trigonometry | Jamb Mathematics

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Get ready to conquer trigonometry in grand style! This exam will test your understanding of angles, identities, and real-world applications, so sharpen your skills and bring your A-game. Dive into your formulas, practice those tricky problems, and prepare to excel with confidence!

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Are you preparing for your JAMB Mathematics exam and feeling a bit uncertain about how to approach the topic of trigonometry? Don’t worry—you’re in the right place! This lesson is here to break it down in a simple, clear, and engaging way, helping you build the strong foundation you need to succeed. Whether you're struggling with complex questions or just seeking a quick refresher, this guide will boost your understanding and confidence. Let’s tackle trigonometry together and move one step closer to achieving your exam success! Blissful learning.

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calculation problem involving trigonometrical ratios of angles;

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Question 1:

Find the value of

\sin 30^\circ

olution:

Using standard trigonometric values,

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

Question 2:

Find the value of

\tan 45^\circ

Solution:

Using standard trigonometric values,

\tan 45^\circ = 1

Question 3:

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

, find

\sin \theta

Solution:

Using the Pythagorean identity:

\sin^2 \theta + \cos^2 \theta = 1

\sin^2 \theta = 1 - \left(\frac{3}{5}\right)^2

\sin^2 \theta = 1 - \frac{9}{25} = \frac{16}{25}

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

Question 4:

Evaluate

\sec 60^\circ

Solution:

Using standard values,

\sec 60^\circ = \frac{1}{\cos 60^\circ} = \frac{1}{\frac{1}{2}} = 2

Question 5:

Find

\cot 30^\circ

Solution:

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

Question 6:

Find

\csc 45^\circ

Solution:

\csc 45^\circ = \frac{1}{\sin 45^\circ} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}

Question 7:

Find

\tan 60^\circ

Solution:

\tan 60^\circ = \sqrt{3}

Question 8:

Evaluate

\cos^2 30^\circ + \sin^2 30^\circ

Solution:

Using the identity

\cos^2 \theta + \sin^2 \theta = 1

\cos^2 30^\circ + \sin^2 30^\circ = 1

Question 9:

Solve for

x

\sin x = 0.5

, where

0^\circ \leq x \leq 360^\circ

Solution:

\sin x = 0.5

occurs at

x = 30^\circ, 150^\circ

Question 10:

Find

\sin 90^\circ

Solution:

\sin 90^\circ = 1

Question 11:

Find

\tan 0^\circ

Solution:

\tan 0^\circ = 0

Question 12:

Evaluate

\cos 180^\circ

Solution:

\cos 180^\circ = -1

Question 13:

Find

\sin 270^\circ

Solution:

\sin 270^\circ = -1

Question 14:

Evaluate

\sec 90^\circ

Solution:

\sec 90^\circ = \frac{1}{\cos 90^\circ} = \frac{1}{0}

, which is undefined.

Question 15:

Find

\tan 135^\circ

Solution:

\tan 135^\circ = -1

Question 16:

Solve

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

, for

0^\circ \leq x \leq 360^\circ

Solution:

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

occurs at

x = 60^\circ, 300^\circ

Question 17:

Find

\tan 225^\circ

Solution:

\tan 225^\circ = 1

Question 18:

Evaluate

\csc 60^\circ

Solution:

\csc 60^\circ = \frac{1}{\sin 60^\circ} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}

Question 19:

Find

\cos 330^\circ

Solution:

\cos 330^\circ = \cos (-30^\circ) = \cos 30^\circ = \frac{\sqrt{3}}{2}

Question 20:

Find

\sin 225^\circ

Solution:

\sin 225^\circ = -\frac{\sqrt{2}}{2}

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Calculation problems on applying special angles to solve problems in trigonometry

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Question 1:

Evaluate

\sin 30^\circ + \cos 60^\circ

Solution:

Using known values,

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

and

\cos 60^\circ = \frac{1}{2}

\sin 30^\circ + \cos 60^\circ = \frac{1}{2} + \frac{1}{2} = 1

Question 2:

Find

x

\tan x = \sqrt{3}

and

0^\circ \leq x \leq 180^\circ

Solution:

From known values,

\tan x = \sqrt{3}

x = 60^\circ, 240^\circ

.
Since

0^\circ \leq x \leq 180^\circ

, the valid solution is

x = 60^\circ

Question 3:

Find

\cos^2 45^\circ + \sin^2 45^\circ

Solution:

Using the Pythagorean identity:

\cos^2 45^\circ + \sin^2 45^\circ = 1

Question 4:

Find

\sec 30^\circ

Solution:

\sec 30^\circ = \frac{1}{\cos 30^\circ} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}

Question 5:

Find

\cot 60^\circ

Solution:

\cot 60^\circ = \frac{1}{\tan 60^\circ} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}

Question 6:

Evaluate

\csc 45^\circ - \sec 45^\circ

Solution:

\csc 45^\circ = \frac{1}{\sin 45^\circ} = \frac{\sqrt{2}}{1} = \sqrt{2}

\sec 45^\circ = \frac{1}{\cos 45^\circ} = \sqrt{2}

\csc 45^\circ - \sec 45^\circ = \sqrt{2} - \sqrt{2} = 0

Question 7:

Find

\sin 60^\circ \times \cos 30^\circ

Solution:

Using known values:

\sin 60^\circ = \frac{\sqrt{3}}{2}, \cos 30^\circ = \frac{\sqrt{3}}{2}

\sin 60^\circ \times \cos 30^\circ = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} = \frac{3}{4}

Question 8:

Find

\tan 45^\circ \times \cot 45^\circ

Solution:

\tan 45^\circ = 1, \cot 45^\circ = 1

\tan 45^\circ \times \cot 45^\circ = 1 \times 1 = 1

Question 9:

Solve for

x

\cos x = -\frac{1}{2}

and

0^\circ \leq x \leq 360^\circ

Solution:

\cos x = -\frac{1}{2}

x = 120^\circ, 240^\circ

Question 10:

Find

\tan 120^\circ

Solution:

\tan 120^\circ = \tan (180^\circ - 60^\circ) = -\tan 60^\circ = -\sqrt{3}

Question 11:

Evaluate

\sin 135^\circ

Solution:

\sin 135^\circ = \sin (180^\circ - 45^\circ) = \sin 45^\circ = \frac{\sqrt{2}}{2}

Question 12:

Find

\cos 225^\circ

Solution:

\cos 225^\circ = \cos (180^\circ + 45^\circ) = -\cos 45^\circ = -\frac{\sqrt{2}}{2}

Question 13:

Find

\tan 315^\circ

Solution:

\tan 315^\circ = \tan (360^\circ - 45^\circ) = -\tan 45^\circ = -1

Question 14:

Solve

2 \cos x = 1

for

0^\circ \leq x \leq 360^\circ

Solution:

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

x = 60^\circ, 300^\circ

Question 15:

Find

\cos 30^\circ + \sin 60^\circ

Solution:

\cos 30^\circ = \frac{\sqrt{3}}{2}, \sin 60^\circ = \frac{\sqrt{3}}{2}

\cos 30^\circ + \sin 60^\circ = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \sqrt{3}

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Calculation problems involving angles of elevation and depression

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Question 1:

A person standing 50 m away from a building observes the top at an angle of elevation of

30^\circ

. Find the height of the building.

Solution:

Using the tangent formula:

\tan \theta = \frac{\text{opposite}}{\text{adjacent}}

\tan 30^\circ = \frac{h}{50}

\frac{1}{\sqrt{3}} = \frac{h}{50}

h = \frac{50}{\sqrt{3}} = \frac{50\sqrt{3}}{3} \approx 28.87

Question 2:

A ladder leans against a wall, making a

60^\circ

angle with the ground. If the foot of the ladder is 4 m from the wall, find the length of the ladder.

Solution:

Using the cosine rule:

\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}

\cos 60^\circ = \frac{4}{L}

\frac{1}{2} = \frac{4}{L}

L = 8

Question 3:

A kite is flying at a height of 30 m. The string makes an angle of

45^\circ

with the ground. Find the length of the string.

Solution:

Using the sine rule:

\sin 45^\circ = \frac{30}{s}

\frac{\sqrt{2}}{2} = \frac{30}{s}

s = \frac{30 \times 2}{\sqrt{2}} = 30\sqrt{2} \approx 42.43

Question 4:

A person at a distance of 20 m from a lamppost observes the top at an elevation of

60^\circ

. Find the height of the lamppost.

Solution:

Using the tangent formula:

\tan 60^\circ = \frac{h}{20}

\sqrt{3} = \frac{h}{20}

h = 20\sqrt{3} \approx 34.64

Question 5:

A ship is spotted at sea from a lighthouse at a height of 50 m. The angle of depression is

20^\circ

. Find the distance of the ship from the base of the lighthouse.

Solution:

Using the tangent formula:

\tan 20^\circ = \frac{50}{d}

d = \frac{50}{\tan 20^\circ} \approx 137.64

Question 6:

A plane is flying at a height of 1000 m. The angle of depression to a point on the ground is

45^\circ

. Find the horizontal distance of the plane from the point.

Solution:

Using the tangent formula:

\tan 45^\circ = \frac{1000}{d}

1 = \frac{1000}{d}

d = 1000

Question 7:

A 6 m tree casts a shadow of 4 m. Find the angle of elevation of the sun.

Solution:

Using the tangent formula:

\tan \theta = \frac{6}{4} = 1.5

\theta = \tan^{-1}(1.5) \approx 56.31^\circ

Question 8:

A drone flies at a height of 120 m above the ground. The angle of depression to an observer is

30^\circ

. Find the horizontal distance between the observer and the drone.

Solution:

Using the tangent formula:

\tan 30^\circ = \frac{120}{d}

d = \frac{120}{\tan 30^\circ} = 120\sqrt{3} \approx 207.85

Question 9:

A balloon rises to a height of 150 m. The angle of elevation from a point 200 m away is

\theta

. Find

\theta

Solution:

Using the tangent formula:

\tan \theta = \frac{150}{200} = 0.75

\theta = \tan^{-1}(0.75) \approx 36.87^\circ

Question 10:

An observer is 30 m from a tower and sees the top at

40^\circ

. Find the tower's height.

Solution:

Using the tangent formula:

\tan 40^\circ = \frac{h}{30}

h = 30 \times \tan 40^\circ \approx 25.21

Question 11:

A person at a height of 50 m observes a boat at an angle of depression of

25^\circ

. Find the boat's distance from the observer.

Solution:

Using the sine rule:

\sin 25^\circ = \frac{50}{d}

d = \frac{50}{\sin 25^\circ} \approx 118.47

Question 12:

A bridge is built over a river. The angle of depression from the bridge to a boat is

35^\circ

. If the bridge is 70 m above water, find the boat's distance from the bridge base.

Solution:

Using the tangent formula:

d = \frac{70}{\tan 35^\circ} \approx 99.86

Question 13:

A ramp is inclined at

25^\circ

and has a length of 10 m. Find the height it reaches.

Solution:

Using the sine rule:

h = 10 \sin 25^\circ \approx 4.23

Question 14:

A 15 m ladder leans against a wall, making a

50^\circ

angle with the ground. Find the height it reaches on the wall.

Solution:

Using the sine rule:

h = 15 \sin 50^\circ \approx 11.49

Question 15:

An observer 50 m away sees the top of a tree at

35^\circ

. Find the tree's height.

Solution:

Using the tangent formula:

h = 50 \tan 35^\circ \approx 35.01

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Calculation problems involving area and solution of triangles

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Question 1:

Find the area of a triangle with base

10

m and height

8

Solution:

Using the area formula:

A = \frac{1}{2} \times \text{base} \times \text{height}

A = \frac{1}{2} \times 10 \times 8 = 40

m².

Question 2:

Find the area of a triangle with sides

a = 7

b = 9

m, and included angle

C = 60^\circ

Solution:

Using the formula:

A = \frac{1}{2} ab \sin C

A = \frac{1}{2} \times 7 \times 9 \times \sin 60^\circ

A = \frac{1}{2} \times 7 \times 9 \times \frac{\sqrt{3}}{2}

A = \frac{63\sqrt{3}}{4} \approx 27.2

m².

Question 3:

A triangle has sides

a = 8

b = 6

m, and

c = 10

m. Find its area using Heron's formula.

Solution:

First, find the semi-perimeter:

s = \frac{8 + 6 + 10}{2} = 12

Using Heron's formula:

A = \sqrt{s(s-a)(s-b)(s-c)}

A = \sqrt{12(12-8)(12-6)(12-10)}

A = \sqrt{12 \times 4 \times 6 \times 2}

A = \sqrt{576} = 24

m².

Question 4:

Solve for angle

B

in a triangle where

A = 50^\circ

C = 70^\circ

Solution:

Using the sum of angles in a triangle:

A + B + C = 180^\circ

50^\circ + B + 70^\circ = 180^\circ

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

Question 5:

Find the third side of a triangle where

a = 5

b = 7

, and included angle

C = 90^\circ

Solution:

Using the Pythagorean theorem:

c^2 = a^2 + b^2

c^2 = 5^2 + 7^2

c^2 = 25 + 49 = 74

c = \sqrt{74} \approx 8.6

Question 6:

Find the missing side in a triangle where

A = 45^\circ

B = 30^\circ

, and

c = 12

using the Law of Sines.

Solution:

Using the Law of Sines:

\frac{a}{\sin A} = \frac{c}{\sin C}

First, find

C

C = 180^\circ - (45^\circ + 30^\circ) = 105^\circ

Now, solving for

a

\frac{a}{\sin 45^\circ} = \frac{12}{\sin 105^\circ}

a = 12 \times \frac{\sin 45^\circ}{\sin 105^\circ}

a \approx 8.56

Question 7:

Find the area of an equilateral triangle with side length

10

Solution:

Using the formula for the area of an equilateral triangle:

A = \frac{\sqrt{3}}{4} s^2

A = \frac{\sqrt{3}}{4} \times 10^2

A = \frac{100\sqrt{3}}{4} \approx 43.3

m².

Question 8:

In a right-angled triangle, the hypotenuse is

10

m, and one of the angles is

30^\circ

. Find the opposite side.

Solution:

Using the sine function:

\sin 30^\circ = \frac{\text{opposite}}{\text{hypotenuse}}

\frac{1}{2} = \frac{x}{10}

x = 10 \times \frac{1}{2} = 5

Question 9:

A triangle has sides

5

6

m, and

7

m. Find the largest angle.

Solution:

Using the Law of Cosines:

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

7^2 = 5^2 + 6^2 - 2(5)(6)\cos C

49 = 25 + 36 - 60\cos C

49 = 61 - 60\cos C

\cos C = \frac{12}{60} = 0.2

C = \cos^{-1}(0.2) \approx 78.46^\circ

Question 10:

Find the area of a triangle with sides

a = 14

b = 20

m, and

c = 24

Solution:

First, find the semi-perimeter:

s = \frac{14 + 20 + 24}{2} = 29

Using Heron's formula:

A = \sqrt{s(s-a)(s-b)(s-c)}

A = \sqrt{29(29-14)(29-20)(29-24)}

A = \sqrt{29 \times 15 \times 9 \times 5}

A = \sqrt{19575} \approx 140

m².

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Calculation problems involving Bearings

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Question 1:

A ship sails 10 km due east and then 24 km due north. Find the bearing of its final position from its starting point.

Solution:

The ship moves 10 km east and 24 km north, forming a right-angled triangle.
The bearing is measured clockwise from the north.
Using the tangent function:

$\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{10}{24}$

$\theta = \tan^{-1} \left( \frac{10}{24} \right) = \tan^{-1} (0.4167) \approx 22.62^\circ$
The bearing is:

$\text{Bearing} = 360^\circ - (90^\circ - 22.62^\circ) = 67.38^\circ$ .

Question 2:

A plane flies from point A on a bearing of

135^\circ

for 50 km, then turns and flies 60 km due east. Find the bearing of its final position from A.

Solution:

The first leg is at $135^\circ$ , meaning it moves southeast.
The second leg is due east.
Using vector addition and trigonometry, find the total displacement and use inverse tangent to determine the final bearing.
Calculate resultant distance and bearing using:

$\tan \theta = \frac{\text{vertical component}}{\text{horizontal component}}$ .
Solve for $\theta$ and adjust to get the correct bearing.

Question 3:

A car moves 30 km on a bearing of

060^\circ

and then 40 km on a bearing of

120^\circ

. Find its resultant displacement and bearing from the starting point.

Solution:

Resolve the movements into horizontal and vertical components.
Use the cosine and sine functions to determine total displacement.
Use inverse tangent to find the resultant angle and adjust it to standard bearings.

Question 4:

An aircraft flies 100 km on a bearing of

045^\circ

and then 80 km on a bearing of

315^\circ

. Find its final position relative to the starting point.

Solution:

Break both legs into x- and y-components using trigonometry.
Sum the components and use the Pythagorean theorem to find the final distance.
Use inverse tangent to determine the bearing.

Question 5:

A person walks 5 km on a bearing of

030^\circ

and then 12 km due north. Find the resultant displacement and bearing.

Solution:

Resolve movements into x- and y-components.
Compute resultant displacement using Pythagoras' theorem.
Find the angle using inverse tangent and adjust for correct bearing.

Question 6:

A ship sails 15 km due north, then 20 km due east. Find the bearing of its final position from the starting point.

Solution:

The ship moves in a right-angled triangle.
Use:

$\tan \theta = \frac{\text{eastward distance}}{\text{northward distance}} = \frac{20}{15}$ .

$\theta = \tan^{-1} (1.3333) \approx 53.13^\circ$ .
The final bearing is:

$90^\circ - 53.13^\circ = 36.87^\circ$ .

Question 7:

An observer sees a lighthouse at a bearing of

300^\circ

. After moving 5 km north, the new bearing is

310^\circ

. Find the distance to the lighthouse.

Solution:

Use trigonometry and the Law of Sines to determine the distance to the lighthouse.

Question 8:

A plane flies 80 km on a bearing of

220^\circ

, then 60 km on a bearing of

280^\circ

. Find its final displacement and bearing from the starting point.

Solution:

Resolve movements into x- and y-components.
Compute the resultant displacement.
Use inverse tangent to determine the bearing.

Question 9:

A ship sails 40 km on a bearing of

150^\circ

, then 30 km on a bearing of

210^\circ

. Find its final position.

Solution:

Resolve each leg into components.
Find the total displacement using the Pythagorean theorem.
Use inverse tangent for the bearing.

Question 10:

A hiker moves 8 km on a bearing of

100^\circ

, then 6 km on a bearing of

200^\circ

. Find the resultant displacement and final bearing.

Solution:

Decompose the movements into components.
Sum the x- and y-components.
Use the Pythagorean theorem and inverse tangent to get the final displacement and bearing.

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I recommend you check my Post on the following:

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- Jamb Mathematics- Lesson notes on trigonometry for utme Success

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This is all we can take on "Jamb Mathematics - Lesson Notes on trigonometry for UTME Candidate"

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