Kuulchat - Senior High School (S.H.S) Mathematics Trigonometry Practice Test Questions

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TRIGONOMETRY

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1.

In the diagram, ABD is a triangle, |AC| = 50 cm and |CD| = d cm.

Find, correct to one decimal place,:

(a)

the value of d;

(b)

|BD|.

Show Solution

(a)

Method I

SOHCAHTOA

SOHCAHTOA is a mnemonic that gives us an easy way to remember the three main trigonometric ratios. They are sine (sin), cosine (cos) and tangent (tan).

SOH

sin(θ) = $\frac{Opposite}{Hypothenuse}$

CAH

cos(θ) = $\frac{Adjacent}{Hypothenuse}$

TOA

tan(θ) = $\frac{Opposite}{Adjacent}$

The longest side (side opposite/facing the right-angle) is the hypotenuse

tan(θ) = $\frac{Opposite}{Adjacent}$

tan 30° = $\frac{|BD|}{50 + d}$

|BD| = tan 30° x (50 + d)

|BD| = 28.87 + 0.58d ----- eqn 1

tan 70° = $\frac{|BD|}{d}$

|BD| = tan 70° x d

|BD| = 2.74d ---- eqn 2

--- eqn 1 = --- eqn 2

2.74d = 28.87 + 0.58d

2.74d - 0.58d = 28.87

2.16d = 28.87

Divide both sides by 2.16

d = $\frac{28.87}{2.16}$ = 13.4 cm (one decimal place)

Method II

∠ ACB + ∠ BCD = 180°(Sum of angles on a straight line)

∠ ACB + 70° = 180°

∠ ACB = 180° - 70°

∠ ACB = 110°

∠ BAC + ∠ ACB + ∠ ABC = 180°(Sum of angles in a triangle)

30° + 110° + ∠ ABC = 180°

140° + ∠ ABC = 180°

∠ ABC = 180° - 140°

∠ ABC = 40°

Sine Rule

$\frac{sin 40°}{50}$ = $\frac{sin 30°}{|BC|}$

Cross multiply

|BC| x sin 40° = 50 x sin 30°

Divide both sides by sin 40°

|BC| = $\frac{50 x sin 30°}{sin 40°}$ = 38.89 cm

SOHCAHTOA

SOHCAHTOA is a mnemonic that gives us an easy way to remember the three main trigonometric ratios. They are sine (sin), cosine (cos) and tangent (tan).

SOH

sin(θ) = $\frac{Opposite}{Hypothenuse}$

CAH

cos(θ) = $\frac{Adjacent}{Hypothenuse}$

TOA

tan(θ) = $\frac{Opposite}{Adjacent}$

The longest side (side opposite/facing the right-angle) is the hypotenuse

d is adjacent to ∠ BCD (70°) and |BC| is hypothenuse.

|BC| = 38.89 cm

CAH

cos(θ) = $\frac{Adjacent}{Hypothenuse}$

cos 70° = $\frac{d}{38.89 cm}$

Multiply both sides by the denominator 38.89 cm

d = 38.89 cm x cos 70°

d = 13.3 cm

(b)

Method I

|BD| = 2.74d ---- eqn 2

But d = 13.4 cm

|BD| = 2.74 x 13.4 cm

|BD| = 36.7 cm (one decimal place)

Method II

|BC| = 38.89 cm

d = 13.3 cm

Pythagorean Theorem

The hypothenuse (c) is the longest side. The side facing the right-angle (∟)

|BD|² + 13.3² = 38.89²

|BD|² = 38.89² - 13.3²

|BD|² = 1335.54

Take square root of both sides.

|BD| = $\sqrt{1335.54}$ = 36.5 cm

2.

In the diagram, MNR is right angle triangle. |MN| = 15 m, |MR| = 10 $\sqrt{3}$ m and ∠ MQN = 72°.

Calculate, correct to the nearest whole number:

(a)

the value of the angle marked x;

(b)

|QR|;

(c)

area of ∆ MQR.

Show Solution

(a)

SOHCAHTOA

SOHCAHTOA is a mnemonic that gives us an easy way to remember the three main trigonometric ratios. They are sine (sin), cosine (cos) and tangent (tan).

SOH

sin(θ) = $\frac{Opposite}{Hypothenuse}$

CAH

cos(θ) = $\frac{Adjacent}{Hypothenuse}$

TOA

tan(θ) = $\frac{Opposite}{Adjacent}$

The longest side (side opposite/facing the right-angle) is the hypotenuse

sin(θ) = $\frac{Opposite}{Hypothenuse}$

Sin ∠ NRM = $\frac{15 m}{10 \sqrt{3}}$

∠ NRM = sin^-1 $(\frac{15 m}{10 \sqrt{3} m})$

∠ NRM = 60°

∠ RQM + ∠ NQM = 180° (sum of angles on a straight line)

∠ NQM = 72°

∠ RQM + 72° = 180°

∠ RQM = 180° - 72°

∠ RQM = 108°

x + ∠ RQM + ∠ NRM = 180° (sum of interior angles of a triangle)

x + 108° + 60° = 180°

x + 168° = 180°

x = 180° - 168°

x = 12°

(b)

|QR| = |NR| - |NQ|

From Pythagorean theorem,

The hypothenuse (c) is the longest side. The side facing the right-angle (∟)

|MR|² = |NR|² + |MN|²

(10 $\sqrt{3}$ )² = |NR|² + 15²

Note:
1. the square affects each term in the bracket
2. when you square a square root, the square root is cancelled

10² x 3 = |NR|² + 225

300 = |NR|² + 225

300 - 225 = |NR|²

75 = |NR|²

Take the square root of both sides.

|NR| = $\sqrt{75}$ = 8.6602540378 ≈ 8.660 m

tan(θ) = $\frac{Opposite}{Adjacent}$

tan 72 = $\frac{15 m}{|QN| m}$

Multiply both sides by |QN|

|QN| x tan 72 = 15

Divide both sides by tan 72

|QN| = $\frac{15 m}{tan 72}$

|QN| = 4.8737954435 ≈ 4.874 m

|QR| = 8.660 m - 4.874 m

|QR| = 3.786 m ≈ 4 m (nearest whole number)

(c)

Area of ∆ MQR = Area of ∆ MRN - Area of ∆ MQN

Area of a triangle = $\frac{1}{2}$ x base x height

Area of ∆ MRN

Base of triangle = 15 m
Height of triangle = 8.660 m

Area of a triangle = $\frac{1}{2}$ x 15 m x 8.660 m = 64.95 m²

Area of ∆ MQN

Base of triangle = 15 m
Height of triangle = 4.874 m

Area of a triangle = $\frac{1}{2}$ x 15 m x 4.874 m = 36.56 m²

Area of ∆ MQR = 64.95 m² - 36.56 m²

Area of ∆ MQR = 28.39 m² ≈ 28 m² (nearest whole number)

3.

Given that sin y = $\frac{5}{13}$ , 0° ≤ 90°, evaluate $\frac{3cos y - 2sin y}{6tan y}$ .

Show Solution

SOHCAHTOA

SOHCAHTOA is a mnemonic that gives us an easy way to remember the three main trigonometric ratios. They are sine (sin), cosine (cos) and tangent (tan).

SOH

sin(θ) = $\frac{Opposite}{Hypothenuse}$

CAH

cos(θ) = $\frac{Adjacent}{Hypothenuse}$

TOA

tan(θ) = $\frac{Opposite}{Adjacent}$

The longest side (side opposite/facing the right-angle) is the hypotenuse

sin(θ) = $\frac{Opposite}{Hypothenuse}$

sin y = $\frac{5}{13}$

Opposite side to y = 5

Hypothenuse side = 13

From pythagoras theorem

5² + x² = 13²

25 + x² = 169

x² = 169 - 25

x² = 144

Take square root of both sides.

x = $\sqrt{144}$ = 12

cos(θ) = $\frac{Adjacent}{Hypothenuse}$

cos y = $\frac{12}{13}$

tan(θ) = $\frac{Opposite}{Adjacent}$

tan y = $\frac{5}{12}$

$\frac{3cos y - 2sin y}{6tan y}$ = [3 x $\frac{12}{13}$ - 2 x $\frac{5}{13}$ ] ÷ [6 x $\frac{5}{12}$

$\frac{3cos y - 2sin y}{6tan y}$ = $\frac{4}{5} or 0.8$

4.

Given that cos (2x + 15°) = sin (x - 30°), 0° ≤ x ≤ 90°, find the value of x.

Show Solution

cos (2x + 15°) = sin (x - 30°) Method I For complementary angles, If cos (θ) = sin (β) then θ + β = 90° 2x + 15° + x - 30° = 90° 3x = 90° + 30° - 15° 3x = 105° Divide both sides by 3. x =

\frac{105°}{3}

= 35°

Method II

Applying sin (90°- A) = cos A or cos (90°- A) = sin A

Applying sin (90°- A) = cos A

sin (x - 30°) = cos (90°- (x - 30°))

cos (2x + 15°) = cos (90°- (x - 30°))

Since cos is at both sides, the angles are also the same.

2x + 15 = 90 - x + 30

2x + x = 120 - 15

3x = 105

Divide both sides by 3

x = $\frac{105}{3}$ = 35°

Applying cos (90°- A) = sin A

cos (2x + 15°) = sin (90° - (2x + 15°))

sin (90° - (2x + 15°)) = sin (x - 30°)

Since both sides has sin, the angles are the same.

90° - (2x + 15°) = x - 30°

90° - 2x - 15° = x - 30°

90° + 30° - 15° = x + 2x

105° = 3x

Divide both sides by 3

x = $\frac{105°}{3}$ = 35°

5.

Given that tan x = 1, 0° ≤ x ≤ 90°, evaluate $\frac{1 - {sin}^{2} x}{cos x}$ .

Show Solution

Method I

SOHCAHTOA

SOHCAHTOA is a mnemonic that gives us an easy way to remember the three main trigonometric ratios. They are sine (sin), cosine (cos) and tangent (tan).

SOH

sin(θ) = $\frac{Opposite}{Hypothenuse}$

CAH

cos(θ) = $\frac{Adjacent}{Hypothenuse}$

TOA

tan(θ) = $\frac{Opposite}{Adjacent}$

The longest side (side opposite/facing the right-angle) is the hypotenuse

TOA

tan(θ) = $\frac{Opposite}{Adjacent}$

tan x = 1 means the opposite side is 1 and adjacent side is also 1

You can draw a right angle to represent the sides and find the hypotenuse

From Pythagorean theorem,

The hypothenuse (c) is the longest side. The side facing the right-angle (∟)

c² = 1² + 1²

c² = 1 + 1

c² = 2

Take square root of both sides.

c = $\sqrt{2}$

SOH

sin(θ) = $\frac{Opposite}{Hypothenuse}$

sin x = $\frac{1}{\sqrt{2}}$

CAH

cos(θ) = $\frac{Adjacent}{Hypothenuse}$

cos x = $\frac{1}{\sqrt{2}}$

$\frac{1 - {sin}^{2} x}{cos x}$ = $\frac{1 - {sin}^{2} x}{cos x}$

sin² x = sin x x sin x

sin² x = $\frac{1}{\sqrt{2}}$ x $\frac{1}{\sqrt{2}}$

sin² x = $\frac{1 x 1}{\sqrt{2} x \sqrt{2}}$

$\sqrt{a}$ x $\sqrt{a}$ = a

$\sqrt{2}$ x $\sqrt{2}$ = 2

sin² x = $\frac{1}{2}$

$\frac{1 - {sin}^{2} x}{cos x}$ = $\frac{1 - \frac{1}{2}}{\frac{1}{\sqrt{2}}}$

$\frac{1 - {sin}^{2} x}{cos x}$ = $\frac{\frac{1 x 2 - 1 x 1}{2}}{\frac{1}{\sqrt{2}}}$

$\frac{1 - {sin}^{2} x}{cos x}$ = $\frac{\frac{2 - 1}{2}}{\frac{1}{\sqrt{2}}}$

$\frac{1 - {sin}^{2} x}{cos x}$ = $\frac{\frac{1}{2}}{\frac{1}{\sqrt{2}}}$

$\frac{1 - {sin}^{2} x}{cos x}$ = $\frac{1}{2}$ ÷ $\frac{1}{\sqrt{2}}$

Reciprocate the fraction at the right of the ÷ and change ÷ to multiplication (x)

Note: reciprocate means the numerator becomes the denominator and the denominator becomes the numerator.

$\frac{1 - {sin}^{2} x}{cos x}$ = $\frac{1}{2}$ x $\frac{\sqrt{2}}{1}$

$\frac{1 - {sin}^{2} x}{cos x}$ = $\frac{1 x \sqrt{2}}{2 x 1}$

$\frac{1 - {sin}^{2} x}{cos x}$ = $\frac{\sqrt{2}}{2}$

Method 2

Finding the value of x and substituting it in the given equation.

tan x = 1

x = tan^-1(1)

x = 45°

$\frac{1 - {sin}^{2} x}{cos x}$ = $\frac{1 - {sin}^{2} 45}{cos 45}$

sin 45° = $\frac{1}{\sqrt{2}} sin$ ² 45° = $\frac{1}{\sqrt{2}}$ x $\frac{1}{\sqrt{2}}$

sin² 45° = $\frac{1 x 1}{\sqrt{2} x \sqrt{2}}$

$\sqrt{2}$ x $\sqrt{2}$ = 2

sin² 45° = $\frac{1}{2}$

cos 45° = $\frac{1}{\sqrt{2}} \frac{1 - {sin}^{2} x}{cos x} = \frac{1 - \frac{1}{2}}{\frac{1}{\sqrt{2}}}$