Angles of Elevation, Depression, and Bearings Practice Questions — JAMB UTME

Question 1Question →

A vessel departs from a harbor $H$ and sails $12\text{ km}$ on a bearing of $040^\circ$ to reach point $A$ . From point $A$ , it changes course and sails $5\text{ km}$ on a bearing of $130^\circ$ to reach point $B$ . What is the bearing of point $B$ from harbor $H$ , to the nearest degree?

063^\circ

073^\circ

017^\circ

107^\circ

Show answer & explanation

Answer: $063^\circ$

Answer

063^\circ

(or

063^\circ

to the nearest degree)

The navigation path forms a right-angled triangle at point

A

because the reverse bearing of

H

from

A

(

220^\circ

) differs from the bearing of

B

from

A

(

130^\circ

) by exactly

90^\circ

. Using the right triangle trigonometric ratio

\tan(\angle AHB) = \frac{5}{12}

, the angle

\angle AHB

is found to be approximately

22.62^\circ

. Adding this angle to the initial bearing of

040^\circ

yields

062.62^\circ

, which rounds to

063^\circ

.

Step-by-Step Solution

1

Determine the back bearing of

H

from

A

and calculate the interior angle at

A

.

Back bearing of

H

from

A = 040^\circ + 180^\circ = 220^\circ

. Interior angle

\angle HAB = 220^\circ - 130^\circ = 90^\circ

.

Knowing the directions of

AH

and

AB

allows us to find the interior angle of

\triangle HAB

at point

A

.

2

Calculate the interior angle

\angle AHB

using right-triangle trigonometry.

\tan(\angle AHB) = \frac{AB}{HA} = \frac{5}{12} \implies \angle AHB = \arctan\left(\frac{5}{12}\right) \approx 22.62^\circ.

Since

\triangle HAB

has a right angle at

A

, tangent relates the opposite side

AB = 5\text{ km}

to the adjacent side

HA = 12\text{ km}

.

3

Calculate the total bearing of

B

from

H

.

\text{Bearing} = 040^\circ + 22.62^\circ = 062.62^\circ \approx 063^\circ.

The line

HB

lies to the right of line

HA

, so the interior angle

\angle AHB

is added to the initial bearing of line

HA

(

040^\circ

).

Key Concept

Bearings and Right-Angled Triangles

Estimated Time:1m 30s

Question 2Question →

A boat sails $10\text{ km}$ from a port $P$ on a bearing of $040^\circ$ to a point $Q$ . From $Q$ , it changes direction and sails $10\sqrt{3}\text{ km}$ on a bearing of $130^\circ$ to reach a point $R$ . What is the bearing of port $P$ from point $R$ ?

280^\circ

100^\circ

250^\circ

160^\circ

Show answer & explanation

Answer: $280^\circ$

Answer

The bearing of port P from point R is

280^\circ

.

By drawing North reference lines at P, Q, and R, the back bearing of P from Q is

040^\circ + 180^\circ = 220^\circ

. The bearing of R from Q is

130^\circ

, creating an interior right angle of

220^\circ - 130^\circ = 90^\circ

at Q. Using right-triangle trigonometry,

\tan(\angle QPR) = \frac{10\sqrt{3}}{10} = \sqrt{3}

, giving

\angle QPR = 60^\circ

. Adding this to the initial bearing of

040^\circ

gives the bearing of R from P as

100^\circ

. Finally, adding

180^\circ

gives the reverse bearing of P from R as

280^\circ

.

Step-by-Step Solution

1

Determine the interior angle

\angle PQR

at vertex

Q

\angle PQR = 90^\circ

The back bearing of P from Q is

040^\circ + 180^\circ = 220^\circ

. The bearing of R from Q is

130^\circ

. The interior angle between QP and QR is

220^\circ - 130^\circ = 90^\circ

.

2

Calculate the angle

\angle QPR

inside right-angled triangle

PQR

\angle QPR = 60^\circ

Since triangle

PQR

is right-angled at

Q

,

\tan(\angle QPR) = \frac{\text{opposite}}{\text{adjacent}} = \frac{QR}{PQ} = \frac{10\sqrt{3}}{10} = \sqrt{3}

. Therefore,

\angle QPR = \arctan(\sqrt{3}) = 60^\circ

.

3

Find the forward bearing of point

R

from port

P

Bearing of

R

from

P = 100^\circ

The bearing of Q from P is

040^\circ

. Since R lies to the right (clockwise) of segment PQ, add

\angle QPR = 60^\circ

to

040^\circ

:

040^\circ + 60^\circ = 100^\circ

.

4

Compute the back bearing of port

P

from point

R

Bearing of

P

from

R = 280^\circ

The back bearing is obtained by adding

180^\circ

to the forward bearing from P to R:

100^\circ + 180^\circ = 280^\circ

.

Key Concept

Three-point bearing calculations using right-triangle trigonometry and back bearings

Estimated Time:2m 30s

Question 3Question →

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 4Question →

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 5Question →

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 6Question →

A ship departs from a port $P$ and sails $10\text{ km}$ on a bearing of $060^\circ$ to reach a position $Q$ . From $Q$ , the ship changes course and sails $10\text{ km}$ on a bearing of $150^\circ$ to arrive at point $R$ . What is the bearing of point $R$ from point $P$ ?

015^\circ

105^\circ

135^\circ

150^\circ

Show answer & explanation

Answer: $105^\circ$

Answer

The bearing of point

R

from point

P

is

105^\circ

.

The back bearing from position

Q

to port

P

is

240^\circ

. The difference between

240^\circ

and the new bearing of

150^\circ

yields an interior angle of

90^\circ

at vertex

Q

. Since both distances

PQ

and

QR

are equal to

10\text{ km}

, triangle

PQR

is a

45^\circ-45^\circ-90^\circ

isosceles right triangle. Adding

\angle QPR = 45^\circ

to the initial bearing of

060^\circ

yields a bearing of

105^\circ

for point

R

from point

P

.

Step-by-Step Solution

1

Determine the back bearing of

P

from

Q

.

The back bearing of

P

from

Q

is

060^\circ + 180^\circ = 240^\circ

.

To find the internal angle at

Q

, we need the direction of

P

relative to

Q

.

2

Calculate the interior angle

\angle PQR

.

\angle PQR = 240^\circ - 150^\circ = 90^\circ

.

The difference between the line back to

P

(

240^\circ

) and the line to

R

(

150^\circ

) forms the interior angle at

Q

.

3

Determine the properties of triangle

PQR

and angle

\angle QPR

.

Since

PQ = QR = 10\text{ km}

and

\angle PQR = 90^\circ

, triangle

PQR

is an isosceles right triangle, so

\angle QPR = 45^\circ

.

The two equal sides subtend equal acute angles in a right-angled triangle:

(180^\circ - 90^\circ) / 2 = 45^\circ

.

4

Calculate the total bearing of

R

from

P

.

Bearing of

R

from

P = 060^\circ + 45^\circ = 105^\circ

.

Point

R

lies clockwise relative to the segment

PQ

, so we add

\angle QPR

to the initial bearing of

PQ

.

Key Concept

Three-point bearing calculations using geometry of parallel lines and right-angled triangles.

Question 7Question →

A forest ranger station $Y$ is located on a bearing of $115^\circ$ from an observation tower $X$ . What is the bearing of the observation tower $X$ from the forest ranger station $Y$ ?

065^\circ

245^\circ

295^\circ

115^\circ

Show answer & explanation

Answer: $295^\circ$

Answer

295^\circ

The bearing of

Y

from

X

is

115^\circ

. To find the bearing of

X

from

Y

(the back bearing), add

180^\circ

to the forward bearing because

115^\circ < 180^\circ

. Calculating

115^\circ + 180^\circ

yields

295^\circ

.

Step-by-Step Solution

1

Identify the given forward bearing

The bearing of

Y

from

X

is

\theta = 115^\circ

.

This is the initial directional angle measured clockwise from True North at point

X

.

2

Calculate the back bearing of

X

from

Y

Back bearing

= 115^\circ + 180^\circ = 295^\circ

.

Because the forward bearing is less than

180^\circ

, the back bearing is obtained by adding

180^\circ

to find the opposite direction.

Key Concept

Back Bearing Calculation

Question 8Question →

A vertical flagpole of height $15\text{ m}$ standing on level ground casts a shadow of length $15\sqrt{3}\text{ m}$ . Calculate the angle of elevation of the sun in degrees.

Show answer & explanation

Answer: 30

Answer

The angle of elevation of the sun is 30 degrees.

The tangent of the angle of elevation is given by the height divided by the shadow length,

\tan\theta = \frac{15}{15\sqrt{3}} = \frac{1}{\sqrt{3}}

, which corresponds to an angle of

30^\circ

.

Step-by-Step Solution

1

Formulate the trigonometric relationship using the right triangle formed by the flagpole, the shadow, and the sunlight ray.

\tan\theta = \frac{\text{height of flagpole}}{\text{length of shadow}} = \frac{15}{15\sqrt{3}}

The tangent of an angle in a right-angled triangle is defined as the ratio of the opposite side to the adjacent side.

2

Simplify the ratio and evaluate the angle.

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

The standard acute angle whose tangent value is

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

is

30^\circ

.

Key Concept

Angle of Elevation and Trigonometric Ratios

Question 9Question →

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 10Question →

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 11Question →

A coastal monitoring station at point $O$ tracks two vessels on horizontal water. Vessel $A$ is located $15\text{ km}$ from $O$ on a bearing of $070^\circ$ , while Vessel $B$ is located $20\text{ km}$ from $O$ on a bearing of $160^\circ$ . What is the direct distance between Vessel $A$ and Vessel $B$ in kilometers?

Show answer & explanation

Answer: 25

Answer

The direct distance between Vessel A and Vessel B is 25 km.

The difference between the two bearings (

160^\circ - 070^\circ = 90^\circ

) establishes that triangle

AOB

is a right-angled triangle at station

O

. Applying Pythagoras' theorem yields

AB = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25\text{ km}

.

Step-by-Step Solution

1

Find the angle between the lines of sight to the two vessels

\angle AOB = 160^\circ - 070^\circ = 90^\circ

Subtracting the smaller bearing angle from the larger bearing angle from the same origin point gives the included angle.

2

Set up the equation for distance AB using Pythagoras' theorem

AB^2 = 15^2 + 20^2 = 225 + 400 = 625

Since the included angle is 90 degrees, the three points form a right-angled triangle where AB is the hypotenuse.

3

Calculate the principal square root of 625

AB = \sqrt{625} = 25\text{ km}

Taking the square root converts the squared distance into the direct linear distance between the vessels.

Key Concept

Calculating the distance between two points using bearings and right-angled triangle properties (Pythagoras' theorem).

Estimated Time:1m 30s

Question 12Question →

A cargo ship departs from port $M$ and sails $24\text{ km}$ on a bearing of $050^\circ$ to reach point $N$ . From point $N$ , the ship changes course and sails $10\text{ km}$ on a bearing of $140^\circ$ to reach point $P$ . What is the direct distance, in kilometers, from port $M$ to point $P$ ?

Show answer & explanation

Answer: 26

Answer

The direct distance from port M to point P is 26 km.

The back bearing from N to M is 230°, and the bearing from N to P is 140°. The interior angle at N is 230° - 140° = 90°. Using the Pythagorean theorem for the right triangle formed by M, N, and P, the direct distance is √(24² + 10²) = √676 = 26 km.

Step-by-Step Solution

1

Calculate the interior angle

\angle MNP

at point

N

\angle MNP = (050^\circ + 180^\circ) - 140^\circ = 230^\circ - 140^\circ = 90^\circ

The back bearing from

N

to

M

is

230^\circ

. Subtracting the forward bearing to

P

(

140^\circ

) gives the enclosed interior angle.

2

Apply the Pythagorean theorem to right-angled triangle

MNP

MP = \sqrt{MN^2 + NP^2} = \sqrt{24^2 + 10^2} = \sqrt{576 + 100} = \sqrt{676} = 26\text{ km}

Because

\angle MNP = 90^\circ

, triangle

MNP

is a right-angled triangle with hypotenuse

MP

.

Key Concept

Bearings and Right-Angled Triangles

Question 13Question →

A field surveyor starts at point $P$ and walks $10\text{ km}$ due East to point $Q$ . From point $Q$ , she changes direction and walks $10\text{ km}$ on a bearing of $210^\circ$ to reach point $R$ . What is the bearing of point $P$ from point $R$ ?

330^\circ

060^\circ

150^\circ

300^\circ

Show answer & explanation

Answer: $330^\circ$

Answer

The bearing of point

P

from point

R

is

330^\circ

.

Point

Q

lies

10\text{ km}

East of

P

(bearing

090^\circ

). From

Q

, moving

10\text{ km}

on bearing

210^\circ

forms an interior angle of

60^\circ

with the line

QP

(which has bearing

270^\circ

). Because

PQ = QR = 10\text{ km}

and the included angle is

60^\circ

,

\triangle PQR

is equilateral, making

\angle QRP = 60^\circ

. The back bearing from

R

to

Q

is

030^\circ

. Subtracting the

60^\circ

interior angle from

030^\circ

yields a bearing of

330^\circ

for point

P

from point

R

.

Step-by-Step Solution

1

Determine the interior angle

\angle PQR

at point

Q

.

\angle PQR = 60^\circ

Since

P

is due West of

Q

, the bearing of

P

from

Q

is

270^\circ

. The bearing of

R

from

Q

is

210^\circ

. The interior angle between these two lines is

270^\circ - 210^\circ = 60^\circ

.

2

Analyze the properties of triangle

PQR

.

Triangle

PQR

is an equilateral triangle with side lengths

10\text{ km}

and interior angles of

60^\circ

.

Given

PQ = 10\text{ km}

and

QR = 10\text{ km}

,

\triangle PQR

is isosceles. Since the vertex angle

\angle PQR = 60^\circ

, the remaining two angles are also

60^\circ

each.

3

Calculate the back bearing of

Q

from

R

.

Bearing of

Q

from

R

is

030^\circ

.

The bearing of

R

from

Q

is

210^\circ

. The back bearing is

210^\circ - 180^\circ = 030^\circ

.

4

Compute the bearing of

P

from

R

.

Bearing of

P

from

R

is

330^\circ

.

From line

RQ

(bearing

030^\circ

), line

RP

lies

60^\circ

counter-clockwise (since

\angle QRP = 60^\circ

). Thus,

030^\circ - 60^\circ = -30^\circ \equiv 330^\circ

.

Key Concept

Bearings and Triangle Geometry

Angles of Elevation, Depression, and Bearings

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