This MCQ module is based on: Trigonometric Ratios of Specific Angles
TOPIC 21 OF 35
Trigonometric Ratios of Specific Angles
🎓 Class 10
Mathematics
CBSE
Theory
Ch 8 — Introduction to Trigonometry
⏱ ~35 min
🌐 Language: [gtranslate]
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This mathematics 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 \(AC=a\sqrt2\).
Fig. 8.14 — Isosceles right triangle for 45° ratios
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 \(\angle BAC=30°,\;\angle ABC=60°\).
Fig. 8.15 — Half of an equilateral triangle gives 30°–60°–90°
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
| ∠A | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin A | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos A | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan A | 0 | 1/√3 | 1 | √3 | Not defined |
| cosec A | Not defined | 2 | √2 | 2/√3 | 1 |
| sec A | 1 | 2/√3 | √2 | 2 | Not defined |
| cot A | Not defined | √3 | 1 | 1/√3 | 0 |
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°.
- For sin, count how many fingers are below the chosen angle, take square root, divide by 2.
- For cos, count fingers above the chosen angle, square-root, divide by 2.
- 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°
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 Analysesin 2A = √3/2 ⇒ 2A = 60° ⇒ A = 30°.
Q3. A student writes "cos 90° + sin 90° = 2". Critique this.
L5 EvaluateWrong. 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 Createcos 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.
R: sin 30° = cos 60° = 1/2.
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.
R: At 90°, the adjacent side vanishes and division by zero is undefined.
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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