Geometry and Trigonometry Practice Questions — JAMB UTME — Page 2

Question 21Question →

If $\sin \theta = \frac{3}{5}$ for an acute angle $\theta$ , what is the exact value of $5 \cos \theta + 4 \tan \theta$ ?

Show answer & explanation

Answer: 7

Answer

The exact value of

5 \cos \theta + 4 \tan \theta

is 7.

For an acute angle

\theta

with

\sin \theta = \frac{3}{5}

, the corresponding right triangle has opposite side = 3, hypotenuse = 5, and adjacent side = 4. Using trig definitions,

\cos \theta = \frac{4}{5}

and

\tan \theta = \frac{3}{4}

. Evaluating

5 \cos \theta + 4 \tan \theta

gives

5\left(\frac{4}{5}\right) + 4\left(\frac{3}{4}\right) = 4 + 3 = 7

.

Step-by-Step Solution

1

Find the adjacent side of the right-angled triangle.

Adjacent side =

\sqrt{5^2 - 3^2} = 4

.

By the Pythagorean theorem (

a^2 + b^2 = c^2

), where opposite = 3 and hypotenuse = 5.

2

Determine the values of

\cos \theta

and

\tan \theta

.

\cos \theta = \frac{4}{5}

and

\tan \theta = \frac{3}{4}

.

Using fundamental trigonometric definitions: cosine is adjacent/hypotenuse and tangent is opposite/adjacent.

3

Substitute these values into

5 \cos \theta + 4 \tan \theta

and simplify.

5\left(\frac{4}{5}\right) + 4\left(\frac{3}{4}\right) = 4 + 3 = 7

.

Performing simple multiplication and addition gives 7.

Key Concept

Basic Trigonometric Ratios in Right Triangles

Question 22Question →

Given that $\theta$ is an acute angle satisfying the relationship $\sec \theta + \tan \theta = 3$ , what is the exact value of $5\sin \theta$ ?

Show answer & explanation

Answer: 4

Answer

The exact value of

5\sin \theta

is 4.

Using the identity

\sec^2 \theta - \tan^2 \theta = 1

, we deduce

(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1

. Given

\sec \theta + \tan \theta = 3

, it follows that

\sec \theta - \tan \theta = \frac{1}{3}

. Solving the system of equations yields

\sec \theta = \frac{5}{3}

and

\tan \theta = \frac{4}{3}

, which gives

\sin \theta = \frac{4}{5}

. Multiplying by 5 gives the final answer of 4.

Step-by-Step Solution

1

Apply the trigonometric Pythagorean identity

\sec^2 \theta - \tan^2 \theta = 1

This relates secant and tangent functions directly.

2

Factorize the identity and solve for

\sec \theta - \tan \theta

(\sec \theta - \tan \theta)(3) = 1 \implies \sec \theta - \tan \tan \theta = \frac{1}{3}

Using the algebraic identity

a^2 - b^2 = (a - b)(a + b)

.

3

Set up a linear system to solve for

\sec \theta

and

\tan \theta

\sec \theta = \frac{5}{3}, \quad \tan \theta = \frac{4}{3}

Adding and subtracting the equations

\sec \theta + \tan \theta = 3

and \sec \theta - \tan \theta = \frac{1}{3} gives the individual function values.

4

Calculate

\sin \theta

and evaluate

5\sin \theta

\sin \theta = \frac{\tan \theta}{\sec \theta} = \frac{4/3}{5/3} = \frac{4}{5} \implies 5\sin \theta = 4

The quotient of tangent and secant gives sine.

Key Concept

Pythagorean Trigonometric Identities

Alternative Method

Draw a right-angled triangle where hypotenuse over adjacent plus opposite over adjacent equals 3:

\frac{c + a}{b} = 3

. By Pythagorean theorem

c^2 - a^2 = b^2

, so

\frac{c - a}{b} = \frac{1}{3}

. Solving yields

a/c = 4/5

, hence

\sin \theta = 4/5

and

5\sin \theta = 4

.

Estimated Time:2m 0s

Question 23Question →

A regular hexagon has a side length of $6\text{ cm}$ . At each vertex of the hexagon, a circular sector of radius $3\text{ cm}$ is formed inside the figure. Taking $\pi = \frac{22}{7}$ and $\sqrt{3} = 1.732$ , what is the area of the remaining region inside the hexagon not covered by the sectors, in $\text{cm}^2$ , correct to two decimal places?

Show answer & explanation

Answer: 36.96

Answer

The area of the remaining region inside the hexagon is

36.96\text{ cm}^2

.

The total area of the regular hexagon is computed by multiplying the area of one equilateral triangle of side

6\text{ cm}

by 6, yielding

54\sqrt{3} = 54 \times 1.732 = 93.528\text{ cm}^2

. Each interior angle of a regular hexagon is

120^\circ

, so each vertex sector has a central angle of

120^\circ

and radius

3\text{ cm}

. The area of one sector is

\frac{120^\circ}{360^\circ} \times \frac{22}{7} \times 3^2 = \frac{66}{7}\text{ cm}^2

. The total area for all six sectors is

6 \times \frac{66}{7} = \frac{396}{7} \approx 56.5714\text{ cm}^2

. Subtracting this from the total area gives

93.528 - 56.5714 = 36.9566\text{ cm}^2

, which rounds to

36.96\text{ cm}^2

.

Step-by-Step Solution

1

Determine the interior angle of the regular hexagon.

Each interior angle is

120^\circ

.

The formula for the interior angle of a regular polygon with

n

sides is

\frac{(n-2) \times 180^\circ}{n}

.

2

Calculate the total area of the 6 circular sectors at the vertices.

Total sector area is

\frac{396}{7} \approx 56.5714\text{ cm}^2

.

Each sector has a central angle of

120^\circ

and radius

3\text{ cm}

. With 6 sectors, the total area is

6 \times \frac{120^\circ}{360^\circ} \times \frac{22}{7} \times 3^2 = 18 \times \frac{22}{7} = \frac{396}{7}\text{ cm}^2

.

3

Calculate the total area of the regular hexagon.

Hexagon area is

93.528\text{ cm}^2

.

A regular hexagon consists of 6 equilateral triangles of side length

6\text{ cm}

. Area =

6 \times \left(\frac{\sqrt{3}}{4} \times 6^2\right) = 54\sqrt{3} = 54 \times 1.732 = 93.528\text{ cm}^2

.

4

Subtract the sector area from the total hexagon area.

93.528 - 56.5714 = 36.9566\text{ cm}^2 \approx 36.96\text{ cm}^2

.

The remaining area is the total area minus the area occupied by the six corner sectors.

Key Concept

Area of regular polygons and circular sectors

Estimated Time:2m 30s

Question 24Question →

In $\Delta ABC$ , side $a = 5\text{ cm}$ , side $b = 5\sqrt{2}\text{ cm}$ , and $\angle A = 30^\circ$ . What are all possible values for $\angle B$ ?

45^\circ

or

135^\circ

45^\circ

only

60^\circ

or

120^\circ

135^\circ

only

Show answer & explanation

Answer: $45^\circ$ or $135^\circ$

Answer

45^\circ

or

135^\circ

Applying the Sine Rule gives

\frac{5}{\sin 30^\circ} = \frac{5\sqrt{2}}{\sin B}

, so

\sin B = \frac{\sqrt{2}}{2}

. The angles whose sine is

\frac{\sqrt{2}}{2}

between

0^\circ

and

180^\circ

are

45^\circ

and

135^\circ

. Checking angle sums:

30^\circ + 45^\circ = 75^\circ < 180^\circ

and

30^\circ + 135^\circ = 165^\circ < 180^\circ

, so both

45^\circ

and

135^\circ

yield valid triangles.

Step-by-Step Solution

1

Set up the Sine Rule formula relating sides

a

,

b

and angles

A

,

B

\frac{a}{\sin A} = \frac{b}{\sin B}

The Sine Rule connects two sides and their opposite angles in any non-right triangle.

2

Substitute given values

a = 5

,

b = 5\sqrt{2}

, and

A = 30^\circ

\frac{5}{\sin 30^\circ} = \frac{5\sqrt{2}}{\sin B} \implies \frac{5}{0.5} = \frac{5\sqrt{2}}{\sin B} \implies 10 = \frac{5\sqrt{2}}{\sin B}

Since

\sin 30^\circ = \frac{1}{2}

, simplifying yields the ratio.

3

Solve for

\sin B

\sin B = \frac{5\sqrt{2}}{10} = \frac{\sqrt{2}}{2}

Isolating

\sin B

gives the principal trigonometric ratio value.

4

Determine all valid values for angle

B

in the range

(0^\circ, 180^\circ)

Acute

B_1 = \arcsin\left(\frac{\sqrt{2}}{2}\right) = 45^\circ

; Obtuse

B_2 = 180^\circ - 45^\circ = 135^\circ

. Both are valid since

30^\circ + 135^\circ = 165^\circ < 180^\circ

.

Since

b > a

, the ambiguous case (SSA) produces two valid distinct triangle solutions.

Key Concept

Sine Rule and the Ambiguous Case (SSA)

Estimated Time:1m 0s

Question 25Question →

In $\Delta ABC$ , side lengths are given as $a = 6\text{ cm}$ and $b = 6\sqrt{3}\text{ cm}$ , and $\angle A = 30^\circ$ . If $\angle B$ is an obtuse angle, what is the length of side $c$ ?

6\text{ cm}

12\text{ cm}

6\sqrt{3}\text{ cm}

3\sqrt{3}\text{ cm}

Show answer & explanation

Answer: $6\text{ cm}$

Answer

The length of side

c

is

6\text{ cm}

.

Using the Sine Rule,

\frac{6}{\sin 30^\circ} = \frac{6\sqrt{3}}{\sin B}

, which yields

\sin B = \frac{\sqrt{3}}{2}

. The two possible values for

\angle B

are

60^\circ

(acute) and

120^\circ

(obtuse). The problem explicitly specifies that

\angle B

is obtuse, so

\angle B = 120^\circ

. Subtracting from

180^\circ

gives

\angle C = 30^\circ

. Since

\angle A = \angle C = 30^\circ

, the triangle is isosceles, making side

c

equal to side

a

, which is

6\text{ cm}

.

Step-by-Step Solution

1

Apply the Sine Rule to find

\sin B

\frac{a}{\sin A} = \frac{b}{\sin B} \implies \frac{6}{\sin 30^\circ} = \frac{6\sqrt{3}}{\sin B} \implies 12 = \frac{6\sqrt{3}}{\sin B} \implies \sin B = \frac{\sqrt{3}}{2}

The Sine Rule relates side lengths to the sines of their opposite angles.

2

Determine the value of

\angle B

using the given condition

\angle B = 60^\circ

or

\angle B = 180^\circ - 60^\circ = 120^\circ

. Since

\angle B

is obtuse,

\angle B = 120^\circ

.

The inverse sine function yields two possible angle solutions between

0^\circ

and

180^\circ

(the ambiguous case).

3

Calculate the third angle

\angle C

\angle C = 180^\circ - (\angle A + \angle B) = 180^\circ - (30^\circ + 120^\circ) = 30^\circ

The sum of interior angles in any triangle is

180^\circ

.

4

Find the length of side

c

Since

\angle A = 30^\circ

and

\angle C = 30^\circ

,

\Delta ABC

is isosceles with side

c = a = 6\text{ cm}

.

Sides opposite to equal angles in a triangle are equal in length.

Key Concept

Ambiguous Case of the Sine Rule (SSA Condition)

Alternative Method

Alternatively, apply the Cosine Rule for angle

A

:

a^2 = b^2 + c^2 - 2bc \cos A \implies 6^2 = (6\sqrt{3})^2 + c^2 - 2(6\sqrt{3})c \cos 30^\circ

. Simplifying gives

36 = 108 + c^2 - 18c \implies c^2 - 18c + 72 = 0

. Factoring yields

(c - 6)(c - 12) = 0

, giving

c = 6\text{ cm}

or

c = 12\text{ cm}

. For

c = 12\text{ cm}

,

b^2 + a^2 = 108 + 36 = 144 = c^2

, making

\angle C = 90^\circ

and

\angle B = 60^\circ

(acute). Thus,

c = 6\text{ cm}

corresponds to the obtuse angle

\angle B = 120^\circ

.

Estimated Time:2m 0s

Question 26Question →

What is the exact numerical value of the trigonometric expression $\frac{4\cos^2 30^\circ + 2\sin^2 45^\circ}{\tan^2 60^\circ - \csc 30^\circ}$ ?

Show answer & explanation

Answer: 4

Answer

4

By evaluating the special angles directly:

4\cos^2 30^\circ = 3

,

2\sin^2 45^\circ = 1

,

\tan^2 60^\circ = 3

, and

\csc 30^\circ = 2

. Substituting these into the expression yields

\frac{3 + 1}{3 - 2} = \frac{4}{1} = 4

.

Step-by-Step Solution

1

Substitute the exact value of

\cos 30^\circ

4\cos^2 30^\circ = 4 \left(\frac{\sqrt{3}}{2}\right)^2 = 4 \left(\frac{3}{4}\right) = 3

The exact value of

\cos 30^\circ

is

\frac{\sqrt{3}}{2}

.

2

Substitute the exact value of

\sin 45^\circ

2\sin^2 45^\circ = 2 \left(\frac{1}{\sqrt{2}}\right)^2 = 2 \left(\frac{1}{2}\right) = 1

The exact value of

\sin 45^\circ

is

\frac{1}{\sqrt{2}}

.

3

Substitute the exact values of

\tan 60^\circ

and

\csc 30^\circ

\tan^2 60^\circ = (\sqrt{3})^2 = 3

and

\csc 30^\circ = \frac{1}{\sin 30^\circ} = 2

The exact value of

\tan 60^\circ

is

\sqrt{3}

and

\csc 30^\circ

is the reciprocal of

\sin 30^\circ = \frac{1}{2}

.

4

Simplify the entire fraction

\frac{3 + 1}{3 - 2} = \frac{4}{1} = 4

Dividing the simplified numerator (4) by the simplified denominator (1) gives 4.

Key Concept

Evaluation of Special Angle Trigonometric Ratios and Reciprocal Functions

Question 27Question →

In $\Delta PQR$ , side $p = 3\text{ cm}$ , side $q = 8\text{ cm}$ , and the included angle $\angle R = 60^\circ$ . What is the length of side $r$ in cm?

Show answer & explanation

Answer: 7

Answer

The length of side

r

is

7\text{ cm}

.

Using the Cosine Rule

r^2 = p^2 + q^2 - 2pq \cos R

with

p=3

,

q=8

, and

\angle R=60^\circ

yields

r^2 = 9 + 64 - 48(0.5) = 49

, which gives

r = 7\text{ cm}

.

Step-by-Step Solution

1

State the Cosine Rule formula for side

r

r^2 = p^2 + q^2 - 2pq \cos R

The Cosine Rule allows calculating the third side of a triangle when two sides and the included angle are given.

2

Substitute the given values into the formula

r^2 = 3^2 + 8^2 - 2(3)(8) \cos 60^\circ

We are given

p = 3

,

q = 8

, and

\angle R = 60^\circ

.

3

Evaluate the trigonometric expression and simplify

r^2 = 9 + 64 - 48(0.5) = 73 - 24 = 49

Since

\cos 60^\circ = 0.5

, multiplying

2 \times 3 \times 8 \times 0.5

yields

24

.

4

Solve for

r

by taking the positive square root

r = 7\text{ cm}

Length must be a positive value, and

\sqrt{49} = 7

.

Key Concept

Cosine Rule for finding an unknown side given two sides and the included angle (SAS).

Question 28Question →

What is the exact simplified value of the trigonometric expression $\frac{\sin 60^\circ + \cos 30^\circ}{\tan 45^\circ + \tan 30^\circ}$ ?

\frac{3(\sqrt{3} - 1)}{2}

\frac{3 - \sqrt{3}}{2}

\frac{3\sqrt{6} - 3\sqrt{2}}{4}

\frac{\sqrt{3} + 1}{3}

Show answer & explanation

Answer: $\frac{3(\sqrt{3} - 1)}{2}$

Answer

\frac{3(\sqrt{3} - 1)}{2}

Substituting the exact values gives a numerator of

\sqrt{3}

and a denominator of

\frac{\sqrt{3}+1}{\sqrt{3}}

. Multiplying by the reciprocal of the denominator gives

\frac{3}{\sqrt{3}+1}

. Rationalizing the denominator by multiplying top and bottom by

(\sqrt{3}-1)

simplifies to

\frac{3(\sqrt{3}-1)}{2}

.

Step-by-Step Solution

1

Substitute the exact standard values for each trigonometric function

\sin 60^\circ = \frac{\sqrt{3}}{2}

,

\cos 30^\circ = \frac{\sqrt{3}}{2}

,

\tan 45^\circ = 1

,

\tan 30^\circ = \frac{1}{\sqrt{3}}

Special angles have precise surd representations that must be used in non-calculator assessments.

2

Simplify the numerator and denominator separately

Numerator:

\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \sqrt{3}

. Denominator:

1 + \frac{1}{\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3}}

.

Combine like terms in the numerator and find a common denominator for the terms in the denominator.

3

Divide the numerator by the denominator

\frac{\sqrt{3}}{\frac{\sqrt{3}+1}{\sqrt{3}}} = \frac{3}{\sqrt{3}+1}

Dividing by a fraction is equivalent to multiplying by its reciprocal.

4

Rationalize the denominator

\frac{3(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{3(\sqrt{3}-1)}{3-1} = \frac{3(\sqrt{3}-1)}{2}

Multiply the numerator and denominator by the conjugate

(\sqrt{3}-1)

to remove the radical from the denominator.

Key Concept

Evaluation and surd simplification of special angle trigonometric expressions

Question 29Question →

Two ships, $P$ and $Q$ , leave a port $O$ at the same time. Ship $P$ sails on a bearing of $040^\circ$ at a constant speed of $25\text{ km/h}$ , while Ship $Q$ sails on a bearing of $100^\circ$ at a constant speed of $40\text{ km/h}$ . What is the distance in kilometers between the two ships after $2$ hours?

Show answer & explanation

Answer: 70

Answer

The distance between the two ships after 2 hours is 70 km.

The distance traveled by Ship P in 2 hours is

50\text{ km}

and by Ship Q is

80\text{ km}

. The angle between their paths is

100^\circ - 040^\circ = 60^\circ

. Applying the Cosine Rule yields

PQ^2 = 50^2 + 80^2 - 2(50)(80)\cos(60^\circ) = 2500 + 6400 - 4000 = 4900

, giving a distance of

\sqrt{4900} = 70\text{ km}

.

Step-by-Step Solution

1

Calculate the distances traveled by Ship P and Ship Q after 2 hours

OP = 50\text{ km}

and

OQ = 80\text{ km}

Distance equals speed multiplied by time.

2

Find the angle between the lines of travel from port O

\angle POQ = 100^\circ - 40^\circ = 60^\circ

The angle between two bearings from a common origin is the difference between their bearing angles.

3

Use the Cosine Rule to calculate the side length PQ

PQ^2 = 50^2 + 80^2 - 2(50)(80)\cos(60^\circ) = 4900

The Cosine Rule

c^2 = a^2 + b^2 - 2ab\cos C

calculates the unknown opposite side given two side lengths and their included angle.

4

Take the square root to find PQ

PQ = 70\text{ km}

Taking the principal square root gives the final linear distance.

Key Concept

Applying the Cosine Rule to solve bearing and distance non-right triangle problems

Estimated Time:2m 0s

Question 30Question →

A forest ranger at a control post $X$ observes a lookout tower at point $Y$ on a bearing of $072^\circ$ . What is the bearing of the control post $X$ from the lookout tower $Y$ ?

252^\circ

108^\circ

288^\circ

162^\circ

Show answer & explanation

Answer: $252^\circ$

Answer

The bearing of the control post

X

from the lookout tower

Y

is

252^\circ

.

The bearing of

Y

from

X

is

072^\circ

. To find the bearing of

X

from

Y

(the back bearing), we add

180^\circ

to the forward bearing because

072^\circ

is less than

180^\circ

. Calculation:

072^\circ + 180^\circ = 252^\circ

.

Step-by-Step Solution

1

Identify the forward bearing of point

Y

from point

X

.

Forward bearing =

072^\circ

.

The question states that

Y

is observed from

X

on a bearing of

072^\circ

.

2

Apply the rule for finding a back bearing when the forward bearing is less than

180^\circ

.

Back bearing =

\text{Forward bearing} + 180^\circ = 072^\circ + 180^\circ = 252^\circ

.

Since

072^\circ < 180^\circ

, we add

180^\circ

to determine the reverse direction from North at point

Y

.

Key Concept

Back Bearing / Reverse Bearing

Estimated Time:45s

Question 31Question →

Find the total number of distinct solutions to the trigonometric equation $2\cos^2(2x) + \sin(2x) - 1 = 0$ in the interval $0^\circ \le x \le 360^\circ$ .

Show answer & explanation

Answer: 6

Answer

The total number of distinct solutions is 6.

Substituting

\cos^2(2x) = 1 - \sin^2(2x)

gives the quadratic

2\sin^2(2x) - \sin(2x) - 1 = 0

, which factors into

(2\sin(2x) + 1)(\sin(2x) - 1) = 0

. For

0^\circ \le x \le 360^\circ

, the angle argument

2x

covers

0^\circ \le 2x \le 720^\circ

. The equation

\sin(2x) = 1

provides 2 values for

x

(

45^\circ, 225^\circ

), while

\sin(2x) = -\frac{1}{2}

provides 4 values for

x

(

105^\circ, 165^\circ, 285^\circ, 345^\circ

). Summing these gives 6 distinct solutions in total.

Step-by-Step Solution

1

Use the Pythagorean trigonometric identity

\cos^2(2x) = 1 - \sin^2(2x)

to express the entire equation in terms of

\sin(2x)

.

2(1 - \sin^2(2x)) + \sin(2x) - 1 = 0 \implies 2 - 2\sin^2(2x) + \sin(2x) - 1 = 0

Converting all trigonometric terms to a single function allows the equation to be solved as a polynomial.

2

Rearrange and factorize the resulting quadratic equation in terms of

\sin(2x)

.

2\sin^2(2x) - \sin(2x) - 1 = 0 \implies (2\sin(2x) + 1)(\sin(2x) - 1) = 0

Factorization splits the quadratic trigonometric equation into two simple linear trigonometric equations.

3

Determine the expanded domain for

2x

given

0^\circ \le x \le 360^\circ

.

0^\circ \le 2x \le 720^\circ

Multiplying the bounds of

x

by 2 accounts for two full rotations in the unit circle.

4

Solve the first linear equation

\sin(2x) = 1

within

0^\circ \le 2x \le 720^\circ

.

2x = 90^\circ, 450^\circ \implies x = 45^\circ, 225^\circ

(2 distinct solutions)

The sine function equals 1 at

90^\circ

in the first revolution and at

90^\circ + 360^\circ = 450^\circ

in the second revolution.

5

Solve the second linear equation

\sin(2x) = -\frac{1}{2}

within

0^\circ \le 2x \le 720^\circ

.

2x = 210^\circ, 330^\circ, 570^\circ, 690^\circ \implies x = 105^\circ, 165^\circ, 285^\circ, 345^\circ

(4 distinct solutions)

The sine function is negative in the 3rd and 4th quadrants of both revolutions.

6

Combine the solution counts from both cases.

Total number of solutions =

2 + 4 = 6

.

Adding the valid solutions from both factor equations gives the complete set of roots.

Key Concept

Solving quadratic trigonometric equations across multiple revolutions

Estimated Time:3m 0s

Question 32Question →

In triangle $LMN$ , the side lengths are given as $l = 7\text{ cm}$ , $m = 8\text{ cm}$ , and $n = 13\text{ cm}$ . What is the measure of angle $N$ in degrees?

Show answer & explanation

Answer: 120

Answer

The measure of angle

N

is

120^\circ

.

Using the Cosine Rule for angle

N

,

\cos N = \frac{l^2 + m^2 - n^2}{2lm} = \frac{7^2 + 8^2 - 13^2}{2(7)(8)} = \frac{-56}{112} = -0.5

. The angle whose cosine is

-0.5

within the interior angles of a triangle (

0^\circ < N < 180^\circ

) is

120^\circ

.

Step-by-Step Solution

1

Set up the Cosine Rule formula for angle

N

\cos N = \frac{l^2 + m^2 - n^2}{2lm}

The Cosine Rule relates the three sides of any triangle to the cosine of one of its interior angles.

2

Substitute the known side lengths into the formula

\cos N = \frac{7^2 + 8^2 - 13^2}{2 \times 7 \times 8} = \frac{49 + 64 - 169}{112}

Side

n = 13\text{ cm}

is opposite to angle

N

and must be subtracted in the numerator.

3

Simplify the fraction

\cos N = \frac{-56}{112} = -0.5

Evaluating the numerical expression gives a negative value, indicating that angle

N

is obtuse.

4

Calculate the principal inverse cosine angle for the triangle

N = 120^\circ

Since

\cos 60^\circ = 0.5

,

\cos(180^\circ - 60^\circ) = -0.5

, giving

N = 120^\circ

.

Key Concept

Applying the Cosine Rule to find an obtuse angle given three side lengths (SSS)

Question 33Question →

A surveyor stands at a point on level ground $50\text{ m}$ away from the base of a vertical transmission tower. If the angle of elevation from the observer's position on the ground to the top of the tower is $45^\circ$ , what is the height of the tower in meters?

Show answer & explanation

Answer: 50

Answer

The height of the transmission tower is

50\text{ meters}

.

In a right-angled triangle, the tangent of the angle of elevation equals the ratio of the height (opposite side) to the horizontal distance (adjacent side). Since

\tan(45^\circ) = 1

, the height of the tower must be equal to the horizontal distance of

50\text{ m}

.

Step-by-Step Solution

1

Formulate the trigonometric relationship using the right triangle formed by the observer, the base of the tower, and the top of the tower.

\tan(45^\circ) = \frac{h}{50}

, where

h

is the height of the tower.

The tangent ratio relates the opposite side (height of tower) to the adjacent side (distance along level ground).

2

Substitute the value of

\tan(45^\circ) = 1

and solve for

h

.

h = 50 \times 1 = 50\text{ m}

.

Multiplying the adjacent side length by

\tan(45^\circ)

yields the exact height.

Key Concept

Angle of elevation using basic right-triangle trigonometry

Question 34Question →

A vertical tower $TC$ of height $h$ metres stands on a horizontal plane with its base at $C$ . Point $A$ on the plane is due West of $C$ , and point $B$ is due South of $A$ , such that the distance $AB = 40\text{ m}$ . If the angles of elevation of the top $T$ of the tower from $A$ and $B$ are $60^\circ$ and $30^\circ$ respectively, what is the height of the tower?

10\sqrt{6}\text{ m}

4\sqrt{30}\text{ m}

20\sqrt{2}\text{ m}

20\sqrt{3}\text{ m}

Show answer & explanation

Answer: $10\sqrt{6}\text{ m}$

Answer

The height of the tower is

10\sqrt{6}\text{ m}

.

Using the trigonometric tangent ratios, the horizontal distances from the base of the tower are

AC = \frac{h}{\sqrt{3}}

and

BC = h\sqrt{3}

. Because point

A

is West of

C

and point

B

is South of

A

, the angle

\angle CAB = 90^\circ

. Applying Pythagoras' theorem

BC^2 = AC^2 + AB^2

yields

3h^2 = \frac{h^2}{3} + 1600

, which simplifies to

h = 10\sqrt{6}\text{ m}

.

Step-by-Step Solution

1

Express the horizontal distances

AC

and

BC

in terms of height

h

.

In vertical right-angled triangle

TAC

,

\tan(60^\circ) = \frac{h}{AC} \implies AC = \frac{h}{\sqrt{3}}

. In vertical right-angled triangle

TBC

, \tan(30^\circ) = \frac{h}{BC} \implies BC = h\sqrt{3}$.

Relate the vertical height to the horizontal ground distances using tangent ratios.

2

Identify the geometry of the ground plane triangle

\triangle CAB

.

Since

A

is due West of

C

and

B

is due South of

A

, the line segments

CA

(East-West) and

AB

(North-South) are perpendicular. Thus,

\triangle CAB

is right-angled at

A

.

Cardinal directions (West and South from

A

) are perpendicular to each other.

3

Apply Pythagoras' theorem to

\triangle CAB

.

BC^2 = AC^2 + AB^2 \implies (h\sqrt{3})^2 = \left(\frac{h}{\sqrt{3}}\right)^2 + 40^2 \implies 3h^2 = \frac{h^2}{3} + 1600

.

Hypotenuse

BC

relates the two legs

AC

and

AB

on the horizontal plane.

4

Solve the equation for

h

.

3h^2 - \frac{h^2}{3} = 1600 \implies \frac{8h^2}{3} = 1600 \implies 8h^2 = 4800 \implies h^2 = 600 \implies h = \sqrt{600} = 10\sqrt{6}\text{ m}

.

Simplify algebraic terms to determine

h

in surd form.

Key Concept

Combining angles of elevation in 3D vertical planes with horizontal plane coordinate geometry and Pythagoras' theorem.

Question 35Question →

What is the gradient of a line that is perpendicular to the line passing through the points $(2, 5)$ and $(4, 11)$ ?

3

-\frac{1}{3}

\frac{1}{3}

-3

Show answer & explanation

Answer: $-\frac{1}{3}$

Answer

The gradient of the perpendicular line is

-\frac{1}{3}

.

The gradient of the line connecting

(2, 5)

and

(4, 11)

is

m = \frac{11-5}{4-2} = 3

. The gradient of a line perpendicular to it must satisfy

m_1 \cdot m_2 = -1

, yielding

-\frac{1}{3}

.

Step-by-Step Solution

1

Calculate the gradient

m_1

of the line passing through

(2, 5)

and

(4, 11)

.

m_1 = \frac{11 - 5}{4 - 2} = \frac{6}{2} = 3

The gradient formula between two points

(x_1, y_1)

and

(x_2, y_2)

is

m = \frac{y_2 - y_1}{x_2 - x_1}

.

2

Apply the condition for perpendicular lines to find the required gradient

m_2

.

m_2 = -\frac{1}{m_1} = -\frac{1}{3}

Two non-vertical lines are perpendicular if and only if the product of their gradients is

-1

(

m_1 \cdot m_2 = -1

).

Key Concept

Perpendicular Line Gradients

Question 36Question →

A straight line $L_1$ passes through the points $(k, 2)$ and $(3, 8)$ . If $L_1$ is perpendicular to the line $L_2$ defined by the equation $2x - 3y + 5 = 0$ , what is the value of $k$ ?

7

-6

-1

12

Show answer & explanation

Answer: $7$

Answer

The value of

k

is

7

.

Rearranging the line equation

2x - 3y + 5 = 0

gives

y = \frac{2}{3}x + \frac{5}{3}

, so its gradient is

\frac{2}{3}

. The line perpendicular to it must have a gradient equal to the negative reciprocal, which is

-\frac{3}{2}

. Equating this to the slope formula

\frac{8 - 2}{3 - k}

gives

\frac{6}{3 - k} = -\frac{3}{2}

. Cross-multiplying yields

12 = -9 + 3k

, which simplifies to

3k = 21

, giving the correct answer

7

.

Step-by-Step Solution

1

Find the gradient (

m_2

) of the line

L_2

Rearrange

2x - 3y + 5 = 0

into

y = mx + c

form:

3y = 2x + 5 \implies y = \frac{2}{3}x + \frac{5}{3}

. Thus,

m_2 = \frac{2}{3}

.

The coefficient of

x

in slope-intercept form gives the gradient of

L_2

.

2

Determine the gradient (

m_1

) of the perpendicular line

L_1

Using the perpendicular condition

m_1 \cdot m_2 = -1

:

m_1 = -\frac{1}{m_2} = -\frac{3}{2}

.

Perpendicular lines have gradients that are negative reciprocals of each other.

3

Express the gradient of

L_1

using the two given points

(k, 2)

and

(3, 8)

m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{8 - 2}{3 - k} = \frac{6}{3 - k}

.

The gradient of a line through

(x_1, y_1)

and

(x_2, y_2)

is given by

\frac{y_2 - y_1}{x_2 - x_1}

.

4

Equate the two gradient expressions and solve for

k

\frac{6}{3 - k} = -\frac{3}{2} \implies 6 \cdot 2 = -3(3 - k) \implies 12 = -9 + 3k \implies 3k = 21 \implies k = 7

.

Solving the linear algebraic equation yields the required value of the unknown coordinate

k

.

Key Concept

Two non-vertical lines are perpendicular if and only if the product of their gradients is

-1

(

m_1 \cdot m_2 = -1

).

Estimated Time:2m 0s

Question 37Question →

In $\Delta PQR$ , side $p = 4\text{ cm}$ , side $q = 4\sqrt{2}\text{ cm}$ , and $\angle P = 30^\circ$ . If $\angle Q$ is an obtuse angle, what is the measure of $\angle Q$ ?

135^\circ

45^\circ

60^\circ

150^\circ

Show answer & explanation

Answer: $135^\circ$

Answer

The measure of angle

Q

is

135^\circ

.

Applying the Sine Rule gives

\sin Q = \frac{4\sqrt{2} \cdot \sin 30^\circ}{4} = \frac{\sqrt{2}}{2}

. The acute reference angle is

45^\circ

, so the obtuse angle is

180^\circ - 45^\circ = 135^\circ

.

Step-by-Step Solution

1

Apply the Sine Rule to relate sides

p, q

and angles

P, Q

\frac{p}{\sin P} = \frac{q}{\sin Q} \implies \frac{4}{\sin 30^\circ} = \frac{4\sqrt{2}}{\sin Q}

The Sine Rule connects pairs of opposite sides and angles in any triangle.

2

Solve for

\sin Q

\sin Q = \frac{4\sqrt{2} \cdot \sin 30^\circ}{4} = \sqrt{2} \cdot \frac{1}{2} = \frac{\sqrt{2}}{2}

Substitute

\sin 30^\circ = \frac{1}{2}

and simplify the numerical expression.

3

Determine the obtuse solution for angle

Q

Q = 180^\circ - 45^\circ = 135^\circ

Since

\sin Q = \frac{\sqrt{2}}{2}

, the principal acute angle is

45^\circ

. Because the problem specifies that angle

Q

is obtuse (

90^\circ < Q < 180^\circ

), we take its supplement.

Key Concept

Ambiguous case of the Sine Rule (SSA condition)

Estimated Time:1m 30s

Question 38Question →

A straight line passing through the points $(1, 3)$ and $(5, k)$ has a gradient of $2$ . What is the value of $k$ ?

Show answer & explanation

Answer: 11

Answer

The value of

k

is

11

.

The gradient

m

of a line passing through

(x_1, y_1)

and

(x_2, y_2)

is given by

m = \frac{y_2 - y_1}{x_2 - x_1}

. Substituting

(1, 3)

,

(5, k)

, and

m = 2

yields

2 = \frac{k - 3}{5 - 1} = \frac{k - 3}{4}

. Multiplying by

4

gives

8 = k - 3

, so

k = 11

.

Step-by-Step Solution

1

Recall the slope/gradient formula for a straight line.

m = \frac{y_2 - y_1}{x_2 - x_1}

The gradient of a straight line passing through two points is the ratio of vertical change to horizontal change.

2

Substitute the known point coordinates and gradient into the formula.

2 = \frac{k - 3}{5 - 1}

We are given

(x_1, y_1) = (1, 3)

,

(x_2, y_2) = (5, k)

, and gradient

m = 2

.

3

Simplify and solve for

k

.

2 = \frac{k - 3}{4} \implies 8 = k - 3 \implies k = 11

Multiplying both sides by

4

clears the fraction, and adding

3

isolates

k

.

Key Concept

Gradient of a straight line passing through two points

Estimated Time:45s

Question 39Question →

In $\Delta ABC$ , the side lengths are given as $a = 3\text{ cm}$ , $b = 5\text{ cm}$ , and $c = 7\text{ cm}$ . What is the measure of angle $C$ in degrees?

Show answer & explanation

Answer: 120

Answer

The measure of angle

C

is

120^\circ

.

Using the Cosine Rule

\cos C = \frac{a^2 + b^2 - c^2}{2ab}

, substituting

a=3

,

b=5

, and

c=7

yields

\cos C = \frac{9 + 25 - 49}{30} = -0.5

. Taking the inverse cosine of

-0.5

gives an angle of

120^\circ

.

Step-by-Step Solution

1

Select the appropriate formula for calculating an interior angle given three sides (SSS).

Use the Cosine Rule rearranged for

\cos C

:

\cos C = \frac{a^2 + b^2 - c^2}{2ab}

.

When all three side lengths of a non-right triangle are known, the Cosine Rule is required to find any of its angles.

2

Substitute side lengths

a=3

,

b=5

, and

c=7

into the Cosine Rule equation.

\cos C = \frac{3^2 + 5^2 - 7^2}{2(3)(5)} = \frac{9 + 25 - 49}{30} = -\frac{15}{30} = -0.5.

Simplifying the numerator and denominator determines the exact trigonometric ratio for angle

C

.

3

Find the inverse cosine of

-0.5

in degrees.

C = \arccos(-0.5) = 120^\circ.

A negative cosine value indicates an obtuse angle in the second quadrant (

90^\circ < C < 180^\circ

).

Key Concept

Using the Cosine Rule to determine obtuse angles in SSS triangles

Question 40Question →

What is the period of the trigonometric function $y = 3\cos(4x)$ ?

90^\circ

120^\circ

180^\circ

1440^\circ

Show answer & explanation

Answer: $90^\circ$

Answer

The period of the given trigonometric function is

90^\circ

.

The period of a function of the form

y = a\cos(bx)

is given by

T = \frac{360^\circ}{b}

. Substituting

b = 4

yields

T = \frac{360^\circ}{4} = 90^\circ

.

Step-by-Step Solution

1

Identify the standard form parameters

For

y = 3\cos(4x)

, amplitude

a = 3

and coefficient of

x

is

b = 4

.

The general form of a cosine function is

y = a\cos(bx + c) + d

.

2

Apply the period formula for trigonometric functions in degrees

Period

T = \frac{360^\circ}{b} = \frac{360^\circ}{4} = 90^\circ

.

The full cycle of a standard cosine graph completes in

360^\circ

, so multiplying the input by

b

compresses the period by a factor of

b

.

Key Concept

Period of a Cosine Function

Estimated Time:45s

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