If sin (θ + 34°) = cos θ and θ + 34° is acute, then θ = ?
Given sin (θ + 340) = cos θ …Equation 1
Since sin θ & cos θ are complementary to each other
so sin θ = cos (900 – θ)
Using the above relations in Equation 1 we get
cos (900 – θ – 340) = cos θ
Since both L.H.S. and R.H.S. are functions of cosine and θ + 340 is acute so we can write
900 – θ – 340 = θ
⇒ 2θ = 560
⇒ θ = 280