Question

In a triangle $ABC,a=5,b=7$ and $\sin A=\frac{3}{4}$, how many such triangles are possible?

• A. 1
• B. 0
• C. 2
• D. infinite

Answer 1

The correct option is B 0

Given, $a=5,b=7,\sin A=\frac{3}{4}$

$\Rightarrow \cos A=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$

By cosine formula, $\cos A=\frac{b^2+c^2-a^2}{2bc}$

$\Rightarrow \frac{\sqrt{7}}{4}=\frac{49+c^2-25}{2\left(7\right)c}\Rightarrow \frac{7\sqrt{7}}{2}=\frac{24+c^2}{c}$

$\Rightarrow 2c^2+48-7\sqrt{7}c=0$

$\Rightarrow D=343-384<0$

$\Rightarrow$ No triangle is possible.