1 decimal place.

Diagram NOT accurately drawn

(Total for Question 6 is 3 marks)

In triangle GHI, angle H = 90 degrees, side GH (adjacent to angle G) = 15.6 cm and side HI (opposite angle G) = 8.3 cm.
Diagram: right-angled triangle GHI, right angle at H, side GH = 15.6 cm, side HI = 8.3 cm, angle G marked but unknown.
Calculate the size of angle G. Give your answer correct to 1 decimal place.

Diagram NOT accurately drawn

(Total for Question 7 is 3 marks)

In triangle JKL, angle K = 90 degrees, angle J = 52 degrees and the hypotenuse JL = 13 cm.
Diagram: right-angled triangle JKL, right angle at K, angle J = 52 degrees, hypotenuse JL = 13 cm, side KL = x cm (opposite angle J) unmarked.
Show that x = 10.2 cm, correct to 1 decimal place.

Diagram NOT accurately drawn

(Total for Question 8 is 3 marks)

Mia stands 30 m from the base of a vertical flagpole. She measures the angle of elevation to the top of the flagpole as 27 degrees.
Diagram: right-angled triangle, horizontal ground distance of 30 m from Mia to the base of the flagpole, vertical flagpole of height h m, angle of elevation of 27 degrees marked at Mia's position.
Calculate the height of the flagpole, h. Give your answer correct to 3 significant figures.

Diagram NOT accurately drawn

(Total for Question 9 is 3 marks)

From the top of a vertical cliff 52 m high, Tomasz observes a boat out at sea. The angle of depression from Tomasz to the boat is 15 degrees.
Diagram: vertical cliff of height 52 m, horizontal line from the top of the cliff, angle of depression of 15 degrees marked between the horizontal and the line of sight to the boat, boat on the sea at horizontal distance d m from the base of the cliff.
Calculate the horizontal distance, d, from the base of the cliff to the boat. Give your answer correct to 3 significant figures.

Diagram NOT accurately drawn

(Total for Question 10 is 3 marks)

A ladder of length 4.8 m leans against a vertical wall. The foot of the ladder is 1.5 m from the base of the wall.
Diagram: right-angled triangle formed by the wall, the ground and the ladder; ladder (hypotenuse) = 4.8 m, distance from the foot of the ladder to the wall = 1.5 m, angle between the ladder and the ground marked θ.

Diagram NOT accurately drawn

(a)Calculate the size of angle θ, the angle the ladder makes with the ground. Give your answer correct to 1 decimal place.(3)

(b)Calculate how far up the wall the ladder reaches. Give your answer correct to 3 significant figures.(2)

(Total for Question 11 is 5 marks)

An isosceles triangle has two equal sides of length 10.4 cm and an apex angle of 48 degrees between them. A line is drawn from the apex perpendicular to the base, meeting the base at its midpoint.
Diagram: isosceles triangle, two equal sides marked 10.4 cm, apex angle of 48 degrees between them, dashed perpendicular line from the apex to the midpoint of the base labelled base = b cm.
Calculate the length of the base, b. Give your answer correct to 3 significant figures.

Diagram NOT accurately drawn

(Total for Question 12 is 4 marks)

A triangular plot of land has a horizontal base of 18 m. At one end of the base, the angle between the base and the sloping boundary is 33 degrees.
Diagram: right-angled triangle representing the plot; horizontal base 18 m, angle of 33 degrees at one end between the base and the hypotenuse (sloping boundary), vertical height h m at the other end, perpendicular to the base.

Diagram NOT accurately drawn

(a)Calculate the height, h, of the triangular plot. Give your answer correct to 3 significant figures.(3)

(b)Hence calculate the area of the triangular plot. Give your answer correct to 3 significant figures.(2)

(Total for Question 13 is 5 marks)

Without using a calculator, write down the exact value of each of the following trigonometric ratios.

(a)sin(30 degrees)(1)

(b)cos(60 degrees)(1)

(c)tan(45 degrees)(1)

(d)sin(45 degrees), giving your answer as an exact surd.(1)

(Total for Question 14 is 4 marks)

A boat sails from a harbour on a bearing of 065 degrees for a distance of 20 km, reaching point B.
Diagram: harbour with a north line drawn vertically upward, bearing of 065 degrees measured clockwise from north to the boat's path, boat's path to point B marked 20 km.

Diagram NOT accurately drawn

(a)Calculate how far north of the harbour point B is. Give your answer correct to 3 significant figures.(2)

(b)Calculate how far east of the harbour point B is. Give your answer correct to 3 significant figures.(2)

(Total for Question 15 is 4 marks)

A wheelchair access ramp has a sloped length of 3.5 m and rises a vertical height of 0.28 m. Building regulations recommend that the angle of a wheelchair ramp should not exceed the angle produced by a gradient of 1 in 12 (a rise of 1 unit for every 12 units travelled horizontally).
Diagram: right-angled triangle representing the ramp; sloped length (hypotenuse) 3.5 m, vertical rise 0.28 m, angle between the ramp and the horizontal marked θ.
Determine, with full working, whether this ramp meets the recommended maximum angle.

Diagram NOT accurately drawn

(Total for Question 16 is 4 marks)

From a point A on level ground, the angle of elevation to the top of a tower is 25 degrees. From a point B, on the same straight line as A and the base of the tower and 20 m closer to the tower, the angle of elevation to the top of the tower is 39 degrees.
Diagram: vertical tower of height h m standing on level ground; points A and B on the ground on the same side of the tower, with B 20 m closer to the tower than A; angle of elevation from A = 25 degrees, angle of elevation from B = 39 degrees.
Calculate the height, h, of the tower. Give your answer correct to 3 significant figures.

Diagram NOT accurately drawn

(Total for Question 17 is 5 marks)

A cuboid ABCDEFGH has a rectangular base ABCD with AB = 15 cm and BC = 6 cm, and vertical edges of height 9 cm (AE = BF = CG = DH = 9 cm). The diagonal AC of the base is drawn, and the space diagonal AG from A to the diagonally opposite top vertex G is also drawn.
Diagram: cuboid ABCDEFGH, base ABCD with AB = 15 cm and BC = 6 cm, vertical height 9 cm, diagonal AC drawn across the base, space diagonal AG drawn from A to G.

Diagram NOT accurately drawn

(a)Calculate the length of the diagonal AC of the base ABCD. Give your answer correct to 3 significant figures.(2)

(b)Calculate the size of the angle between the diagonal AG and the base ABCD (angle GAC). Give your answer correct to 1 decimal place.(3)

(Total for Question 18 is 5 marks)

The diagram shows the triangular cross-section of a roof truss, made from two right-angled triangles PQR and QRS sharing the vertical ridge support QR. The horizontal base PS is divided at R into PR = 12 m and RS = 8 m. The angle between PQ and the base at P is 50 degrees.
Diagram: triangular roof cross-section, horizontal base PS with R between P and S so that PR = 12 m and RS = 8 m, vertical ridge support QR drawn from R up to the apex Q, angle at P (between PQ and PR) = 50 degrees.

Diagram NOT accurately drawn

(a)Calculate the height of the ridge support, QR. Give your answer correct to 3 significant figures.(3)

(b)Calculate the size of the angle between QS and the base at S (angle QSR). Give your answer correct to 1 decimal place.(3)

(Total for Question 19 is 6 marks)

Mark scheme · 5.19 SOHCAHTOA (Trigonometry)

Question 1

(a) B1 AC cao
(a) Answer: AC
(b) B1 BC cao
(b) Answer: BC
(c) B1 AB cao
(c) Answer: AB

Question 2

(a) B1 Sine (SOH) cao
(a) Answer: Sine
(b) B1 Cosine (CAH) cao
(b) Answer: Cosine
(c) B1 Tangent (TOA) cao
(c) Answer: Tangent

Question 3

M1 sin(34) = x / 9.6 oe
M1 x = 9.6 x sin(34)
A1 awrt 5.37 cm
Answer: 5.37 cm (3 s.f.)

Question 4

M1 cos(47) = 7.3 / y oe
M1 y = 7.3 / cos(47)
A1 awrt 10.7 cm
Answer: 10.7 cm (3 s.f.)

Question 5

M1 tan(38) = z / 11.2 oe
M1 z = 11.2 x tan(38)
A1 awrt 8.75 cm
Answer: 8.75 cm (3 s.f.)

Question 6

M1 sin(D) = 6.4 / 10.5 oe
M1 D = sin^-1(6.4 / 10.5)
A1 awrt 37.6 degrees
Answer: 37.6 degrees (1 d.p.)

Question 7

M1 tan(G) = 8.3 / 15.6 oe
M1 G = tan^-1(8.3 / 15.6)
A1 awrt 28.0 degrees
Answer: 28.0 degrees (1 d.p.)

Question 8

M1 x = 13 x sin(52) oe
M1 x = 10.244... seen (at least 3 s.f. before rounding)
A1 10.2 cm cso, correctly rounded to 1 d.p. as required
Answer: x = 10.2 cm (as required)

Question 9

M1 tan(27) = h / 30 oe
M1 h = 30 x tan(27)
A1 awrt 15.3 m
Answer: 15.3 m (3 s.f.)

Question 10

M1 tan(15) = 52 / d oe (using alternate angle, angle of elevation from boat = 15)
M1 d = 52 / tan(15)
A1 awrt 194 m
Answer: 194 m (3 s.f.)

Question 11

(a) M1 cos(θ) = 1.5 / 4.8 oe
(a) M1 θ = cos^-1(1.5 / 4.8)
(a) A1 awrt 71.8 degrees
(a) Answer: 71.8 degrees (1 d.p.)
(b) M1 √4.8² - 1.5² oe, or 4.8 x sin(their θ)
(b) A1 awrt 4.56 m
(b) Answer: 4.56 m (3 s.f.)

Question 12

M1 half apex angle = 48 / 2 = 24 (the perpendicular bisects the apex angle by symmetry)
M1 half base = 10.4 x sin(24) oe
dM1 b = 2 x (their half base)
A1 awrt 8.46 cm
Answer: 8.46 cm (3 s.f.)

Question 13

(a) M1 tan(33) = h / 18 oe
(a) M1 h = 18 x tan(33)
(a) A1 awrt 11.7 m
(a) Answer: 11.7 m (3 s.f.)
(b) M1 0.5 x 18 x (their h) oe
(b) A1 awrt 105 m² ft from part a
(b) Answer: 105 m² (3 s.f.)

Question 14

(a) B1 1/2 oe (0.5) cao
(a) Answer: 1/2
(b) B1 1/2 oe (0.5) cao
(b) Answer: 1/2
(c) B1 1 cao
(c) Answer: 1
(d) B1 √2/2 oe (e.g. 1/√2) cao, exact form required
(d) Answer: √2/2

Question 15

(a) M1 20 x cos(65) oe
(a) A1 awrt 8.45 km
(a) Answer: 8.45 km (3 s.f.)
(b) M1 20 x sin(65) oe
(b) A1 awrt 18.1 km
(b) Answer: 18.1 km (3 s.f.)

Question 16

M1 sin(θ) = 0.28 / 3.5 oe
A1 θ = awrt 4.6 degrees
M1 maximum angle = tan^-1(1/12) oe (= awrt 4.8 degrees)
C1 correct conclusion with comparison, e.g. since 4.6 < 4.8 (or equivalent gradient/tangent comparison), the ramp meets the recommended maximum angle cso
Answer: The ramp's angle is awrt 4.6 degrees, which is less than the maximum recommended angle of awrt 4.8 degrees, so the ramp does meet the recommended maximum.

Question 17

P1 distance from A to base of tower = h / tan(25) oe
P1 distance from B to base of tower = h / tan(39) oe
P1 forms equation h / tan(25) - h / tan(39) = 20 oe (using AB = 20 m)
M1 h = 20 / (1/tan(25) - 1/tan(39)) oe, correctly rearranged and solved
A1 awrt 22.0 m
Answer: 22.0 m (3 s.f.)

Question 18

(a) M1 AC = √15² + 6² oe
(a) A1 awrt 16.2 cm
(a) Answer: 16.2 cm (3 s.f.)
(b) M1 recognises right angle at C (CG perpendicular to base), so tan(GAC) = GC / AC oe
(b) M1 angle GAC = tan^-1(9 / (their AC))
(b) A1 awrt 29.1 degrees
(b) Answer: 29.1 degrees (1 d.p.)

Question 19

(a) M1 tan(50) = QR / 12 oe
(a) M1 QR = 12 x tan(50)
(a) A1 awrt 14.3 m
(a) Answer: 14.3 m (3 s.f.)
(b) M1 tan(QSR) = (their QR) / 8 oe
(b) M1 QSR = tan^-1((their QR) / 8)
(b) A1 awrt 60.8 degrees
(b) Answer: 60.8 degrees (1 d.p.)

Question 19

6 marks

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SOHCAHTOA (Trigonometry) - Worksheets, Questions and Revision

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5.19 SOHCAHTOA (Trigonometry)

Revision notes: SOHCAHTOA (Trigonometry)

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Question 12

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Question 15

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Question 17

Question 18

Question 19

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