Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If $\Delta$ $ABC$ $b=3$ cm,$c=4$ cm, and the length of the perpendicular from $A$ to the circle $BC$ is $2$ cm, then the number of solutions of the triangle is?

• A. $1$
• B. $0$
• C. $3$
• D. $2$

Answer 1

Answer: (D)

$2$

Given $b=3,c=4,$ and length of perpendicular from A to BC $=2$
If D is foot of perpendicular,then we know $AD=c\sin B=2\Rightarrow \sin B=\frac{1}{2}\Rightarrow \cos B=\frac{\sqrt{3}}{2}$
Now using cosine rule $\cos B=\frac{a^2+c^2-b^2}{2ac}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{a^2+7}{8a}$ $\Rightarrow a^2-4\sqrt{3}a+7=0$
$\Rightarrow$ Discriminant $=48-28=20>0$
Hence there will be two distinct real value of $c$. exists. $\Rightarrow$ Two triangle is possible with the given information