Question

If sinθ and –cosθ are the roots of the equation

ax² – bx – c = 0, where a, b, c are the sides of a triangle ABC, then cosB is equal to

• A. 1 – $\frac{c}{2a}$
• B. 1 – $\frac{c}{a}$
• C. 1 + $\frac{c}{2a}$
• D. 1 + $\frac{c}{3a}$

Answer 1

Answer: (C)

1 + $\frac{c}{2a}$

Answer 2

Here sin θ – cos θ  = $\frac{b}{a}$ and sinθcosθ = $\frac{c}{a}$

⇒ 1 – 2sinθcosθ = $\frac{b^{2}}{a^{2}}$

or 1 – $\frac{2c}{a} = \frac{b^{2}}{a^{2}}$

⇒ a² – b² = 2ac.

Hence cosB=$\frac{a^{2} + c^{2}–b^{2}}{2ac}$=$\frac{2ac + c^{2}}{2ac}$= 1 + $\frac{c}{2a}$.