Question Bank - Mathematics

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In a triangle ABC, sin A - cos B - cos C = 0. What is angle B equal to?

A.
\(\frac{\pi}{6}\)
B.
\(\frac{\pi}{4}\)
C.
\(\frac{\pi}{3}\)
D.
\(\frac{\pi}{2}\)

Solution:

Formula used :sin 2? = 2 sin ? cos ? cos C + cos D = 2cos(\(\frac{C+D}{2}\)) cos(\(\frac{C-D}{2}\))Calculaton :sin A - cos B - cos C = 0? sin A = cos B + cos CUsing the formula given above? 2 sin(A/2) cos(A/2) = 2cos(\(\frac{B+C}{2}\)) cos(\(\frac{B-C}{2}\))We know that, for a ?ABC, ? A + ? B + ? C = ? ? sin(A/2) cos(A/2) = 2cos(\(\frac{? - A}{2}\)) cos(\(\frac{B-C}{2}\))Since, cos(?/2 - ?) = sin ? ? sin(A/2) cos(A/2) = sin(\(\frac{ A}{2}\)) cos(\(\frac{B-C}{2}\))? cos(A/2) = cos(\(\frac{B-C}{2}\))? A/2 = (B - C)/2? A = B - C? B = A + CBut ? A + ? B + ? C = ? ? B = ? - B? 2B = ? ? B = ?/2? Angle B is equal to \(\frac{\pi}{2}\).

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