Chapter: 2021· JEE

In △ABC , cos B= 3/5 , AB=5 and R=5 . find the area of △ABC

Doubtly Asked 20 Jul 2021 Updated 23 Oct 2026 Student doubt Save WhatsApp Report

2 answers

Best answer

correct answer is 6+8√3

Doubtly · 20 Jul 2021 Report

Was this answer helpful?

Answer: \(6 + 8\sqrt3\) square units

\(\cos B = \tfrac35\), so \(\sin B = \tfrac45\). By the sine rule, \(b = 2R\sin B = 2(5)\left(\tfrac45\right) = 8\). Also \(c = AB = 5\).

Use the cosine rule for side \(a = BC\):

\[b^{2} = a^{2} + c^{2} - 2ac\cos B\]

\[64 = a^{2} + 25 - 6a\]

\[a^{2} - 6a - 39 = 0 \Rightarrow a = 3 \pm 4\sqrt3\]

Since a side is positive, \(a = 3 + 4\sqrt3\).

\[\text{Area} = \tfrac12\,ac\sin B = \tfrac12(3+4\sqrt3)(5)\left(\tfrac45\right)\]

\[= 2(3 + 4\sqrt3) = 6 + 8\sqrt3\]

Doubtly AI · 23 Oct 2026 AI-verified · Oct 2026 Report

Was this answer helpful?