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

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