SSC CGL Preparation: Concepts and Practice

Learning outcome

Derive exact trigonometric values at standard acute angles and use them in calculations and right-triangle measurements.

Concepts and assumptions

All angles here are in degrees. A reference angle inside a non-degenerate right triangle is strictly between 0° and 90°. Side lengths are positive; answers remain exact unless explicitly approximated.

For 45°, use a right isosceles triangle with equal legs 1. Pythagoras gives hypotenuse √2. Thus sin 45° = cos 45° = 1/√2 = √2/2 and tan 45° = 1.

For 30° and 60°, bisect an equilateral triangle of side 2 by its altitude. This splits one 60° angle into two 30° angles. Each resulting right triangle has hypotenuse 2, short leg 1 and height √(4 − 1) = √3. The side opposite 30° is 1; the side opposite 60° is √3. Dividing appropriate sides gives the table below.

The unit-circle extension defines cos θ and sin θ as horizontal and vertical coordinates. At 0° the point is (1, 0); at 90° it is (0, 1). These endpoints are not acute reference angles of ordinary right triangles.

Ratio	0°	30°	45°	60°	90°
sin	0	1/2	√2/2	√3/2	1
cos	1	√3/2	√2/2	1/2	0
tan	0	√3/3	1	√3	Undefined

For acute angles, cosec = 1/sin, sec = 1/cos and cot = 1/tan. In the endpoint extension, tan 90° and sec 90° are undefined; cosec 0° and cot 0° are undefined. Division by zero does not give infinity.

The notation sin² θ means (sin θ)², not sin(θ²).

Worked examples

Example 1 — Direct substitution. Evaluate (2 sin 30° + cos 60°)/tan 45°.

The denominator is 1, so division is valid. Substitute: [2(1/2) + 1/2]/1 = (1 + 1/2)/1 = 3/2.

Example 2 — Side lengths. In ABC, ∠C = 90°, ∠A = 30° and AB = 18 cm.

BC is opposite A: BC/18 = sin 30° = 1/2, so BC = 9 cm. AC/18 = cos 30° = √3/2, so AC = 9√3 cm. Check: 9² + (9√3)² = 81 + 243 = 324 = 18².

Example 3 — Reciprocals and products. Evaluate tan 60° × cot 30° − sec 45°/cosec 45°.

cot 30° = 1/(√3/3) = √3. sec 45° = cosec 45° = √2, a nonzero denominator. Thus the expression is √3 × √3 − √2/√2 = 3 − 1 = 2.

Common mistakes

Swapping 30° and 60° values; using a leg as the hypotenuse; forgetting to square coefficients; treating undefined endpoint ratios as numbers.

Practice questions

1. Evaluate 4 sin 30° − 2 cos 60° + tan 45°.
2. An acute angle is 45° and its adjacent leg is 9 cm. Find the opposite leg and hypotenuse.
3. Evaluate sin² 60° + cos² 45°.
4. Evaluate sec² 30° + cosec² 60° − cot² 45°.

Worked solutions

1. Substitute: 4(1/2) − 2(1/2) + 1 = 2 − 1 + 1 = 2.
2. Opposite = 9 tan 45° = 9 cm. Hypotenuse = 9/cos 45° = 9/(√2/2) = 9√2 cm.
3. (√3/2)² + (√2/2)² = 3/4 + 1/2 = 5/4.
4. sec 30° = cosec 60° = 2/√3; cot 45° = 1. Therefore 4/3 + 4/3 − 1 = 5/3.