Question
In a triangle ABC, if 5∠A = 3∠B = 15∠C, then find the
value of cosA + sin(C + 25 ° )Solution
Given, 5∠A = 3∠B = 15∠C
Or, ∠A:∠B:∠C = (1/5):(1/3):(1/15) = 3:5:1
Let ∠A, ∠B and ∠C be '3x', '5x' and 'x', respectively
Therefore, 3x + 5x + x = 180 o (sum of the interior angles of a triangle is 180 o )
Or, 9x = 180 o
Or, x = 20 o
Therefore, cosA + sin(C + 25 o ) = cos60 o + sin45 o = (1/2) + (1/√2) = {(√2 + 2)/2√2}
Rationalising by multiplying numerator and denominator by √2, we get= (√2 + 1)/2
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