# Extra Questions For Class 10 Maths Chapter 8 #4

#### Trigonometry | Trigonometric Values of Complementary Angles

This CBSE class 10 Maths practice question is from the topic Trigonometry. It tests the concept of trigonometric values of complementary angles.

Question 4 : In ΔABC right angled at B, sin C = $$frac{\text{5}}{\text{13}}$. Find$i) sin A
(ii) cos A
(iii) cos C

## NCERT Solution to Class 10 Maths

### Explanatory Answer | Trigonometry Extra Question 4

sin C = $$frac{\text{side opposite to ∠C}}{\text{hypotenuse}}$=$\frac{\text{AB}}{\text{AC}}$= $\frac{\text{5}}{\text{13}}$ ......$1)
Let AB = 5k and AC = 13k

#### Calculate Sine and Cosine values of the angles

Using Pythagoras theorem, BC = $$sqrt{AC^2− AB^2}$; BC = $\sqrt{$13k$^2− (5k)^2 }) = $$sqrt{169 k^2− 25 k^2 }$ = $\sqrt{144k^2 }$ = 12k sin A = $\frac{\text{side opposite to ∠A}}{\text{hypotenuse}}$= $\frac{\text{BC}}{\text{AC}}$= $\frac{\text{12k}}{\text{13k}}$= $\frac{\text{12}}{\text{13}}$ ......$2)

cos A = $$frac{\text{side adjcent to ∠ A}}{\text{hypotenuse}}$ = $\frac{\text{AB}}{\text{AC}}$= $\frac{\text{5k}}{\text{13k}}$= $\frac{\text{5}}{\text{13}}$ ......$3)

cos C = $$frac{\text{side adjcent to ∠ C}}{\text{hypotenuse}}$= $\frac{\text{BC}}{\text{AC}}$ = $\frac{\text{12k}}{\text{13k}}$ = $\frac{\text{12}}{\text{13}}$ ......$4)

Inference
From (1) and (3): sin A = cos C
From (2) and (4): sin C = cos A

#### Why?

If ABC is right angled at B, side adjacent to ∠ A will be the side opposite to ∠C.
Similarly side opposite to ∠A will be the side adjacent to ∠C.
i.e., ∠A = 90° - ∠C and ∠C = 90° - ∠A

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