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NDA 2016 Maths · Mathematics · Properties of Triangles · easy

In a triangle ABC with the usual notation, bcos C+ccos B is equal to

Single correct option

In a triangle ABCABC with the usual notation, bcos⁡C+ccos⁡Bb\cos C+c\cos B is equal to
  1. A. b+cb+c
  2. B. 2a2a
  3. C. aa
  4. D. a2\dfrac a2
Show answer and solution

Answer: (C) aa

This is the projection formula. Dropping the perpendicular from AA to BCBC, the foot divides BCBC into segments ccos⁡Bc\cos B and bcos⁡Cb\cos C, whose sum is the whole side aa. It can also be verified using the cosine rule: b⋅a2+b2−c22ab+c⋅a2+c2−b22ac=ab\cdot\dfrac{a^2+b^2-c^2}{2ab}+c\cdot\dfrac{a^2+c^2-b^2}{2ac}=a.
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Original question written in the NDA 2016 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake