JEE Mains · Maths · STD 11 - Trigonometrical equations
If in a triangle \(\mathrm{ABC}, \mathrm{AB}=5\) units, \(\angle \mathrm{B}=\cos ^{-1}\left(\frac{3}{5}\right)\) and radius of circum circle of \(\triangle \mathrm{ABC}\) is \(5\) units, then the area (in sq. units) of \(\triangle \mathrm{ABC}\) is:
- A \(6+8 \sqrt{3}\)
- B \(8+2 \sqrt{2}\)
- C \(4+2 \sqrt{3}\)
- D \(10+6 \sqrt{2}\)
Answer & Solution
Correct Answer
(A) \(6+8 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
\(\cos B=\frac{3}{5} \Rightarrow \sin B=\frac{4}{5}\) Now, \(\frac{b}{\sin B}=2 R \Rightarrow b=2(5)\left(\frac{4}{5}\right)=8\) Now, by cosine formula \(\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}\) \(\Rightarrow \frac{3}{5}=\frac{a^{2}+25-64}{2(5) a}\)…
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