A) \[\frac{s(s-a)(s-b)(s-c)}{2}\]
B) \[\frac{{{b}^{2}}{{\sin }^{2}}\sin A}{\sin B}\]
C) \[ab\sin C\]
D) \[\frac{1}{2}\frac{{{a}^{2}}\sin B\sin C}{\sin A}\]
Correct Answer: D
Solution :
Key Idea: We know that \[\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k\] Area\[=\frac{1}{2}ac\,\,\sin B=\frac{1}{2}ab\,\,\sin C=\frac{1}{2}bc\,\,\sin A\] Given, the vertex angles are \[A,\,\,\,B,\,\,\,C\] and side\[BC\]. \[\therefore \]Area\[=\frac{1}{2}a(k\sin C)\sin B\] \[=\frac{1}{2}\frac{{{a}^{2}}k\sin C\sin B}{a}=\frac{1}{2}\frac{{{a}^{2}}k\sin C\sin B}{k\sin A}\] \[=\frac{1}{2}{{a}^{2}}\frac{\sin B\sin C}{\sin A}\]
You need to login to perform this action.
You will be redirected in
3 sec
