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Trigonometric Ratios of Specific Angles

🎓 Class 10 Mathematics CBSE Theory Ch 8 — Introduction to Trigonometry ⏱ ~35 min 📢 Join Class 10 Channel
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This MCQ module is based on: Trigonometric Ratios of Specific Angles

This assessment will be based on: Trigonometric Ratios of Specific Angles
Targeting Class 10 level in Trigonometry, with Intermediate difficulty.

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8.3 Trigonometric Ratios of Some Specific Angles

From geometry you already know shapes like the isosceles right trianglei (45°-45°-90°) and the equilateral trianglei (60°-60°-60°). We now compute the trigonometric ratios of 0°, 30°, 45°, 60° and 90°.

Ratios of 45°

Consider an isosceles right triangle ABC, right-angled at B, with \(AB=BC=a\). Then \(\angle A=\angle C=45°\) and by Pythagoras \(AC=a\sqrt2\).

A B C a a a√2 45° 45°
Fig. 8.14 — Isosceles right triangle for 45° ratios
\[\sin 45°=\frac{BC}{AC}=\frac{a}{a\sqrt2}=\frac{1}{\sqrt2},\quad \cos 45°=\frac{1}{\sqrt2},\quad \tan 45°=1.\] \[\csc 45°=\sqrt2,\quad \sec 45°=\sqrt2,\quad \cot 45°=1.\]

Ratios of 30° and 60°

Consider an equilateral triangle ABD with each side \(2a\). Drop the perpendicular AC from A to BD. Then BC = CD = \(a\), AC = \(a\sqrt3\), and \(\angle BAC=30°,\;\angle ABC=60°\).

A B D C a a 2a a√3 30° 60°
Fig. 8.15 — Half of an equilateral triangle gives 30°–60°–90°
\[\sin 30°=\frac{BC}{AB}=\frac{a}{2a}=\frac{1}{2},\;\cos 30°=\frac{AC}{AB}=\frac{\sqrt3}{2},\;\tan 30°=\frac{1}{\sqrt3}.\] \[\sin 60°=\frac{\sqrt3}{2},\;\cos 60°=\frac{1}{2},\;\tan 60°=\sqrt3.\] \[\csc 30°=2,\;\sec 30°=\frac{2}{\sqrt3},\;\cot 30°=\sqrt3;\quad\csc 60°=\frac{2}{\sqrt3},\;\sec 60°=2,\;\cot 60°=\frac{1}{\sqrt3}.\]

Ratios of 0° and 90°

When angle A is made smaller and smaller in a right triangle (with BC held fixed shrinking, or with angle approaching the base), side BC approaches 0 and AC approaches AB. Taking limits:

\[\sin 0°=0,\;\cos 0°=1,\;\tan 0°=0,\;\sec 0°=1,\;\csc 0°\text{ and }\cot 0°\text{ are not defined.}\]

When A is close to 90°, AB approaches 0 and BC approaches AC:

\[\sin 90°=1,\;\cos 90°=0,\;\tan 90°\text{ and }\sec 90°\text{ are not defined,}\;\csc 90°=1,\;\cot 90°=0.\]
Table 8.1 — Standard Values
∠A0°30°45°60°90°
sin A01/21/√2√3/21
cos A1√3/21/√21/20
tan A01/√31√3Not defined
cosec ANot defined2√22/√31
sec A12/√3√22Not defined
cot ANot defined√311/√30

Remark. As A increases from 0° to 90°, sin A increases from 0 to 1, while cos A decreases from 1 to 0.

Activity — Memorise the Table with Fingers
Trick: Hold up your left hand. Number fingers 0–4 for 0°, 30°, 45°, 60°, 90°.
  1. For sin, count how many fingers are below the chosen angle, take square root, divide by 2.
  2. For cos, count fingers above the chosen angle, square-root, divide by 2.
  3. Try sin 45°: two fingers below. √2/2 = 1/√2. ✓
sin: 0, √1/2, √2/2, √3/2, √4/2 = 0, 1/2, 1/√2, √3/2, 1. cos: reverse. A quick mnemonic that students never forget.

Example 6

In \(\triangle ABC\) right-angled at B, AB = 5 cm and \(\angle ACB=30°\). Find BC and AC.

Solution. AB is opposite C. So \(\tan C=\dfrac{AB}{BC}\Rightarrow \tan 30°=\dfrac{5}{BC}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{5}{BC}\Rightarrow BC=5\sqrt3\) cm.

Also \(\sin 30°=\dfrac{AB}{AC}\Rightarrow \dfrac{1}{2}=\dfrac{5}{AC}\Rightarrow AC=10\) cm. Alternatively \(AC=\sqrt{25+75}=10\) cm.

Example 7

In \(\triangle PQR\) right-angled at Q, PQ = 3 cm, PR = 6 cm. Determine \(\angle QPR\) and \(\angle PRQ\).

Solution. \(\sin R=\dfrac{PQ}{PR}=\dfrac{3}{6}=\dfrac{1}{2}\Rightarrow \angle R=30°\). Hence \(\angle P=60°\).

Example 8

If \(\sin(A-B)=\tfrac{1}{2}\) and \(\cos(A+B)=\tfrac{1}{2}\), with \(0°B\), find A and B.

Solution. \(\sin(A-B)=1/2\Rightarrow A-B=30°\); \(\cos(A+B)=1/2\Rightarrow A+B=60°\). Solving: A = 45°, B = 15°.

Competency-Based Questions
Q1. Evaluate: sin 60° cos 30° + sin 30° cos 60°.
L3 Apply
= (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1. (This is sin(60°+30°) = sin 90° = 1.)
Q2. If 2 sin 2A = √3, find the value of A for 0° ≤ A ≤ 90°.
L4 Analyse
sin 2A = √3/2 ⇒ 2A = 60° ⇒ A = 30°.
Q3. A student writes "cos 90° + sin 90° = 2". Critique this.
L5 Evaluate
Wrong. cos 90° = 0 and sin 90° = 1, so the sum is 1 not 2. The student may have confused cos 90° with cos 0°.
Q4. Construct a right triangle where angle A = 60° and the adjacent side to A is 4 cm. Compute all sides and verify using Table 8.1.
L6 Create
cos 60° = adjacent/hypotenuse ⇒ 1/2 = 4/hyp ⇒ hyp = 8. Opposite = √(64−16) = √48 = 4√3. Verify tan 60° = 4√3/4 = √3 ✓.
Assertion & Reason
A: sin 30° + cos 60° = 1.
R: sin 30° = cos 60° = 1/2.
A) Both true; R explains A
B) Both true; R does not explain A
C) A true, R false
D) A false, R true
Answer: A. Each equals 1/2, sum 1. R gives the values needed to compute A.
A: tan 90° is not defined.
R: At 90°, the adjacent side vanishes and division by zero is undefined.
A) Both true; R explains A
B) Both true; R does not explain A
C) A true, R false
D) A false, R true
Answer: A. tan = opposite/adjacent; the adjacent side has length 0 at 90°.

Frequently Asked Questions

Why is sin 30 degrees equal to 1/2?
In a 30-60-90 right triangle, the side opposite the 30-degree angle is half the hypotenuse. So sin 30 = opposite/hypotenuse = 1/2.
Why are sin 45 and cos 45 equal?
In a 45-45-90 right triangle, the two legs are equal and the hypotenuse is root 2 times a leg. Both sine and cosine equal leg/hypotenuse = 1/root 2.
What are the values at 0 and 90 degrees?
sin 0 = 0, cos 0 = 1, tan 0 = 0. sin 90 = 1, cos 90 = 0, tan 90 is undefined because cos 90 is 0.
Why is tan 90 undefined?
Because tan A = sin A / cos A, and cos 90 = 0 makes the denominator zero, so tan 90 is undefined (approaches infinity).
How do these values help in problems?
They allow exact computation in geometry, physics and engineering without a calculator, especially for problems involving 30-60-90 or 45-45-90 right triangles.
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