Trigonometric ratios of specific angles

Every trigonometry problem in Class 10 leans on the same five angles: $0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$ . Rather than looking these values up, you are expected to know $\sin 30^\circ = \frac12$ or $\tan 60^\circ = \sqrt3$ instantly, the way you know your multiplication tables. The good news is that none of it needs memorising by brute force — every entry falls out of two very simple triangles you can redraw in under a minute: an equilateral triangle cut in half, and an isosceles right triangle. Once you can rebuild the table on demand, you are ready for the numerical evaluations this chapter asks for, and for the Heights and Distances chapter later, where angles of elevation and depression are always restricted to these same five values.

The core insight

The exact trigonometric values at $0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$ come from two elementary triangles — half an equilateral triangle for $30^\circ$ and $60^\circ$ , an isosceles right triangle for $45^\circ$ — with $0^\circ$ and $90^\circ$ read off as the limiting cases where one side shrinks to zero.

The core idea

The five standard angles are not arbitrary — they are exactly the angles that appear when you cut two of the simplest possible triangles in half.

The equilateral route ( $30^\circ, 60^\circ$ ). Drop a perpendicular from one vertex of an equilateral triangle to the opposite side. That single altitude splits the $60^\circ$ vertex angle into two $30^\circ$ halves and produces a right triangle whose three sides are in the ratio $1 : \sqrt3 : 2$ — a ratio worth remembering on sight.

The isosceles route ( $45^\circ$ ). A right triangle with both legs equal has two $45^\circ$ acute angles by symmetry, and Pythagoras fixes the hypotenuse at $\sqrt2$ times a leg, giving sides in the ratio $1 : 1 : \sqrt2$ .

The limiting angles ( $0^\circ, 90^\circ$ ). These are not built from a triangle at all — imagine the angle in a right triangle shrinking toward $0^\circ$ : the opposite side collapses to nothing while the adjacent side grows to match the hypotenuse, so $\sin 0^\circ = 0$ and $\cos 0^\circ = 1$ . Swap the roles as the angle grows toward $90^\circ$ and the reverse happens.

The standard table, the one thing in this topic worth having by heart:

$\theta$	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$
$\sin\theta$	$0$	$\frac{1}{2}$	$\frac{1}{\sqrt2}$	$\frac{\sqrt3}{2}$	$1$
$\cos\theta$	$1$	$\frac{\sqrt3}{2}$	$\frac{1}{\sqrt2}$	$\frac{1}{2}$	$0$
$\tan\theta$	$0$	$\frac{1}{\sqrt3}$	$1$	$\sqrt3$	undefined
$\csc\theta$	undefined	$2$	$\sqrt2$	$\frac{2}{\sqrt3}$	$1$
$\sec\theta$	$1$	$\frac{2}{\sqrt3}$	$\sqrt2$	$2$	undefined
$\cot\theta$	undefined	$\sqrt3$	$1$	$\frac{1}{\sqrt3}$	$0$

A quick shape-check that helps you spot a swapped value at a glance: as $\theta$ climbs from $0^\circ$ to $90^\circ$ , $\sin\theta$ only ever increases and $\cos\theta$ only ever decreases.

\sin^2\theta + \cos^2\theta = 1 \qquad \text{(holds for every row of the table above)}

where: $\theta$ is any of the five standard angles. This identity is a handy check — square and add any $\sin$ – $\cos$ pair from the table and the result must land on exactly $1$ ; if it does not, one of the two entries was copied wrong.

Figure

Two triangles side by side, an equilateral triangle split by an altitude into two 30 60 90 triangles with sides in ratio 1 root 3 2, and an isosceles right triangle with legs 1 and hypotenuse root 2 giving the two 45 degree angles

Derivation ADeriving the exact 30 degree and 60 degree ratios from an equilateral triangle

The goal is to show, from nothing more than the definition of an equilateral triangle and the Pythagoras theorem, exactly why $\sin 30^\circ = \frac12$ , $\cos 30^\circ = \frac{\sqrt3}{2}$ , $\tan 30^\circ = \frac{1}{\sqrt3}$ and the matching $60^\circ$ values are true, rather than asking you to accept the standard table on faith.

Step 1. Draw equilateral triangle $ABC$ with every side equal to $2$ units — the exact length does not matter, but $2$ keeps the arithmetic clean since it is about to be halved. Every angle of an equilateral triangle is $60^\circ$ , so angle $A =$ angle $B =$ angle $C = 60^\circ$ .

Step 2. Drop a perpendicular from vertex $A$ down to side $BC$ , meeting it at point $D$ . Because triangle $ABC$ is equilateral (hence isosceles from any vertex), this altitude does two things at once: it bisects the base, so $BD = DC = 1$ , and it bisects angle $A$ , splitting the $60^\circ$ angle into two $30^\circ$ halves.

Step 3. Focus on right triangle $ABD$ alone. It has a right angle at $D$ , angle $ABD = 60^\circ$ (unchanged from the original triangle), angle $BAD = 30^\circ$ (half of angle $A$ ), hypotenuse $AB = 2$ , and base $BD = 1$ . Find the third side $AD$ using the Pythagoras theorem: $AD = \sqrt{AB^2 - BD^2} = \sqrt{2^2 - 1^2} = \sqrt{4-1} = \sqrt3$ .

Step 4. Read the ratios straight off triangle $ABD$ for angle $BAD = 30^\circ$ : the side opposite is $BD = 1$ , the side adjacent is $AD = \sqrt3$ , and the hypotenuse is $AB = 2$ . This gives $\sin 30^\circ = \frac{1}{2}$ , $\cos 30^\circ = \frac{\sqrt3}{2}$ , and $\tan 30^\circ = \frac{1}{\sqrt3}$ .

Step 5. Read the same triangle from the other acute angle, $ABD = 60^\circ$ , where the opposite and adjacent sides swap roles: opposite is now $AD = \sqrt3$ , adjacent is $BD = 1$ , hypotenuse is still $AB = 2$ . This gives $\sin 60^\circ = \frac{\sqrt3}{2}$ , $\cos 60^\circ = \frac{1}{2}$ , and $\tan 60^\circ = \sqrt3$ — exactly the values in the standard table, and notice $\sin 30^\circ = \cos 60^\circ$ and $\cos 30^\circ = \sin 60^\circ$ , a pattern that always holds for complementary angles.

Worked example

Problem. Evaluate $2\sin^2 30^\circ + 3\cos^2 45^\circ - \tan^2 60^\circ$ .

Step 1. Pull the needed values from the standard table: $\sin 30^\circ = \frac12$ , $\cos 45^\circ = \frac{1}{\sqrt2}$ , $\tan 60^\circ = \sqrt3$ .

Step 2. Square each value before substituting, since the expression asks for squares, not the values themselves.

\sin^2 30^\circ = \frac14, \qquad \cos^2 45^\circ = \frac12, \qquad \tan^2 60^\circ = 3

Step 3. Substitute into the original expression and simplify the coefficients.

2\left(\frac14\right) + 3\left(\frac12\right) - 3 = \frac12 + \frac32 - 3

Step 4. Combine the fractions first, then subtract: $\frac12 + \frac32 = 2$ , and $2 - 3 = -1$ . The value of the expression is $\mathbf{-1}$ .

Competitive stretch

Take $A = 30^\circ$ and $B = 60^\circ$ and check that $\sin(A+B) = \sin A\cos B + \cos A \sin B$ using only the standard table, without knowing the general angle-sum formula. The left side: $A+B = 90^\circ$ , so $\sin(A+B) = \sin 90^\circ = 1$ . The right side: $\sin A \cos B + \cos A \sin B = \left(\frac12\right)\left(\frac12\right) + \left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt3}{2}\right) = \frac14 + \frac34 = 1$ . Both sides equal $1$ , confirming the relation for this specific pair of angles — the general proof of the sum formula itself belongs to a later class and is not required here.

Common traps

Swapping $\sin 30^\circ = \frac12$ with $\sin 60^\circ = \frac{\sqrt3}{2}$ — use the rule that $\sin$ increases and $\cos$ decreases from $0^\circ$ to $90^\circ$ to catch a swap instantly.
Writing $\tan 90^\circ$ or $\cot 0^\circ$ as if they had a finite value — both are undefined, since they correspond to dividing by a side of length zero.
Squaring a surd incorrectly, e.g. treating $\left(\frac{1}{\sqrt2}\right)^2$ as $\frac{1}{2\sqrt2}$ instead of the correct $\frac12$ .
Rounding to decimals like $0.5$ or $0.707$ instead of keeping exact surd form — CBSE answers must stay exact ( $\frac{1}{\sqrt2}$ , $\sqrt3$ , and so on) unless a question explicitly asks for a decimal.

Ask the tutor: If the table keeps slipping your mind under exam pressure, ask the AI tutor for a fast way to redraw the two triangles and rebuild every entry from scratch.

Do I really have to memorise the whole table, or can I rebuild it every time?

Both skills matter, but rebuilding is the safety net. In a fast-moving exam you want the table on instant recall, yet if your memory ever slips, you should be able to redraw the equilateral triangle for $30^\circ$ and $60^\circ$ and the isosceles right triangle for $45^\circ$ in under a minute and read the ratios straight off the sides. Practising the derivation a few times makes the memorised version stick far better than rote repetition alone.

Why are tan 90 degrees and cot 0 degrees undefined instead of just very large or zero?

Because $\tan\theta = \frac{\sin\theta}{\cos\theta}$ , and at $\theta = 90^\circ$ the denominator $\cos 90^\circ = 0$ , so the ratio is a division by zero — mathematics leaves this undefined rather than assigning it any number. The same thing happens to $\cot\theta = \frac{\cos\theta}{\sin\theta}$ at $\theta = 0^\circ$ , where $\sin 0^\circ = 0$ . It is not that these values are simply "very large"; the ratio genuinely has no defined value at that exact angle.

I keep confusing which side is opposite and which is adjacent when I read the triangle. Any trick?

Always ask "opposite and adjacent to which angle?" first — the same triangle gives different opposite and adjacent sides depending on which acute angle you are standing at. In the equilateral-triangle derivation, the side $BD$ is opposite to angle $30^\circ$ but adjacent to angle $60^\circ$ in the very same triangle. Label the angle you are working with clearly before you write any ratio, and the opposite-versus-adjacent mix-up mostly disappears.

Trigonometric ratios of specific angles

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads