MathsCBSE Class 10 · Introduction to Trigonometry

Trigonometric Ratios of 30°, 45° and 60° for Class 10

Q: Do I need to learn the derivation of these values or just the table?

Learn both. The syllabus includes finding these values, so a question can ask you to derive them from an equilateral or isosceles right triangle. Knowing the derivation also lets you rebuild any value you forget.

Q: Should I rationalise answers like 1 by root 3?

Both forms are correct. Rationalise only if the question asks, or if it makes the next step easier. Do not waste time on it otherwise.

Q: Is sin 60 equal to 2 times sin 30?

No. Sin 60 is root 3 by 2, while 2 times sin 30 is 1. Trigonometric ratios do not double when the angle doubles.

Q: Are complementary angle results like sin of 90 minus A equal to cos A in the syllabus?

Trigonometric ratios of complementary angles as a topic are not in the 2026-27 syllabus.

Updated 14 October 2026 · Fact-checked

The ratios of 30, 45 and 60 degrees are exact values such as sin 30 = 1/2, tan 45 = 1 and cos 30 = root 3 by 2. You derive them from an isosceles right triangle and a halved equilateral triangle. The board asks you to evaluate expressions, find unknown angles, solve for a variable and derive the values.

Every value comes from these two triangles: the side ratios 1 : 1 : √2 and 1 : √3 : 2 give all sin, cos and tan values, and 'a' always cancels.

Illustration drawn for QuestionBuddy.

Understand Trigonometric Ratios of 30°, 45° and 60°

Most angles have trigonometric ratios that are long decimals. Three angles, $3 0^{\circ}$ , $4 5^{\circ}$ and $6 0^{\circ}$ , are special. Their ratios are exact numbers built from $1$ , $2$ , $2$ and $3$ . You do not need a calculator for them, and you do not need to memorise them blindly, because two simple triangles give every value.

**The $4 5^{\circ}$ triangle.** In $△ A B C$ right-angled at $B$ with $∠ A = 4 5^{\circ}$ , the third angle $∠ C$ is also $4 5^{\circ}$ . Equal angles face equal sides, so $A B = B C = a$ . By Pythagoras, $A C = a^{2} + a^{2} = a 2$ . So $sin 4 5^{\circ} = \frac{a}{a 2} = \frac{1}{2}$ , $cos 4 5^{\circ} = \frac{1}{2}$ and $tan 4 5^{\circ} = 1$ .

**The $3 0^{\circ}$ and $6 0^{\circ}$ triangle.** Take an equilateral triangle $A B C$ of side $2 a$ . Drop the perpendicular $A D$ to $B C$ . The two triangles $A B D$ and $A C D$ are congruent (RHS), so $B D = D C = a$ and $∠ B A D = 3 0^{\circ}$ , while $∠ B = 6 0^{\circ}$ . By Pythagoras, $A D = 4 a^{2} - a^{2} = a 3$ . In $△ A B D$ : $sin 3 0^{\circ} = \frac{B D}{A B} = \frac{1}{2}$ , $cos 3 0^{\circ} = \frac{A D}{A B} = \frac{3}{2}$ , $tan 3 0^{\circ} = \frac{1}{3}$ . Standing at $B$ instead, the opposite side is $A D$ , so $sin 6 0^{\circ} = \frac{3}{2}$ , $cos 6 0^{\circ} = \frac{1}{2}$ , $tan 6 0^{\circ} = 3$ .

Why does the value of $a$ not matter? Because the ratios depend only on the angle, not on the size of the triangle (similar triangles). The $a$ always cancels. The other three ratios are just reciprocals: for example $cosec 3 0^{\circ} = 2$ and $sec 4 5^{\circ} = 2$ .

These values hold only for these exact angles. You use them to evaluate expressions, to find an unknown angle (if $tan 2 x = 1$ with $2 x$ acute, then $2 x = 4 5^{\circ}$ ), and to check identities. Along with the values at $0^{\circ}$ and $9 0^{\circ}$ they make the full table, and they are the backbone of the next chapter on heights and distances.

Trigonometric Ratios of 30°, 45° and 60°: formulas to remember

Values at 30°: $sin 3 0^{\circ} = \frac{1}{2}, cos 3 0^{\circ} = \frac{3}{2}, tan 3 0^{\circ} = \frac{1}{3}, cosec 3 0^{\circ} = 2, sec 3 0^{\circ} = \frac{2}{3}, cot 3 0^{\circ} = 3$; From the half equilateral triangle, standing at the 30° vertex.
Values at 45°: $sin 4 5^{\circ} = cos 4 5^{\circ} = \frac{1}{2}, tan 4 5^{\circ} = cot 4 5^{\circ} = 1, sec 4 5^{\circ} = cosec 4 5^{\circ} = 2$; From the isosceles right triangle with sides a, a, a√2.
Values at 60°: $sin 6 0^{\circ} = \frac{3}{2}, cos 6 0^{\circ} = \frac{1}{2}, tan 6 0^{\circ} = 3, cosec 6 0^{\circ} = \frac{2}{3}, sec 6 0^{\circ} = 2, cot 6 0^{\circ} = \frac{1}{3}$; Same half equilateral triangle, standing at the 60° vertex.
Squared values (most used): $sin^{2} 3 0^{\circ} = \frac{1}{4}, sin^{2} 4 5^{\circ} = \frac{1}{2}, sin^{2} 6 0^{\circ} = \frac{3}{4}, tan^{2} 3 0^{\circ} = \frac{1}{3}, tan^{2} 6 0^{\circ} = 3$; Square the whole fraction, numerator and denominator.
Side ratios of the two special triangles: $3 0^{\circ} : 6 0^{\circ} : 9 0^{\circ} \Rightarrow 1 : 3 : 2, 4 5^{\circ} : 4 5^{\circ} : 9 0^{\circ} \Rightarrow 1 : 1 : 2$; Sides opposite the listed angles. Rebuild any value from these.

How to solve Trigonometric Ratios of 30°, 45° and 60° questions

Almost every question on this topic is either an evaluation, an equation for an unknown angle, or a derivation. This method covers all three.

1Read the angles carefully and note any squares, products or sums of angles such as $sin (A + B)$ .
2If an angle is written as a sum or multiple, find that angle first; never split $sin (A + B)$ into $sin A + sin B$ .
3Replace each ratio by its exact value from the two special triangles, writing each substitution on its own line.
4Square whole fractions: $(\frac{3}{2})^{2} = \frac{3}{4}$ .
5Simplify the numerator and denominator separately using an LCM, then divide.
6For an unknown angle, reduce the equation to one ratio equal to a standard value, read off the acute angle, then solve for the variable.
7Check by substituting back, or by a quick identity such as $sin^{2} A + cos^{2} A = 1$ .

Trigonometric Ratios of 30°, 45° and 60° in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Values of the ratios at 30°, 45° and 60°

Imagine you and your friend are flying kites at a Makar Sankranti festival on flat ground, both standing at the same point $O$ . Both kites happen to be directly above the same point $P$ on the ground, one above the other. Your string makes $3 0^{\circ}$ with the ground and your friend's string makes $6 0^{\circ}$ . Suppose your friend's kite is 20 m higher than yours. Assume both strings are taut, so each is a straight line, and ignore the height of your hands.

Find the distance $O P$ , the height of each kite and the length of each string.

Try it first, then show the solutionSolution

Let $O P = d$ m. Each kite, $P$ and $O$ form a right triangle with the right angle at $P$ , because the kite is directly above $P$ .
Your kite: $tan 3 0^{\circ} = \frac{h _{1}}{d}$ , so $h_{1} = \frac{d}{3}$ .
Friend's kite: $tan 6 0^{\circ} = \frac{h _{2}}{d}$ , so $h_{2} = d 3$ .
$h_{2} - h_{1} = 20$ , so $d 3 - \frac{d}{3} = \frac{3 d - d}{3} = \frac{2 d}{3} = 20$ .
$d = 103$ m, so $h_{1} = \frac{10 3}{3} = 10$ m and $h_{2} = 103 \times 3 = 30$ m.
Your string $= \frac{h _{1}}{s i n 3 0 ^{\circ}} = \frac{10}{1/2} = 20$ m.
Friend's string $= \frac{h _{2}}{s i n 6 0 ^{\circ}} = \frac{30 \times 2}{3} = 203$ m.
Check: $20 cos 3 0^{\circ} = 103$ and $203 cos 6 0^{\circ} = 103$ , both equal to $d$ ; and $30 - 10 = 20$ m.

Answer: $O P = 103$ m; heights 10 m (your kite) and 30 m (friend's kite); strings 20 m and $203$ m.

Memory hook: Same ground distance, so tan decides the height: tan 60° is three times tan 30°, so the 60° kite flies three times as high.

Tricks to remember Trigonometric Ratios of 30°, 45° and 60°

Short memory aids. Say each one aloud once, then use it on a question the same day.

The square-root ladder
Write $0, 1, 2, 3, 4$ and divide each by 2. You get $sin 0^{\circ}, sin 3 0^{\circ}, sin 4 5^{\circ}, sin 6 0^{\circ}, sin 9 0^{\circ}$ : $0, \frac{1}{2}, \frac{1}{2}, \frac{3}{2}, 1$ . Read it backwards for $cos$ .
Use it when: Rebuild the sine and cosine rows in seconds before the exam starts.
Tan is sin over cos
$tan 3 0^{\circ} = \frac{1/2}{3 /2} = \frac{1}{3}$ and $tan 6 0^{\circ} = 3$ . Small angle, small tan; $4 5^{\circ}$ sits in the middle at 1.
Use it when: Never confuse $tan 3 0^{\circ}$ and $tan 6 0^{\circ}$ .
1, root 3, 2
In the $3 0^{\circ}$ - $6 0^{\circ}$ - $9 0^{\circ}$ triangle the smallest side faces the smallest angle: 1 faces $3 0^{\circ}$ , $3$ faces $6 0^{\circ}$ , 2 is the hypotenuse.
Use it when: Read any ratio of 30° or 60° straight from the picture.

Trigonometric Ratios of 30°, 45° and 60°: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Evaluating a mixed expression · 2 marks · Very short answerBoard level

Evaluate: $\frac{2 s i n ^{2} 6 0 ^{\circ} + 3 s e c ^{2} 4 5 ^{\circ} - c o t ^{2} 3 0 ^{\circ}}{4 c o s ^{2} 4 5 ^{\circ} - t a n ^{2} 6 0 ^{\circ}}$

Try it first, then show the solutionSolution

$sin^{2} 6 0^{\circ} = \frac{3}{4}$ , $sec^{2} 4 5^{\circ} = 2$ , $cot^{2} 3 0^{\circ} = 3$ , $cos^{2} 4 5^{\circ} = \frac{1}{2}$ , $tan^{2} 6 0^{\circ} = 3$ . (½ mark)
Numerator $= 2 \times \frac{3}{4} + 3 \times 2 - 3 = \frac{3}{2} + 3 = \frac{9}{2}$ . (½ mark)
Denominator $= 4 \times \frac{1}{2} - 3 = - 1$ . (½ mark)
Value $= \frac{9/2}{- 1} = - \frac{9}{2}$ . (½ mark)

Answer: $- \frac{9}{2}$

Example 2: Finding an unknown angle · 3 marks · Short answerBoard level

Find the acute angle $x$ if $3 tan 2 x = cos 6 0^{\circ} + sin 4 5^{\circ} cos 4 5^{\circ}$ , where $2 x$ is acute. Verify your answer.

Try it first, then show the solutionSolution

$cos 6 0^{\circ} = \frac{1}{2}$ and $sin 4 5^{\circ} cos 4 5^{\circ} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2}$ . (1 mark)
RHS $= \frac{1}{2} + \frac{1}{2} = 1$ , so $tan 2 x = \frac{1}{3}$ . (½ mark)
$tan 3 0^{\circ} = \frac{1}{3}$ , so $2 x = 3 0^{\circ}$ . (½ mark)
$x = 1 5^{\circ}$ . (½ mark)
Check: $3 tan 3 0^{\circ} = 3 \cdot \frac{1}{3} = 1 =$ RHS. (½ mark)

Answer: $x = 1 5^{\circ}$

Example 3: Two conditions on A + B and A − B · 3 marks · Short answerHarder board level

If $cos (A + B) = \frac{1}{2}$ and $sin (A - B) = \frac{1}{2}$ , where $0^{\circ} < A + B < 9 0^{\circ}$ and $A > B$ , find $A$ and $B$ . Hence find $sec^{2} A + cot^{2} 2 B$ .

Try it first, then show the solutionSolution

$cos 6 0^{\circ} = \frac{1}{2}$ , so $A + B = 6 0^{\circ}$ . (½ mark)
$sin 3 0^{\circ} = \frac{1}{2}$ , so $A - B = 3 0^{\circ}$ . (½ mark)
Adding: $2 A = 9 0^{\circ}$ , so $A = 4 5^{\circ}$ ; then $B = 1 5^{\circ}$ . (1 mark)
$sec^{2} 4 5^{\circ} = 2$ and $2 B = 3 0^{\circ}$ , so $cot^{2} 3 0^{\circ} = 3$ . (½ mark)
$sec^{2} A + cot^{2} 2 B = 2 + 3 = 5$ . (½ mark)

Answer: $A = 4 5^{\circ}$ , $B = 1 5^{\circ}$ , value $= 5$

Example 4: Deriving the values and verifying two results · 5 marks · Long answerBoard level

(i) $A B C$ is an equilateral triangle of side $2 a$ , and $A D$ is perpendicular to $B C$ . Using this figure, find $sin 3 0^{\circ}$ , $cos 3 0^{\circ}$ , $tan 3 0^{\circ}$ , $sin 6 0^{\circ}$ , $cos 6 0^{\circ}$ and $tan 6 0^{\circ}$ .
(ii) Hence verify that $sin 6 0^{\circ} = \frac{2 t a n 3 0 ^{\circ}}{1 + t a n ^{2} 3 0 ^{\circ}}$ and $cos 6 0^{\circ} = \frac{1 - t a n ^{2} 3 0 ^{\circ}}{1 + t a n ^{2} 3 0 ^{\circ}}$ .

Try it first, then show the solutionSolution

$△ A B D ≅ △ A C D$ by RHS ( $A B = A C$ , $A D$ common, $∠ A D B = ∠ A D C = 9 0^{\circ}$ ), so $B D = D C = a$ and $∠ B A D = 3 0^{\circ}$ ; $∠ B = 6 0^{\circ}$ . (1 mark)
By Pythagoras in $△ A B D$ : $A D = 4 a^{2} - a^{2} = a 3$ . (½ mark)
At $∠ B A D = 3 0^{\circ}$ : $sin 3 0^{\circ} = \frac{a}{2 a} = \frac{1}{2}$ , $cos 3 0^{\circ} = \frac{a 3}{2 a} = \frac{3}{2}$ , $tan 3 0^{\circ} = \frac{a}{a 3} = \frac{1}{3}$ . (1 mark)
At $∠ B = 6 0^{\circ}$ : $sin 6 0^{\circ} = \frac{3}{2}$ , $cos 6 0^{\circ} = \frac{1}{2}$ , $tan 6 0^{\circ} = 3$ . (½ mark)
$1 + tan^{2} 3 0^{\circ} = 1 + \frac{1}{3} = \frac{4}{3}$ . (½ mark)
$\frac{2 t a n 3 0 ^{\circ}}{1 + t a n ^{2} 3 0 ^{\circ}} = \frac{2/ 3}{4/3} = \frac{6}{4 3} = \frac{3}{2} = sin 6 0^{\circ}$ . (1 mark)
$\frac{1 - t a n ^{2} 3 0 ^{\circ}}{1 + t a n ^{2} 3 0 ^{\circ}} = \frac{2/3}{4/3} = \frac{1}{2} = cos 6 0^{\circ}$ . (½ mark)

Answer: (i) $sin 3 0^{\circ} = \frac{1}{2}$ , $cos 3 0^{\circ} = \frac{3}{2}$ , $tan 3 0^{\circ} = \frac{1}{3}$ , $sin 6 0^{\circ} = \frac{3}{2}$ , $cos 6 0^{\circ} = \frac{1}{2}$ , $tan 6 0^{\circ} = 3$ . (ii) Both results verified.

Example 5: Case study: two set squares · 4 marks · Case-basedBoard level

A geometry box has two right-angled set squares. Set square P has angles $3 0^{\circ}$ , $6 0^{\circ}$ , $9 0^{\circ}$ and its longest edge (hypotenuse) is 10 cm. Set square Q has angles $4 5^{\circ}$ , $4 5^{\circ}$ , $9 0^{\circ}$ and its hypotenuse is $82$ cm.
(i) Find the length of the shortest edge of P. (1 mark)
(ii) Find the length of each shorter edge of Q. (1 mark)
(iii) Find the areas of P and Q, and say which is larger. Take $3 \approx 1.73$ for comparison. (2 marks)

Try it first, then show the solutionSolution

(i) The shortest edge faces $3 0^{\circ}$ : $10 sin 3 0^{\circ} = 10 \times \frac{1}{2} = 5$ cm. (1 mark)
(ii) Each leg $= 82 sin 4 5^{\circ} = 82 \times \frac{1}{2} = 8$ cm. (1 mark)
(iii) Other leg of P $= 10 cos 3 0^{\circ} = 53$ cm, so area of P $= \frac{1}{2} \times 5 \times 53 = \frac{25 3}{2}$ cm $^{2}$ . (1 mark)
Area of Q $= \frac{1}{2} \times 8 \times 8 = 32$ cm $^{2}$ ; area of P $\approx \frac{25 \times 1.73}{2} \approx 21.6$ cm $^{2}$ , so Q is larger. (1 mark)

Answer: (i) 5 cm (ii) 8 cm (iii) P: $\frac{25 3}{2}$ cm $^{2}$ (about 21.6 cm $^{2}$ ), Q: 32 cm $^{2}$ ; Q is larger.

Example 6: MCQ with an unknown angle · 1 mark · MCQBoard level

If $3 tan^{2} θ - 1 = 0$ and $θ$ is acute, then $sin 2 θ + cos 2 θ$ equals
(a) $\frac{3 + 1}{2}$
(b) $3$
(c) $1$
(d) $\frac{3 - 1}{2}$

Try it first, then show the solutionSolution

$tan^{2} θ = \frac{1}{3}$ , and $tan θ > 0$ for an acute angle, so $tan θ = \frac{1}{3}$ and $θ = 3 0^{\circ}$ . (½ mark)
$2 θ = 6 0^{\circ}$ , so the value is $sin 6 0^{\circ} + cos 6 0^{\circ} = \frac{3}{2} + \frac{1}{2} = \frac{3 + 1}{2}$ . (½ mark)

Answer: (a) $\frac{3 + 1}{2}$

Example 7: Assertion-reason with a hidden angle · 1 mark · Assertion-reasonHarder board level

Assertion (A): If $θ$ is acute and $sec θ = \frac{2 t a n θ}{3}$ , then $θ = 6 0^{\circ}$ .
Reason (R): $sec θ \geq 1$ for every acute angle $θ$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

Write both sides over $cos θ$ : $\frac{1}{c o s θ} = \frac{2 s i n θ}{3 c o s θ}$ , so $sin θ = \frac{3}{2}$ and $θ = 6 0^{\circ}$ . A is true. (½ mark)
R is true, since $0 < cos θ \leq 1$ . But the proof of A never uses R; it uses $tan θ = \frac{s i n θ}{c o s θ}$ . So R does not explain A. (½ mark)

Answer: (b) Both A and R are true but R is not the correct explanation of A.

Example 8: Given sin θ + cos θ = √2, find tan θ + cot θ · 3 marks · Short answerHarder board level

If $θ$ is acute and $sin θ + cos θ = 2$ , find $θ$ and the value of $tan θ + cot θ$ .

Try it first, then show the solutionSolution

Square both sides: $sin^{2} θ + 2 sin θ cos θ + cos^{2} θ = 2$ . (½ mark)
Using $sin^{2} θ + cos^{2} θ = 1$ : $1 + 2 sin θ cos θ = 2$ , so $sin θ cos θ = \frac{1}{2}$ . (½ mark)
$(sin θ - cos θ)^{2} = 1 - 2 sin θ cos θ = 1 - 1 = 0$ , so $sin θ = cos θ$ , $tan θ = 1$ and $θ = 4 5^{\circ}$ . (1 mark)
$tan θ + cot θ = tan 4 5^{\circ} + cot 4 5^{\circ} = 1 + 1 = 2$ . (½ mark)
Check: $sin 4 5^{\circ} + cos 4 5^{\circ} = \frac{2}{2} = 2$ ; also $tan θ + cot θ = \frac{s i n ^{2} θ + c o s ^{2} θ}{s i n θ c o s θ} = \frac{1}{1/2} = 2$ . (½ mark)

Answer: $θ = 4 5^{\circ}$ and $tan θ + cot θ = 2$

Example 9: Two right triangles sharing a side · 5 marks · Long answerHarder board level

In $△ A B C$ , $A D$ is perpendicular to $B C$ , with $D$ between $B$ and $C$ . $∠ B = 4 5^{\circ}$ , $∠ C = 3 0^{\circ}$ and $B C = 10 (3 + 1)$ cm. Find $A D$ , $A B$ , $A C$ and the area of $△ A B C$ .

Try it first, then show the solutionSolution

Let $A D = h$ cm. In $△ A D B$ , right-angled at $D$ : $tan 4 5^{\circ} = \frac{A D}{B D}$ , so $1 = \frac{h}{B D}$ and $B D = h$ . (1 mark)
In $△ A D C$ , right-angled at $D$ : $tan 3 0^{\circ} = \frac{A D}{D C}$ , so $\frac{1}{3} = \frac{h}{D C}$ and $D C = h 3$ . (1 mark)
$B D + D C = B C$ (D lies between B and C): $h + h 3 = 10 (3 + 1)$ , so $h (3 + 1) = 10 (3 + 1)$ and $h = 10$ cm. (1 mark)
$A B = \frac{A D}{s i n 4 5 ^{\circ}} = 102$ cm and $A C = \frac{A D}{s i n 3 0 ^{\circ}} = \frac{10}{1/2} = 20$ cm. (1 mark)
Area $= \frac{1}{2} \times B C \times A D = \frac{1}{2} \times 10 (3 + 1) \times 10 = 50 (3 + 1)$ cm $^{2}$ . Check: $D C = 103$ and $A C^{2} = 100 + 300 = 400 = 2 0^{2}$ . (1 mark)

Answer: $A D = 10$ cm, $A B = 102$ cm, $A C = 20$ cm, area $= 50 (3 + 1)$ cm $^{2}$

Example 10: A quadratic in tan with two acute roots · 3 marks · Short answerHOTS / competency

The acute angles $α$ and $β$ , with $α < β$ , both satisfy $tan^{2} x - (1 + 3) tan x + 3 = 0$ . Find $α$ and $β$ , and hence evaluate $sin^{2} α + cos^{2} β + sec^{2} (2 α - β)$ .

Try it first, then show the solutionSolution

Treat $tan x$ as the unknown. The constant term $3 = 1 \times 3$ and $1 + 3$ is their sum, so $(tan x - 1) (tan x - 3) = 0$ . (1 mark)
$tan x = 1$ gives $4 5^{\circ}$ and $tan x = 3$ gives $6 0^{\circ}$ . Both are acute, so $α = 4 5^{\circ}$ and $β = 6 0^{\circ}$ . (1 mark)
$2 α - β = 3 0^{\circ}$ . Value $= sin^{2} 4 5^{\circ} + cos^{2} 6 0^{\circ} + sec^{2} 3 0^{\circ} = \frac{1}{2} + \frac{1}{4} + \frac{4}{3} = \frac{6 + 3 + 16}{12} = \frac{25}{12}$ . (1 mark)

Answer: $α = 4 5^{\circ}$ , $β = 6 0^{\circ}$ , value $= \frac{25}{12}$

Where students lose marks in Trigonometric Ratios of 30°, 45° and 60°

The slips examiners see most often, and what to write instead.

Mistake: Squaring only the numerator
Students square the surd and forget the denominator.
Wrong working
$cos^{2} 3 0^{\circ} = \frac{3}{2}$
Right working
$cos^{2} 3 0^{\circ} = (\frac{3}{2})^{2} = \frac{3}{4}$
Do this: Put brackets round the value before squaring.
Usually costs: The final answer and usually ½ to 1 mark.
Mistake: Splitting a ratio over a sum or multiple
Treating sin like a number that multiplies.
Wrong working
$sin (3 0^{\circ} + 3 0^{\circ}) = sin 3 0^{\circ} + sin 3 0^{\circ} = 1$
Right working
$sin (3 0^{\circ} + 3 0^{\circ}) = sin 6 0^{\circ} = \frac{3}{2}$
Do this: Always add the angles first, then take the ratio.
Usually costs: The whole question in an MCQ or the method marks in a long answer.
Mistake: Swapping the 30° and 60° values
Rote memory without the triangle picture.
Wrong working
$tan 3 0^{\circ} = 3$
Right working
$tan 3 0^{\circ} = \frac{1}{3}$ , since the side opposite $3 0^{\circ}$ is the shortest.
Do this: Sketch the 1 : √3 : 2 triangle in the margin.
Usually costs: Every later step built on the wrong value.
Mistake: Forgetting that the angle found is 2x, not x
Forgetting that the found angle is $2 x$ , not $x$ .
Wrong working
$tan 2 x = 1 \Rightarrow x = 4 5^{\circ}$
Right working
$tan 2 x = 1 \Rightarrow 2 x = 4 5^{\circ} \Rightarrow x = 22. 5^{\circ}$
Do this: Write the line ' $2 x = \dots$ ' before dividing.
Usually costs: The final ½ to 1 mark.

Exam tips

Write each substitution on a separate line before simplifying; examiners give marks for correct values even if arithmetic slips later.
Keep answers in surd form like $\frac{3}{2}$ unless the question asks for a decimal or rationalised form.
In derivation questions, state why $B D = D C$ (congruent triangles by RHS) and show Pythagoras; these steps carry marks.
For unknown-angle questions, always substitute your answer back to check; it takes a few seconds.
In MCQs, test options with the exact values rather than trusting a pattern.

Practice questions on Trigonometric Ratios of 30°, 45° and 60°

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The value of $\frac{1 - t a n ^{2} 3 0 ^{\circ}}{1 + t a n ^{2} 3 0 ^{\circ}}$ is
(a) $0$
(b) $\frac{1}{2}$
(c) $1$
(d) $\frac{3}{2}$
Hint
Use $tan^{2} 3 0^{\circ} = \frac{1}{3}$ .
Answer
(b) $\frac{1}{2}$ . Working: $\frac{1 - 1/3}{1 + 1/3} = \frac{2/3}{4/3} = \frac{1}{2}$ .
Q2 · 3 marks · Short answerHarder board level
If $sin (A + 2 B) = \frac{3}{2}$ and $cos (A + 4 B) = 0$ , where $0^{\circ} < A + 2 B < A + 4 B \leq 9 0^{\circ}$ and $A > B$ , find $A$ and $B$ . Hence evaluate $sin^{2} (A + B) + tan^{2} 2 A$ .
Hint
$cos 9 0^{\circ} = 0$ . Subtract the two angle equations.
Answer
$A + 2 B = 6 0^{\circ}$ and $A + 4 B = 9 0^{\circ}$ . Subtracting: $2 B = 3 0^{\circ}$ , so $B = 1 5^{\circ}$ and $A = 3 0^{\circ}$ ( $A > B$ , as required). Then $A + B = 4 5^{\circ}$ and $2 A = 6 0^{\circ}$ , so the value is $\frac{1}{2} + 3 = \frac{7}{2}$ .
Q3 · 3 marks · Short answerHarder board level
If $tan (A + B) = 3$ and $tan (A - B) = \frac{1}{3}$ , where $A + B$ and $A - B$ are acute and $A > B$ , find $A$ and $B$ . Hence evaluate $\frac{s i n ^{2} A + c o s ^{2} 2 B}{t a n ^{2} A - s e c ^{2} 2 B}$ and explain why the answer is negative.
Hint
Find $A + B$ and $A - B$ from the table, then add and subtract.
Answer
$A + B = 6 0^{\circ}$ , $A - B = 3 0^{\circ}$ , so $A = 4 5^{\circ}$ , $B = 1 5^{\circ}$ and $2 B = 3 0^{\circ}$ . Numerator $= \frac{1}{2} + \frac{3}{4} = \frac{5}{4}$ . Denominator $= 1 - \frac{4}{3} = - \frac{1}{3}$ . Value $= \frac{5}{4} \times (- 3) = - \frac{15}{4}$ . It is negative because $sec^{2} 3 0^{\circ} = \frac{4}{3}$ is more than $tan^{2} 4 5^{\circ} = 1$ .
Q4 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a ladder of length $L$ m stands with its foot at a point $P$ on the floor of a lane between two parallel vertical walls. Leaned against the left wall, it makes $6 0^{\circ}$ with the ground. Keeping its foot at $P$ , it is turned over to rest on the right wall, where it makes $3 0^{\circ}$ with the ground. The lane is $3 (3 + 1)$ m wide.
(i) Write the width of the lane in terms of $L$ . (1 mark)
(ii) Find $L$ . (1 mark)
(iii) Find the height reached on each wall and the difference between them. Take $3 \approx 1.73$ . (2 marks)
Hint
The width is the sum of the two horizontal distances from $P$ to the walls.
Answer
(i) Distance to left wall $= L cos 6 0^{\circ} = \frac{L}{2}$ ; to right wall $= L cos 3 0^{\circ} = \frac{3 L}{2}$ . Width $= \frac{L ( 1 + 3 )}{2}$ . (1)
(ii) $\frac{L ( 1 + 3 )}{2} = 3 (3 + 1)$ , so $L = 6$ m. (1)
(iii) Left wall: $6 sin 6 0^{\circ} = 33 \approx 5.19$ m. Right wall: $6 sin 3 0^{\circ} = 3$ m. Difference $= 3 (3 - 1) \approx 2.19$ m. (2)
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If $θ$ is acute and $2 sin θ = tan θ$ , then $θ = 6 0^{\circ}$ .
Reason (R): $tan θ = \frac{s i n θ}{c o s θ}$ , and $sin θ \neq = 0$ for an acute angle.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Replace $tan θ$ using R and divide by $sin θ$ .
Answer
(a). Using R, $2 sin θ = \frac{s i n θ}{c o s θ}$ . Since $sin θ \neq = 0$ , divide by it: $cos θ = \frac{1}{2}$ , so $θ = 6 0^{\circ}$ . Both are true and R is exactly what gives A.
Q6 · 2 marks · Very short answerHarder board level
If $3 sin θ = cos θ$ and $θ$ is acute, find $θ$ and the value of $\frac{3 c o s ^{2} θ + 2 c o s θ}{3 c o s θ + 2}$ .
Hint
Factorise the numerator before substituting.
Answer
$tan θ = \frac{1}{3}$ , so $θ = 3 0^{\circ}$ (1 mark). The expression is $\frac{c o s θ ( 3 c o s θ + 2 )}{3 c o s θ + 2} = cos θ = \frac{3}{2}$ (1 mark).
Q7 · 3 marks · Short answerBoard level
Find $x$ if $x sin 3 0^{\circ} cos^{2} 4 5^{\circ} = \frac{c o t ^{2} 3 0 ^{\circ} s e c 6 0 ^{\circ} t a n 4 5 ^{\circ}}{cosec ^{2} 4 5 ^{\circ} cosec 3 0 ^{\circ}}$ .
Hint
Simplify each side to a number first; the left side becomes a multiple of $x$ .
Answer
$x = 6$ . RHS $= \frac{3 \times 2 \times 1}{2 \times 2} = \frac{3}{2}$ . LHS $= x \times \frac{1}{2} \times \frac{1}{2} = \frac{x}{4}$ . So $\frac{x}{4} = \frac{3}{2}$ , $x = 6$ .
Q8 · 3 marks · Short answerBoard level
If $tan (2 A - B) = 1$ and $sin (A + B) = \frac{3}{2}$ , where $2 A - B$ and $A + B$ are acute, find $A$ and $B$ .
Hint
Turn each condition into a linear equation in $A$ and $B$ , then add.
Answer
$A = 3 5^{\circ}$ , $B = 2 5^{\circ}$ . $2 A - B = 4 5^{\circ}$ and $A + B = 6 0^{\circ}$ . Adding: $3 A = 10 5^{\circ}$ , $A = 3 5^{\circ}$ , $B = 2 5^{\circ}$ . Check: $2 (3 5^{\circ}) - 2 5^{\circ} = 4 5^{\circ}$ .
Q9 · 3 marks · Short answerBoard level
In $△ A B C$ , right-angled at $B$ , $∠ A = 3 0^{\circ}$ and $A C = 12$ cm. Point $D$ lies on $A B$ such that $∠ B D C = 4 5^{\circ}$ . Find $B C$ , $A B$ and $A D$ .
Hint
First solve $△ A B C$ using $3 0^{\circ}$ , then use $tan 4 5^{\circ}$ in $△ D B C$ .
Answer
$B C = 12 sin 3 0^{\circ} = 6$ cm, $A B = 12 cos 3 0^{\circ} = 63$ cm. In $△ D B C$ , $tan 4 5^{\circ} = \frac{B C}{B D} = 1$ , so $B D = 6$ cm. $A D = 63 - 6 = 6 (3 - 1)$ cm.
Q10 · 3 marks · Short answerHarder board level
Riya writes: " $sin 6 0^{\circ} = sin (2 \times 3 0^{\circ}) = 2 sin 3 0^{\circ} = 1$ ."
(i) Show that her answer is wrong.
(ii) Verify that $sin 6 0^{\circ} = 2 sin 3 0^{\circ} cos 3 0^{\circ}$ instead.
(iii) Does $tan 6 0^{\circ} = 2 tan 3 0^{\circ}$ ? Justify.
Hint
Compare each claim with the exact table values.
Answer
(i) $sin 6 0^{\circ} = \frac{3}{2} \neq = 1$ , so doubling the angle does not double the sine. (ii) $2 \times \frac{1}{2} \times \frac{3}{2} = \frac{3}{2}$ , true. (iii) No: $2 tan 3 0^{\circ} = \frac{2}{3}$ but $tan 6 0^{\circ} = 3 = \frac{3}{3}$ .
Q11 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a painter rests a straight ladder 6 m long against a vertical wall on level ground, making $6 0^{\circ}$ with the ground. While he works, its foot slips straight outwards by $3 (3 - 1)$ m, and its top stays on the wall.
(i) Find the first distance of the foot from the wall and the first height of the top. (1 mark)
(ii) Find the new distance of the foot from the wall, and hence the new angle the ladder makes with the ground. (1 mark)
(iii) How far does the top slide down? What length of ladder, at this new angle, would reach the original height? Take $3 \approx 1.73$ . (2 marks)
Hint
In (ii), find the cosine of the new angle from the new foot distance and the ladder length.
Answer
(i) Foot $= 6 cos 6 0^{\circ} = 3$ m; height $= 6 sin 6 0^{\circ} = 33$ m. (1)
(ii) New foot distance $= 3 + 3 (3 - 1) = 33$ m. $cos θ = \frac{3 3}{6} = \frac{3}{2}$ , so the new angle is $3 0^{\circ}$ . (1)
(iii) New height $= 6 sin 3 0^{\circ} = 3$ m, so the top slides down $33 - 3 = 3 (3 - 1) \approx 2.19$ m. A ladder at $3 0^{\circ}$ reaching $33$ m must have length $\frac{3 3}{s i n 3 0 ^{\circ}} = 63 \approx 10.38$ m. (2)
Q12 · 5 marks · Long answerHarder board level
$θ$ is an acute angle ( $0^{\circ} < θ < 9 0^{\circ}$ ) such that $2 sin^{2} θ - 3 sin θ + 1 = 0$ .
(i) Find $θ$ , giving a reason for rejecting any other value.
(ii) Hence evaluate $4 cos^{2} θ + 3 tan^{2} θ - cosec^{2} 2 θ$ .
(iii) Verify that $sin^{2} θ + cos^{2} θ = 1$ for this $θ$ .
Hint
Treat $sin θ$ as the unknown and factorise the quadratic. Remember $2 θ$ is a new angle.
Answer
(i) $(2 sin θ - 1) (sin θ - 1) = 0$ , so $sin θ = \frac{1}{2}$ or $sin θ = 1$ (1 mark). $sin θ = 1$ gives $θ = 9 0^{\circ}$ , which is not acute, so $θ = 3 0^{\circ}$ (1 mark). (ii) $4 cos^{2} 3 0^{\circ} = 4 \times \frac{3}{4} = 3$ , $3 tan^{2} 3 0^{\circ} = 3 \times \frac{1}{3} = 1$ , and $2 θ = 6 0^{\circ}$ so $cosec^{2} 6 0^{\circ} = \frac{4}{3}$ (1 mark). Value $= 3 + 1 - \frac{4}{3} = \frac{8}{3}$ (1 mark). (iii) $sin^{2} 3 0^{\circ} + cos^{2} 3 0^{\circ} = \frac{1}{4} + \frac{3}{4} = 1$ (1 mark).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. $sin 3 0^{\circ}$ = ?Show answer
$\frac{1}{2}$
2. $cos 4 5^{\circ}$ = ?Show answer
$\frac{1}{2}$
3. $tan 6 0^{\circ}$ = ?Show answer
$3$
4. $sec 3 0^{\circ}$ = ?Show answer
$\frac{2}{3}$
5.Side ratio of a $3 0^{\circ}$ - $6 0^{\circ}$ - $9 0^{\circ}$ triangle?Show answer
$1 : 3 : 2$
6. $sin^{2} 6 0^{\circ}$ = ?Show answer
$\frac{3}{4}$

Trigonometric Ratios of 30°, 45° and 60°: frequently asked questions

Do I need to learn the derivation of these values or just the table?

Learn both. The syllabus includes finding these values, so a question can ask you to derive them from an equilateral or isosceles right triangle. Knowing the derivation also lets you rebuild any value you forget.

Should I rationalise answers like 1 by root 3?

Both forms are correct. Rationalise only if the question asks, or if it makes the next step easier. Do not waste time on it otherwise.

Is sin 60 equal to 2 times sin 30?

No. Sin 60 is root 3 by 2, while 2 times sin 30 is 1. Trigonometric ratios do not double when the angle doubles.

Are complementary angle results like sin of 90 minus A equal to cos A in the syllabus?

Trigonometric ratios of complementary angles as a topic are not in the 2026-27 syllabus.