Question

Find the number of triangles $ABC$ that can be formed with $a=3,b=8$ and $\sin A=\frac{5}{13}$

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (A)

$0$

Given, $a=3,b=8,\sin A=\frac{5}{13}$

$\because \cos A=\frac{b^2+c^2-a^2}{2bc}$

$\Rightarrow c^2-\left(2b\cos A\right)c+(b^2-a^2)=0$

$\Rightarrow c=\frac{2b\cos A\pm \sqrt{4b^2{\cos }^{2}A-4(b^2-a^2)}}{2}$

$\Rightarrow c=b\cos A\pm \sqrt{b^2{\cos }^{2}A-b^2+a^2}$

$\Rightarrow c=b\cos A\pm \sqrt{a^2-b^2{\sin }^{2}A}$

Now $a^2-b^2{\sin }^{2}A=9-64\left(\frac{25}{169}\right)=\frac{1521-1600}{169}<0$

$\Rightarrow c$ has imaginary values

$\Rightarrow$ No triangle is possible