Question

Consider triangle ABC such that AB = 24, BC = 13 and Angle C = 2 x Angle A

What are sin(Angle A) and sin (Angle C)?

Answer 1

Explanation:

We know that a/sin A = b/sin B = c/sin C

ie. 13/sin A = 24/Sin C

=> 13/sin theeta = 24/sin 2theeta

=> 13/sin theeta = 24/2sin theeta.cos theeta

=> 13 = 12/cos theeta => cos theeta = 12/13

So sin theeta = √(1-cos^2 theeta)=√(1 - 144/169)

= √(25/169)= 5/13

So sin A = sin theeta = 5/13

sin C = sin 2theeta = 2sin theeta.coe theeta

= 2×(5)13)(12/13) = 120/169