Question

If in a triangle $ABC,\sin \left(A\right)=\frac{1}{5},\cos \left(B\right)=\frac{2}{7}$, then $\cos \left(C\right)=$

• A. $\frac{(\sqrt{45}-2\sqrt{24})}{35}$
• B. $\frac{(\sqrt{45}+2\sqrt{24})}{35}$
• C. $\frac{(7\sqrt{24}+10)}{35}$
• D. $0.85$
• E. $1$

Answer 1

Answer: (A)

$\frac{(\sqrt{45}-2\sqrt{24})}{35}$

Given $\sin A=\frac{1}{5}$, which implies $\cos A=\frac{\sqrt{24}}{5}$

Also given $\cos B=\frac{2}{7}$, which implies $\sin B=\frac{\sqrt{45}}{7}$

$\Rightarrow \cos C=\cos (180-A-B)$

$=-\cos (A+B)$

$=\sin A\sin B-\cos A\cos B$

$=\frac{(\sqrt{45}-2\sqrt{24})}{35}$