Trigonometry: Practice Questions & Notes

Trigonometry is the study of the relationships between angles and sides of right-angled triangles using trigonometric ratios.

At its core, it’s about answering:

"Given an angle, how are the sides of a triangle related?"
"Given sides, can we find the angle?"

These relationships are fundamental to geometry, physics, architecture, and real-life measurements like navigation and satellite tracking.

Trigonometric Ratios

For a right-angled triangle, with angle $\theta$ , and sides:

Ratio	Definition	Formula
Sine	Opposite / Hypotenuse	$\sin \theta = \frac{\text{Opp}}{\text{Hyp}}$
Cosine	Adjacent / Hypotenuse	$\cos \theta = \frac{\text{Adj}}{\text{Hyp}}$
Tangent	Opposite / Adjacent	$\tan \theta = \frac{\text{Opp}}{\text{Adj}}$
Cosecant	Reciprocal of sine	$\csc \theta = \frac{1}{\sin \theta}$
Secant	Reciprocal of cosine	$\sec \theta = \frac{1}{\cos \theta}$
Cotangent	Reciprocal of tangent	$\cot \theta = \frac{1}{\tan \theta}$

$\theta$	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$
$\sin \theta$	0	$\frac{1}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{\sqrt{3}}{2}$	1
$\cos \theta$	1	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{2}$	0
$\tan \theta$	0	$\frac{1}{\sqrt{3}}$	1	$\sqrt{3}$	Not defined

Remember with the "SinCos Table Trick": √0/2, √1/2, √2/2, √3/2, √4/2 for sin values from 0° to 90°. Reverse for cos.

These are always true for any angle $\theta$ :

Tip: Derive the second and third by dividing the first identity by $\cos^2 \theta$ or $\sin^2 \theta$ respectively.

\sin(90^\circ - \theta) = \cos \theta,\quad \tan(90^\circ - \theta) = \cot \theta, \quad \sec(90^\circ - \theta) = \csc \theta

For a point $P(x, y)$ and angle $\theta$ made with x-axis:

\tan \theta = \frac{y}{x}

The value of sin, cos, etc., depends on the quadrant the point lies in.

Quadrant	Sign of sin	cos	tan
I	+	+	+
II	+	–	–
III	–	–	+
IV	–	+	–

Mnemonic: "All Students Take Calculus"
I: All +
II: Sin +
III: Tan +
IV: Cos +

Trigonometry is used to find height or distance using angles and a known side.

Example:
If the angle of elevation to a tower is $30^\circ$ and the observer is 100 m away horizontally:

\tan 30^\circ = \frac{h}{100} \Rightarrow h = 100 \times \frac{1}{\sqrt{3}} \approx 57.7 \text{ m}

Mistake	Tip
Confusing tan and sin	tan is Opp/Adj, sin is Opp/Hyp
Using wrong quadrant signs	Use ASTC rule
Forgetting that sec, csc, cot are reciprocals	Memorize in pairs (sin–csc, cos–sec, tan–cot)
Not converting angles properly	Make sure angles are in degrees unless stated otherwise
Using wrong triangle side	Carefully label sides w.r.t. angle in question

Find $\tan \theta$ if $\sin \theta = \frac{3}{5}$ , $\theta \in (0^\circ, 90^\circ)$

Then $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$
So $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{3/5}{4/5} = \frac{3}{4}$

Simplify $\frac{1 - \cos^2 \theta}{\sin^2 \theta}$

= \frac{\sin^2 \theta}{\sin^2 \theta} = 1

Find the height of a tree if it casts a 20 m shadow and the angle of elevation of the sun is $45^\circ$

\tan 45^\circ = \frac{h}{20} \Rightarrow 1 = \frac{h}{20} \Rightarrow h = 20 \text{ m}

A tower is 100 m tall. From the top, the angle of depression to a car on the ground is $30^\circ$ . Find distance of the car from the tower base.

\tan 30^\circ = \frac{100}{x} \Rightarrow x = \frac{100}{\tan 30^\circ} = 100 \sqrt{3} \approx 173.2 \text{ m}