Question

In a triangle ABC, ∠B = π/ 3 and ∠C = π/4 and D divides BC internally in the ratio 1:3. Then, find the value of  sin∠BAD/sin∠CAD .

Answer 1

Let ∠BAD = α ∠CAD = β

In ∆ADB, applying sine formulae, we get

x/sinα = AD/sin(π/3)

In ∆ADC, applying sine formulae, we get

3x/sinβ = AD/sin(π/4)

Dividing Eq. (1) by Eq. (2), we get