Exams › SSC CGL (Prelims) › General
If cos B = p/q where B lies in the interval (0, pi/2), express sin B in terms of p and q.
- sqrt(q^2 - p^2) / q
- sqrt(q^2 + p^2) / p
- sqrt(q^2 - p^2) / p
- p / sqrt(q^2 - p^2)
Correct answer: sqrt(q^2 - p^2) / q
Solution
From sin^2 B = 1 - cos^2 B = 1 - (p/q)^2 = (q^2 - p^2)/q^2, taking the positive root (first quadrant) gives sin B = sqrt(q^2 - p^2)/q.
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