Geometry and Trigonometry Practice Questions — JAMB UTME — Page 7

Question 121Question →

An isosceles triangle has a perimeter of $36\text{ cm}$ and a base of length $16\text{ cm}$ . Calculate the area of the triangle in $\text{cm}^2$ .

Show answer & explanation

Answer: 48

Answer

The area of the isosceles triangle is

48\text{ cm}^2

.

The two equal sides of the isosceles triangle measure

\frac{36 - 16}{2} = 10\text{ cm}

each. An altitude dropped perpendicularly to the base bisects the

16\text{ cm}

base into two

8\text{ cm}

segments. By the Pythagorean theorem, the perpendicular height is

h = \sqrt{10^2 - 8^2} = 6\text{ cm}

. Therefore, the area is

\frac{1}{2} \times 16 \times 6 = 48\text{ cm}^2

.

Step-by-Step Solution

1

Determine the length of the two equal sides

Each equal side is

10\text{ cm}

Subtract the base length from the total perimeter (

36 - 16 = 20\text{ cm}

) and divide by

2

.

2

Calculate the perpendicular height (altitude) to the base

Height

h = 6\text{ cm}

The perpendicular altitude bisects the base into two

8\text{ cm}

segments, creating right-angled triangles with hypotenuse

10\text{ cm}

. Use Pythagoras' theorem:

h = \sqrt{10^2 - 8^2} = 6\text{ cm}

.

3

Calculate the area of the triangle

Area =

48\text{ cm}^2

Multiply half the base by the perpendicular height:

\frac{1}{2} \times 16 \times 6 = 48\text{ cm}^2

.

Key Concept

Perimeter and Area of Isosceles Triangles using Pythagorean Theorem

Estimated Time:1m 30s

Question 122Question →

A straight line $L_1$ passes through the points $A(-2, 1)$ and $B(4, 5)$ . A second line $L_2$ is perpendicular to $L_1$ and passes through the midpoint of the line segment $AB$ . If $L_2$ intersects the $x$ -axis at point $R$ , and $R$ divides the line segment joining $P(1, -2)$ and $Q(k, 4)$ internally in the ratio $1 : 2$ , what is the value of $k$ ?

7

4

-5

5

Show answer & explanation

Answer: $7$

Answer

The value of

k

is

7

.

The midpoint of

AB

is

(1, 3)

and the gradient of

L_1

is

\frac{2}{3}

. The gradient of the perpendicular line

L_2

is

-\frac{3}{2}

, yielding the equation

3x + 2y - 9 = 0

. Setting

y = 0

gives the

x

-intercept

R(3, 0)

. Applying the internal section formula for ratio

1 : 2

on the

x

-coordinates yields

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

, which solves directly to

k = 7

.

Step-by-Step Solution

1

Find the midpoint

M

of line segment

AB

and the gradient of line

L_1

.

Midpoint

M = \left(\frac{-2 + 4}{2}, \frac{1 + 5}{2}\right) = (1, 3)

. Gradient

m_1 = \frac{5 - 1}{4 - (-2)} = \frac{4}{6} = \frac{2}{3}

.

Line

L_2

passes through the midpoint

M

and its orientation depends on the gradient of

L_1

.

2

Determine the gradient

m_2

of

L_2

and its equation.

Since

L_2 \perp L_1

,

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

. Equation of

L_2

:

y - 3 = -\frac{3}{2}(x - 1) \implies 3x + 2y - 9 = 0

.

Perpendicular lines have gradients whose product is

-1

(

m_1 \cdot m_2 = -1

).

3

Find the coordinates of point

R

, the

x

-intercept of

L_2

.

Set

y = 0

in

3x + 2y - 9 = 0 \implies 3x - 9 = 0 \implies x = 3

. Thus,

R = (3, 0)

.

The

x

-intercept occurs where

y = 0

on the coordinate plane.

4

Apply the section formula to find

k

.

Point

R(3, 0)

divides

P(1, -2)

and

Q(k, 4)

in ratio

1 : 2

. The

x

-coordinate is given by

x_R = \frac{1(k) + 2(1)}{1 + 2} = \frac{k + 2}{3}

. Setting

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

gives

k + 2 = 9 \implies k = 7

.

The internal section formula states that a point dividing

(x_1, y_1)

and

(x_2, y_2)

in ratio

m : n

has coordinates

\left(\frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n}\right)

.

Key Concept

Perpendicular lines, midpoints, intercepts, and section formula in coordinate geometry

Question 123Question →

Two straight paths diverge from a junction $J$ at an angle of $120^\circ$ . A person walks $5\text{ km}$ along the first path to point $A$ , and another person walks $16\text{ km}$ along the second path to point $B$ . What is the direct distance between $A$ and $B$ in kilometres?

Show answer & explanation

Answer: 19

Answer

The direct distance between points

A

and

B

is

19\text{ km}

.

Using the Cosine Rule

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

with sides

5\text{ km}

and

16\text{ km}

and included angle

120^\circ

yields

c^2 = 5^2 + 16^2 - 2(5)(16)(-0.5) = 25 + 256 + 80 = 361

. Taking the square root gives

19\text{ km}

.

Step-by-Step Solution

1

Identify given values and setup the Cosine Rule formula

Side

a = 5

, side

b = 16

, and included angle

\theta = 120^\circ

The scenario provides two sides and the included angle (SAS configuration), which requires the Cosine Rule to find the third side.

2

Substitute the values into

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

c^2 = 5^2 + 16^2 - 2(5)(16) \cos(120^\circ)

Populating the formula allows evaluation of the unknown distance squared.

3

Evaluate the trigonometric term and simplify

c^2 = 25 + 256 - 160(-0.5) = 281 + 80 = 361

The cosine of an obtuse angle in the second quadrant (

120^\circ

) is negative:

\cos(120^\circ) = -0.5

.

4

Take the square root to find the distance

c

c = \sqrt{361} = 19\text{ km}

Taking the positive square root gives the physical distance between the two points.

Key Concept

Applying the Cosine Rule to find the length of an unknown side in a non-right triangle given two sides and the included angle (SAS).

Question 124Question →

A surveyor at station $A$ observes two landmarks, $B$ and $C$ . Landmark $B$ is located at a distance of $14\text{ km}$ from $A$ on a bearing of $025^\circ$ . Landmark $C$ is located at a distance of $6\sqrt{2}\text{ km}$ from $A$ on a bearing of $070^\circ$ . What is the direct distance between landmark $B$ and landmark $C$ ?

10\text{ km}

7\sqrt{2}\text{ km}

8\text{ km}

2\sqrt{46}\text{ km}

Show answer & explanation

Answer: $10\text{ km}$

Answer

The direct distance between landmark

B

and landmark

C

is

10\text{ km}

.

The included angle

\angle BAC

between the two bearings is

070^\circ - 025^\circ = 45^\circ

. Applying the Cosine Rule

a^2 = b^2 + c^2 - 2bc \cos A

gives

a^2 = (6\sqrt{2})^2 + 14^2 - 2(6\sqrt{2})(14)\cos 45^\circ = 72 + 196 - 168 = 100

. Taking the square root gives

10\text{ km}

.

Step-by-Step Solution

1

Determine the interior angle

\angle BAC

from the given bearings.

\angle BAC = 070^\circ - 025^\circ = 45^\circ

The difference between two bearings measured clockwise from North from the same point gives the included angle between the lines of sight.

2

Identify the side lengths adjacent to angle

A

in

\Delta ABC

.

c = AB = 14\text{ km}

,

b = AC = 6\sqrt{2}\text{ km}

, and included angle

A = 45^\circ

We have a Side-Angle-Side (SAS) triangle configuration, requiring the Cosine Rule to find the opposite side

a = BC

.

3

Apply the Cosine Rule

a^2 = b^2 + c^2 - 2bc \cos A

.

a^2 = (6\sqrt{2})^2 + 14^2 - 2(6\sqrt{2})(14)\cos 45^\circ

Substituting known values into the Cosine Rule formula.

4

Simplify the terms and solve for

a

.

a^2 = 72 + 196 - 168\sqrt{2}\left(\frac{1}{\sqrt{2}}\right) = 268 - 168 = 100 \implies a = \sqrt{100} = 10\text{ km}

Squaring

6\sqrt{2}

gives

36 \times 2 = 72

,

14^2 = 196

, and simplifying the cosine term yields

168

.

Key Concept

Cosine Rule for SAS non-right triangles in bearing contexts

Estimated Time:2m 0s

Question 125Question →

Find the value of $x$ , in degrees, for $0^\circ \le x \le 90^\circ$ that satisfies the trigonometric equation $\sin 2x = \cos(x + 30^\circ)$ .

Show answer & explanation

Answer: 20

Answer

The value of

x

in the interval

0^\circ \le x \le 90^\circ

satisfying the equation is

20^\circ

.

Using the co-function identity

\cos \alpha = \sin(90^\circ - \alpha)

, we convert the right-hand side to

\sin(90^\circ - (x + 30^\circ)) = \sin(60^\circ - x)

. Equating the arguments gives

2x = 60^\circ - x

, which simplifies to

3x = 60^\circ

, yielding

x = 20^\circ

.

Step-by-Step Solution

1

Apply the co-function trigonometric identity

\cos(x + 30^\circ) = \sin(90^\circ - (x + 30^\circ)) = \sin(60^\circ - x)

Converting cosine to sine allows direct comparison of sine functions on both sides of the equation.

2

Set up the equation equating the angle expressions

2x = 60^\circ - x

Since

\sin(2x) = \sin(60^\circ - x)

and

x \in [0^\circ, 90^\circ]

, equating the principal angle arguments gives the primary solution.

3

Solve the linear equation for

x

3x = 60^\circ \implies x = 20^\circ

Adding

x

to both sides gives

3x = 60^\circ

, and dividing by 3 yields

x = 20^\circ

.

Key Concept

Co-function identities and simple trigonometric equations

Question 126Question →

For the domain $0^\circ \le x \le 180^\circ$ , what is the complete solution set to the trigonometric equation $4\sin^2 x - 3 = 0$ ?

\{60^\circ\}

\{60^\circ, 120^\circ\}

\{30^\circ, 150^\circ\}

\{60^\circ, 240^\circ\}

Show answer & explanation

Answer: $\{60^\circ, 120^\circ\}$

Answer

\{60^\circ, 120^\circ\}

Solving the equation

4\sin^2 x - 3 = 0

gives

\sin^2 x = \frac{3}{4}

, which simplifies to

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

within the interval

0^\circ \le x \le 180^\circ

. The principal angle is

60^\circ

. Since sine is positive in both the first and second quadrants, the second valid angle in the domain is

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

, giving the solution set

\{60^\circ, 120^\circ\}

.

Step-by-Step Solution

1

Isolate the trigonometric term

\sin^2 x = \frac{3}{4}

Rearrange

4\sin^2 x - 3 = 0

by adding 3 to both sides and dividing by 4.

2

Take the square root of both sides

\sin x = \pm \frac{\sqrt{3}}{2}

Taking the square root yields both positive and negative ratios.

3

Apply the domain restriction

0^\circ \le x \le 180^\circ

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

The sine function is non-negative in the first and second quadrants (

0^\circ \le x \le 180^\circ

), so the negative root has no solutions in this interval.

4

Determine the angles for

x

x = 60^\circ \text{ and } x = 180^\circ - 60^\circ = 120^\circ

The reference angle is

60^\circ

because

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

. In Quadrant II, the corresponding angle is

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

.

Key Concept

Solving Quadratic Trigonometric Equations

Question 127Question →

A point $P(x, y)$ moves in a Cartesian plane such that the square of its distance from $A(3, 0)$ exceeds the square of its distance from $B(-1, 2)$ by $4$ units. Which of the following equations represents the locus of $P$ ?

2x - y = 0

2x + y - 4 = 0

x - 2y + 1 = 0

x^2 + y^2 - 4x + 2y - 4 = 0

Show answer & explanation

Answer: $2x - y = 0$

Answer

The equation representing the locus of

P

is

2x - y = 0

.

Using the distance formula,

PA^2 = (x - 3)^2 + y^2 = x^2 - 6x + 9 + y^2

and

PB^2 = (x + 1)^2 + (y - 2)^2 = x^2 + 2x + 1 + y^2 - 4y + 4

. Subtracting

PB^2

from

PA^2

yields

(x^2 + y^2 - 6x + 9) - (x^2 + y^2 + 2x - 4y + 5) = 4

, which simplifies to

-8x + 4y = 0

, or

2x - y = 0

.

Step-by-Step Solution

1

Express the square of the distance from

P(x, y)

to

A(3, 0)

using the distance formula

PA^2 = (x - 3)^2 + (y - 0)^2 = x^2 - 6x + 9 + y^2

The distance formula gives

PA^2 = (x - x_A)^2 + (y - y_A)^2

.

2

Express the square of the distance from

P(x, y)

to

B(-1, 2)

PB^2 = (x - (-1))^2 + (y - 2)^2 = (x + 1)^2 + (y - 2)^2 = x^2 + 2x + 1 + y^2 - 4y + 4 = x^2 + y^2 + 2x - 4y + 5

Expand both algebraic squares for

PB^2

.

3

Set up the locus condition

PA^2 - PB^2 = 4

and simplify

(x^2 + y^2 - 6x + 9) - (x^2 + y^2 + 2x - 4y + 5) = 4 \implies -8x + 4y + 4 = 4

Subtracting

PB^2

from

PA^2

cancels the quadratic

x^2

and

y^2

terms.

4

Rearrange into standard linear equation form

-8x + 4y = 0 \implies 2x - y = 0

Divide the entire equation by

-4

to simplify.

Key Concept

Locus defined by difference of squares of distances to two fixed points

Estimated Time:1m 30s

Question 128Question →

The parallel lines $L_1: 3x - 4y + 25 = 0$ and $L_2: 3x - 4y = 0$ are intersected by a straight line $L_3$ with gradient $m > 1$ . If the length of the line segment of $L_3$ intercepted between $L_1$ and $L_2$ is $5\sqrt{5}$ units, find the value of $m$ .

Show answer & explanation

Answer: 2

Answer

The value of the gradient

m

is 2.

The perpendicular distance between the parallel lines

L_1: 3x - 4y + 25 = 0

and

L_2: 3x - 4y = 0

is

d = \frac{25}{\sqrt{3^2 + (-4)^2}} = 5

units. The acute angle

\phi

between

L_3

and the parallel lines satisfies

\sin(\phi) = \frac{5}{5\sqrt{5}} = \frac{1}{\sqrt{5}}

, which gives

\tan(\phi) = \frac{1}{2}

. The gradient of

L_1

and

L_2

is

m_1 = \frac{3}{4}

. Using the tangent formula for the angle between two lines,

\tan(\phi) = \left|\frac{m - m_1}{1 + m m_1}\right| \implies \frac{1}{2} = \left|\frac{4m - 3}{4 + 3m}\right|

, which yields

m = 2

or

m = \frac{2}{11}

. Under the constraint

m > 1

, the unique value of

m

is

2

.

Step-by-Step Solution

1

Calculate the perpendicular distance

d

between the parallel lines

L_1

and

L_2

d = \frac{|25 - 0|}{\sqrt{3^2 + (-4)^2}} = \frac{25}{5} = 5

units

The perpendicular distance between two parallel lines

Ax + By + C_1 = 0

and

Ax + By + C_2 = 0

is given by

\frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

.

2

Determine the trigonometric relationship between the perpendicular distance, intercepted segment, and intersection angle

\phi

\sin(\phi) = \frac{\text{perpendicular distance}}{\text{intercepted segment}} = \frac{5}{5\sqrt{5}} = \frac{1}{\sqrt{5}}

The perpendicular distance forms the opposite side of a right triangle whose hypotenuse is the intercepted segment.

3

Calculate

\tan(\phi)

using right-triangle trigonometry

Since

\sin(\phi) = \frac{1}{\sqrt{5}}

,

\cos(\phi) = \sqrt{1 - \sin^2(\phi)} = \frac{2}{\sqrt{5}}

, so

\tan(\phi) = \frac{1}{2}

The angle between lines formula requires

\tan(\phi)

.

4

Apply the angle between two lines formula and solve for

m

\tan(\phi) = \left| \frac{m - m_1}{1 + m m_1} \right| \implies \frac{1}{2} = \left| \frac{4m - 3}{4 + 3m} \right|

where

m_1 = \frac{3}{4}

. Case 1:

\frac{4m - 3}{4 + 3m} = \frac{1}{2} \implies 8m - 6 = 4 + 3m \implies 5m = 10 \implies m = 2

. Case 2:

\frac{4m - 3}{4 + 3m} = -\frac{1}{2} \implies 8m - 6 = -4 - 3m \implies 11m = 2 \implies m = \frac{2}{11}

.

Evaluating the absolute value produces two potential solutions.

5

Apply the domain constraint

m > 1

m = 2

The problem restricts

m > 1

, which excludes

m = \frac{2}{11}

.

Key Concept

Distance between parallel lines and angle of intersection between straight lines

Question 129Question →

Three of the exterior angles of a convex polygon are each $50^\circ$ , while the remaining exterior angles are each $35^\circ$ . How many sides does the polygon have?

6

8

9

10

Show answer & explanation

Answer: 9

Answer

The total number of sides of the polygon is 9.

The sum of all exterior angles of a convex polygon is

360^\circ

. The sum of the first three angles is

3 \times 50^\circ = 150^\circ

. Subtracting this from

360^\circ

leaves

210^\circ

for the remaining angles. Dividing

210^\circ

by

35^\circ

yields

6

remaining angles. Adding the initial

3

angles gives

3 + 6 = 9

sides in total.

Step-by-Step Solution

1

State the formula for the sum of exterior angles of a convex polygon.

The sum of exterior angles is always

360^\circ

.

This fundamental geometric property applies to all convex polygons regardless of the number of sides.

2

Set up an algebraic equation using the given exterior angle measures.

Let

k

be the number of remaining exterior angles measuring

35^\circ

. Then

3(50^\circ) + k(35^\circ) = 360^\circ

.

The total sum is composed of three

50^\circ

angles and

k

remaining

35^\circ

angles.

3

Solve for

k

.

150^\circ + 35^\circ k = 360^\circ \implies 35^\circ k = 210^\circ \implies k = 6

.

Subtracting

150^\circ

from both sides isolates the term containing

k

.

4

Calculate the total number of sides

n

.

n = 3 + 6 = 9

.

The polygon has a total number of sides equal to the total number of exterior angles (

3

initial angles

+ 6

remaining angles).

Key Concept

The sum of the exterior angles of any convex polygon is equal to 360 degrees.

Estimated Time:1m 15s

Question 130Question →

In triangle $PQR$ , the side length $p = 12\text{ cm}$ , side length $q = 18\text{ cm}$ , and $\sin P = 0.4$ . What is the exact value of $\sin Q$ ?

Show answer & explanation

Answer: 0.6

Answer

The value of

\sin Q

is 0.6.

Applying the Sine Rule

\frac{p}{\sin P} = \frac{q}{\sin Q}

with

p = 12\text{ cm}

,

q = 18\text{ cm}

, and

\sin P = 0.4

gives

\frac{12}{0.4} = \frac{18}{\sin Q}

. Evaluating

\frac{12}{0.4} = 30

leads to

30 = \frac{18}{\sin Q}

, which yields

\sin Q = \frac{18}{30} = 0.6

.

Step-by-Step Solution

1

State the Sine Rule equation for the given triangle sides and angles.

\frac{p}{\sin P} = \frac{q}{\sin Q}

The Sine Rule relates the lengths of the sides of a triangle to the sines of its opposite angles.

2

Substitute the known numerical values into the Sine Rule equation.

\frac{12}{0.4} = \frac{18}{\sin Q}

Substituting

p = 12

,

q = 18

, and

\sin P = 0.4

sets up an equation with a single unknown.

3

Simplify the left side of the equation and solve for

\sin Q

.

\sin Q = \frac{18}{30} = 0.6

Dividing 12 by 0.4 yields 30, so rearranging gives

\sin Q = \frac{18}{30} = 0.6

.

Key Concept

Using the Sine Rule to calculate an unknown sine ratio

Question 131Question →

In $\Delta PQR$ , side $p = 6\text{ cm}$ , side $q = 6\sqrt{2}\text{ cm}$ , and $\angle P = 30^\circ$ . If $\angle Q$ is an acute angle, what is the measure of $\angle Q$ ?

45^\circ

135^\circ

60^\circ

30^\circ

Show answer & explanation

Answer: $45^\circ$

Answer

45^\circ

Using the Sine Rule

\frac{\sin Q}{q} = \frac{\sin P}{p}

, we substitute the given values to find

\sin Q = \frac{6\sqrt{2} \cdot \sin 30^\circ}{6} = \frac{\sqrt{2}}{2}

. Since

\angle Q

is specified as an acute angle,

\angle Q = 45^\circ

.

Step-by-Step Solution

1

Set up the Sine Rule formula relating sides

p, q

and their opposite angles

P, Q

.

\frac{p}{\sin P} = \frac{q}{\sin Q}

The Sine Rule allows finding an unknown angle when two side lengths and one opposite angle are known.

2

Substitute the given values

p = 6

,

q = 6\sqrt{2}

, and

P = 30^\circ

into the formula.

\frac{6}{\sin 30^\circ} = \frac{6\sqrt{2}}{\sin Q}

Populating known quantities permits solving for

\sin Q

.

3

Solve for

\sin Q

.

\sin Q = \frac{6\sqrt{2} \cdot \sin 30^\circ}{6} = \sqrt{2} \cdot \frac{1}{2} = \frac{\sqrt{2}}{2}

Simplifying the algebraic fraction yields the exact sine ratio for angle

Q

.

4

Determine the acute angle

Q

corresponding to

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

.

\angle Q = 45^\circ

The principal acute angle with sine equal to

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

is

45^\circ

.

Key Concept

Applying the Sine Rule to calculate an unknown acute angle in a non-right-angled triangle

Question 132Question →

In $\Delta ABC$ , side $a = 4\text{ cm}$ , side $b = 4\sqrt{2}\text{ cm}$ , and $\angle A = 30^\circ$ . If $\angle B$ is an obtuse angle, what is the measure of $\angle B$ ?

45^\circ

135^\circ

105^\circ

150^\circ

Show answer & explanation

Answer: $135^\circ$

Answer

135^\circ

Applying the Sine Rule gives

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

, which simplifies to

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

. The inverse sine operation yields an acute angle of

45^\circ

and an obtuse angle of

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

. Since the stem specifies that angle B is obtuse, the correct value is

135^\circ

.

Step-by-Step Solution

1

Apply the Sine Rule formula relating sides

a

,

b

and their opposite angles

A

,

B

.

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

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

2

Substitute the known values into the equation:

a = 4

,

b = 4\sqrt{2}

, and

A = 30^\circ

.

\frac{4}{\sin 30^\circ} = \frac{4\sqrt{2}}{\sin B} \implies \frac{4}{0.5} = \frac{4\sqrt{2}}{\sin B} \implies 8 = \frac{4\sqrt{2}}{\sin B}

\sin 30^\circ = 0.5

.

3

Solve for

\sin B

.

\sin B = \frac{4\sqrt{2}}{8} = \frac{\sqrt{2}}{2}

Rearranging the equation yields the value for

\sin B

.

4

Find the obtuse angle whose sine is

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

.

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

Sine is positive in both the first and second quadrants. The acute reference angle is

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

, so the supplementary obtuse angle is

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

.

Key Concept

Sine Rule and the Ambiguous Case

Question 133Question →

A search-and-rescue helicopter leaves a central station $P$ and flies $16\text{ km}$ on a bearing of $050^\circ$ to reach a waypoint $Q$ . It then changes course and flies $12\text{ km}$ on a bearing of $140^\circ$ to reach a mountain rescue site $R$ . From the central station $P$ , the angle of elevation to the helicopter hovering vertically above point $R$ is $45^\circ$ . What is the vertical height of the helicopter above the horizontal plane of station $P$ , in kilometers?

Show answer & explanation

Answer: 20

Answer

The vertical height of the helicopter above the horizontal plane of station

P

is

20\text{ km}

.

The horizontal journey forms a right-angled triangle

PQR

with side lengths

16\text{ km}

and

12\text{ km}

, yielding a hypotenuse (horizontal distance

PR

) of

20\text{ km}

. Since the angle of elevation from

P

to the hovering helicopter is

45^\circ

, the vertical height is equal to

20 \times \tan(45^\circ) = 20\text{ km}

.

Step-by-Step Solution

1

Find the back bearing of station

P

from waypoint

Q

Back bearing =

050^\circ + 180^\circ = 230^\circ

To determine the enclosed interior angle at point

Q

, the reverse direction from

Q

to

P

must be calculated.

2

Calculate the interior angle

\angle PQR

\angle PQR = 230^\circ - 140^\circ = 90^\circ

Subtracting the forward bearing of

R

from the back bearing of

P

yields the right angle between the two paths.

3

Calculate the horizontal displacement distance

PR

PR = \sqrt{16^2 + 12^2} = 20\text{ km}

Since

\triangle PQR

is right-angled at

Q

, the distance

PR

is obtained using the Pythagorean theorem.

4

Determine the vertical altitude using trigonometry

\text{Height} = PR \times \tan(45^\circ) = 20 \times 1 = 20\text{ km}

In the vertical right-angled triangle formed by

P

, the ground projection of

R

, and the helicopter,

\tan(45^\circ) = \frac{\text{Height}}{\text{Horizontal Distance}}

.

Key Concept

Combining 3-point bearings in 2D with right-triangle trigonometry for 3D angles of elevation.

Question 134Question →

An observer standing at the top of a vertical lighthouse observes two boats, $X$ and $Y$ , on the surrounding horizontal sea surface. Boat $X$ lies due South of the lighthouse at an angle of depression of $30^\circ$ , while boat $Y$ lies due East of the lighthouse at an angle of depression of $45^\circ$ . If the straight-line distance between boat $X$ and boat $Y$ is $80\text{ m}$ , what is the height of the lighthouse?

40\text{ m}

40\sqrt{3}\text{ m}

40\sqrt{2}\text{ m}

20\sqrt{3}\text{ m}

Show answer & explanation

Answer: $40\text{ m}$

Answer

The height of the lighthouse is

40\text{ m}

.

The height of the lighthouse is

40\text{ m}

. Since boat

X

is due South and boat

Y

is due East, the line segments connecting the base of the lighthouse to each boat form a right angle (

90^\circ

). Using basic trigonometry, the distance to boat

X

is

h\sqrt{3}

and to boat

Y

is

h

. Applying Pythagoras' theorem to the right triangle formed on the sea surface gives

(h\sqrt{3})^2 + h^2 = 80^2

, which simplifies to

4h^2 = 6400

, giving

h = 40\text{ m}

.

Step-by-Step Solution

1

Express the horizontal distance from the lighthouse base

L

to boat

X

in terms of height

h

.

LX = \frac{h}{\tan 30^\circ} = h\sqrt{3}\text{ m}

The angle of elevation from boat

X

to the top of the lighthouse is equal to the angle of depression (

30^\circ

).

2

Express the horizontal distance from the lighthouse base

L

to boat

Y

in terms of height

h

.

LY = \frac{h}{\tan 45^\circ} = h\text{ m}

The angle of elevation from boat

Y

to the top of the lighthouse is

45^\circ

.

3

Set up Pythagoras' theorem for right-angled triangle

XLY

on the horizontal plane.

XY^2 = LX^2 + LY^2 \implies 80^2 = (h\sqrt{3})^2 + h^2

Boat

X

is due South and boat

Y

is due East of the lighthouse base, making

\angle XLY = 90^\circ

.

4

Solve the algebraic equation for

h

.

6400 = 3h^2 + h^2 = 4h^2 \implies h^2 = 1600 \implies h = 40\text{ m}

Dividing

6400

by

4

gives

1600

, whose square root is

40

.

Key Concept

3D Geometry combining Angles of Elevation/Depression with Perpendicular Bearings

Question 135Question →

Find the sum of all values of $x$ (in degrees) in the interval $0^\circ \le x \le 180^\circ$ that satisfy the trigonometric equation $\cos(3x - 45^\circ) = -\frac{\sqrt{2}}{2}$ .

Show answer & explanation

Answer: 330

Answer

The sum of all values of

x

satisfying the equation in the domain

0^\circ \le x \le 180^\circ

is 330.

Transforming the domain

0^\circ \le x \le 180^\circ

gives

-45^\circ \le 3x - 45^\circ \le 495^\circ

. The angles within this range where the cosine value equals

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

are

135^\circ

,

225^\circ

, and

495^\circ

. Solving

3x - 45^\circ

for each of these angles gives

x = 60^\circ

,

90^\circ

, and

180^\circ

. Summing these three roots yields

330^\circ

.

Step-by-Step Solution

1

Determine the interval of the transformed angle

\theta = 3x - 45^\circ

.

-45^\circ \le 3x - 45^\circ \le 495^\circ

Applying the linear transformation

3x - 45^\circ

to the given domain

0^\circ \le x \le 180^\circ

establishes the exact boundaries for the argument of the cosine function.

2

Find all values of

\theta

within

[-45^\circ, 495^\circ]

satisfying

\cos \theta = -\frac{\sqrt{2}}{2}

.

\theta \in \{135^\circ, 225^\circ, 495^\circ\}

Cosine is negative in Quadrants II and III. The reference angle is

45^\circ

, giving base solutions

135^\circ

and

225^\circ

. Adding

360^\circ

to

135^\circ

gives

495^\circ

, which lies exactly on the upper boundary.

3

Solve for

x

by setting

3x - 45^\circ

equal to each valid

\theta

.

x \in \{60^\circ, 90^\circ, 180^\circ\}

Isolating

x

yields

x = \frac{\theta + 45^\circ}{3}

. All three resulting values lie within

[0^\circ, 180^\circ]

.

4

Sum the valid solution values.

60^\circ + 90^\circ + 180^\circ = 330^\circ

The problem asks specifically for the sum of all solution angles in degrees.

Key Concept

Solving multi-angle trigonometric equations with phase shifts across a specified domain

Question 136Question →

A solid rectangular wooden block measures $10\text{ cm}$ by $14\text{ cm}$ by $12\text{ cm}$ . A cylindrical hole of radius $3.5\text{ cm}$ is drilled straight through the block along its height of $12\text{ cm}$ . Taking $\pi = \frac{22}{7}$ , what is the volume of the remaining solid in $\text{cm}^3$ ?

Show answer & explanation

Answer: 1218

Answer

The volume of the remaining solid is

1218\text{ cm}^3

.

The initial total volume of the rectangular block is

10 \times 14 \times 12 = 1680\text{ cm}^3

. The volume of the cylindrical hole drilled through it is

\frac{22}{7} \times (3.5)^2 \times 12 = 462\text{ cm}^3

. Subtracting the removed cylindrical volume from the block gives

1680 - 462 = 1218\text{ cm}^3

.

Step-by-Step Solution

1

Calculate the total volume of the rectangular block.

Volume of block =

1680\text{ cm}^3

The initial volume of the cuboid before drilling is calculated by multiplying its length, width, and height:

10 \times 14 \times 12 = 1680\text{ cm}^3

.

2

Calculate the volume of the cylindrical hole removed from the block.

Volume of cylinder =

462\text{ cm}^3

The cylindrical hole has radius

r = 3.5\text{ cm} = \frac{7}{2}\text{ cm}

and height

h = 12\text{ cm}

. Using

V = \pi r^2 h

, we get

\frac{22}{7} \times \frac{49}{4} \times 12 = 462\text{ cm}^3

.

3

Subtract the volume of the cylindrical hole from the total volume of the block.

Remaining volume =

1218\text{ cm}^3

Because material is removed by drilling, the remaining volume is

1680 - 462 = 1218\text{ cm}^3

.

Key Concept

Volume of a composite solid (cuboid with a cylindrical cavity)

Question 137Question →

What is the period, in degrees, of the trigonometric function $y = 7\sin(5x) - 2$ ?

Show answer & explanation

Answer: 72

Answer

The period of the trigonometric function is 72 degrees.

For any function of the form

y = A\sin(Bx) + D

, the period

T

in degrees is calculated using

T = \frac{360^\circ}{|B|}

. For the given equation

y = 7\sin(5x) - 2

, the value of

B

is 5. Substituting this into the formula gives

T = \frac{360^\circ}{5} = 72^\circ

.

Step-by-Step Solution

1

Identify the coefficient

B

of the variable

x

in the given function

y = 7\sin(5x) - 2

.

Here,

A = 7

,

B = 5

, and

D = -2

.

The period of a sine function depends on the angular frequency parameter

B

multiplying the input variable

x

.

2

Apply the standard formula for finding the period

T

of a sine function in degrees:

T = \frac{360^\circ}{|B|}

.

T = \frac{360^\circ}{5} = 72^\circ

.

Dividing the standard full revolution of

360^\circ

by the multiplier

5

determines the angle needed for one full cycle.

Key Concept

Period of a Trigonometric Graph

Estimated Time:45s

Question 138Question →

The interior angles of a convex polygon form an arithmetic progression. If the smallest interior angle is $120^\circ$ and the common difference between consecutive interior angles is $5^\circ$ , how many sides does the polygon have?

Show answer & explanation

Answer: 9

Answer

The polygon has 9 sides.

Equating the interior angle sum formula

(n-2) \times 180^\circ

with the sum formula for an arithmetic progression

\frac{n}{2}[2(120^\circ) + (n-1)5^\circ]

yields the quadratic equation

n^2 - 25n + 144 = 0

. Solving gives

n = 9

and

n = 16

. For

n = 16

, the largest angle would be

195^\circ

, which is impossible for a convex polygon. Thus, the only valid number of sides is 9.

Step-by-Step Solution

1

Formulate expressions for the sum of the interior angles.

The interior angle sum of an

n

-sided polygon is

S = (n-2) \times 180^\circ

. As an arithmetic sequence with

a = 120^\circ

and

d = 5^\circ

, the sum is

S = \frac{n}{2}[2(120^\circ) + (n-1)5^\circ]

.

Both methods express the total sum of all interior angles of the polygon.

2

Equate the two sum formulas and simplify into a quadratic equation.

(n-2) \times 180 = \frac{n}{2}(235 + 5n) \implies n^2 - 25n + 144 = 0

.

Expanding and dividing by 5 reduces the equality to standard quadratic form.

3

Solve for

n

and test the validity of the roots based on polygon convexity.

(n - 9)(n - 16) = 0 \implies n = 9

or

n = 16

. Testing

n = 16

gives a largest angle of

120^\circ + 15(5^\circ) = 195^\circ

(invalid as it exceeds

180^\circ

). Testing

n = 9

gives a largest angle of

120^\circ + 8(5^\circ) = 160^\circ

(valid).

A convex polygon cannot have any interior angle greater than or equal to

180^\circ

.

Key Concept

Polygon interior angle sum and arithmetic progressions

Question 139Question →

Find the smallest positive value of $\theta$ , in degrees, that satisfies the trigonometric equation $2\sin(3\theta - 30^\circ) = \sqrt{3}$ .

Show answer & explanation

Answer: 30

Answer

The smallest positive angle

\theta

is

30^\circ

.

To find the smallest positive value of

\theta

, first isolate the sine function by dividing both sides by 2 to obtain

\sin(3\theta - 30^\circ) = \frac{\sqrt{3}}{2}

. The smallest positive angle with a sine of

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

is

60^\circ

. Setting

3\theta - 30^\circ = 60^\circ

yields

3\theta = 90^\circ

, which gives

\theta = 30^\circ

.

Step-by-Step Solution

1

Isolate the trigonometric ratio

\sin(3\theta - 30^\circ) = \frac{\sqrt{3}}{2}

Dividing both sides of

2\sin(3\theta - 30^\circ) = \sqrt{3}

by 2 simplifies the equation into standard form.

2

Determine the primary angle solution

3\theta - 30^\circ = 60^\circ

The smallest positive angle whose sine equals

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

is

60^\circ

.

3

Solve the linear equation for

\theta

\theta = 30^\circ

Adding

30^\circ

to both sides gives

3\theta = 90^\circ

, and dividing by 3 yields

\theta = 30^\circ

.

Key Concept

Solving Trigonometric Equations with Linear Argument Transformations

Question 140Question →

In $\Delta ABC$ , side $a = 10\text{ cm}$ , side $b = 16\text{ cm}$ , and $\sin A = \frac{3}{8}$ . If $\angle B$ is an obtuse angle, what is the exact value of $\cos B$ ?

-\frac{4}{5}

\frac{4}{5}

\frac{3}{5}

-\frac{3}{5}

Show answer & explanation

Answer: $-\frac{4}{5}$

Answer

-\frac{4}{5}

Applying the Sine Rule gives

\sin B = \frac{b \sin A}{a} = \frac{16 \times \frac{3}{8}}{10} = \frac{3}{5}

. Since

\angle B

is an obtuse angle, it lies in the second quadrant where cosine is negative. Using

\cos B = -\sqrt{1 - \sin^2 B}

, we get

\cos B = -\sqrt{1 - \left(\frac{3}{5}\right)^2} = -\frac{4}{5}

.

Step-by-Step Solution

1

Apply the Sine Rule to calculate

\sin B

.

\sin B = \frac{b \sin A}{a} = \frac{16 \times \frac{3}{8}}{10} = \frac{6}{10} = \frac{3}{5}

The Sine Rule states that

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

.

2

Determine the sign of

\cos B

based on the given angle type.

Since

\angle B

is an obtuse angle (

90^\circ < B < 180^\circ

),

\cos B < 0

.

Cosine is negative in the second quadrant.

3

Calculate

\cos B

using the Pythagorean trigonometric identity.

\cos B = -\sqrt{1 - \sin^2 B} = -\sqrt{1 - \left(\frac{3}{5}\right)^2} = -\sqrt{\frac{16}{25}} = -\frac{4}{5}

Substitute

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

into

\cos B = -\sqrt{1 - \sin^2 B}

.

Key Concept

Sine Rule and Trigonometric Ratios of Obtuse Angles

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