Q11 of 268 Page 4

In a ∆ABC prove that

In ∆ABC,

Sum of angles of a triangle = 180°

A + B + C = 180°

⇒ A + B = 180° - C

…(3)

Taking LHS

(from eq (3))

[∵ cos (90° - θ) = sin θ]

=RHS

Hence Proved

In a ∆ABC prove that

If sin 3A = cos(A – 26^o), where 3A is an acute angle, find the value of A.

Find θ if cos(2 θ +54^o)= sin θ, where (2θ +54^o) is an acute angle.