If the lengths of the sides of a right-angled triangle $ABC,$ right angled at $C,$ are in arithmetic progression, then the value of $5\left(\sin A+\sin B\right)$ is

Solution

${\left(a+d\right)}^2={\left(a-d\right)}^2+a^2$ (Pythagoras theorem)
$\cancel{a^2}+\cancel{d^2}+2ad=\cancel{a^2}+\cancel{d^2}-2ad+a^2\Rightarrow 4ad=a^2\Rightarrow a=4d$
$\sin A=\frac{a-d}{a+d}=\frac{4d-d}{4d+d}=\frac{3}{5}$
$\sin B=\frac{a}{a+d}=\frac{4d}{5d}=\frac{4}{5}$
$\Rightarrow 5\sin A+5\sin B=3+4=7$
$\Rightarrow 5\left(\sin A+\sin B\right)=7$

Hence, the correct answer is 7.