If sin(2A-3B)=cos(3A+8B) where (2A-3B) (3A+8B) are acute angles, then what is the value of (A+B) a)25 b)23 - Maths - Trigonometry

Question

If sin(2A-3B)=cos(3A+8B) where (2A-3B) & (3A+8B) are acute
angles, then what is the value of (A+B)

a)25
b)23
c)18
d)12

Varun.Rawat · Accepted Answer

We have,     sin2A - 3B = cos3A + 8B⇒sin2A - 3B = sin90° - 3A + 8B    as, sin 90°-θ = cos θ⇒sin2A - 3B = sin90°-3A-8B⇒2A - 3B = 90°-3A-8B⇒2A + 3A - 3B + 8B = 90°⇒5A + 5B = 90°⇒5A + B = 90°⇒A + B = 90°5⇒A + B = 18°

If sin(2A-3B)=cos(3A+8B) where (2A-3B) & (3A+8B) are acute angles, then what is the value of (A+B) a)25 b)23 c)18 d)12

If sin(2A-3B)=cos(3A+8B) where (2A-3B) & (3A+8B) are acute angles, then what is the value of (A+B)

a)25
b)23
c)18
d)12