Question

In $△ABC$, $\angle C={90}^{\circ }$. If the sides lengths of the traingle are in A.P., then the value of $5(\sin A+\sin B)$ is

Answer 1

Given $\angle C={90}^{\circ }\Rightarrow A+B={90}^{\circ }$
Let the side lengths be $(a-d),a,(a+d)$

Using Pythagoras theorem, we get
$(a+d{)}^2=a^2+(a-d{)}^2\Rightarrow 4ad-a^2=0\Rightarrow a=4d(\because a\ne 0)$
Therefore, the sides lengths of the triangle are $3d,4d,5d$

Hence,
$\sin A+\sin B=\frac{3}{5}+\frac{4}{5}\Rightarrow 5(\sin A+\sin B)=7$