Question

In the adjoining figure, ∆ABC is right-angled at B, D is mid-point of BC, AC = 5 units, BC = 4 units and ∠BAD = θ, find the values of:
(i) tan θ (ii) sin θ (iii) sin²θ + cos²θ.

Answer 1

Answer:

1. tan tetha=2/3

2. sin tetha= 2/√20=√20/10

3. sin2 tetha+ cos2tetha= 1

Answer 2

(i) tan θ = 4 / 3

(ii) sin θ = 4 / 5

(iii) sin²θ + cos²θ = 1

Explanation:

Given: ABC is a right-angled triangle with <B as 90 degrees, AC = 5 (hypotenuse) and BC = 4 (height).

In a right angled triangle, hypotenuse^2 = base^2 + height^2

5^2 = base^2 * 4^2

25 - 16 = base^2

9 = base^2

So base AB = root of 9 = 3 units.

(i) tan θ = height / base = 4 / 3

(ii) sin θ = height / hypotenuse = 4 / 5

(iii) sin²θ + cos²θ = (4 / 5)^2 + (3 / 5)^2  = 16 / 25 + 9 / 25 = 25 / 25 = 1

(as Cos = base / hypotenuse)