Question

If the lengths of the sides of a right angled triangle ABC right angled at C are in A.P., find $5(\sin A+\sin B)$.

Answer 1

Here $\angle C={90}^o,A+B={90}^o$
$c^2=a^2+b^2$ & $2b=a+c$
Since $c=2b-a$ & $c^2=a^2+b^2$
$\Rightarrow (2b-a{)}^2=a^2+b^2$
or $\frac{b}{a}=\frac{4}{3}$
$\Rightarrow \frac{sinB}{sinA}=\frac{4}{3}$ or $\frac{sinB+sinA}{sinB-sinA}=\frac{7}{1}$ or $cot\frac{B-A}{2}=\frac{1}{7}\Rightarrow cos\frac{A-B}{2}=\frac{7}{5\sqrt{2}}$
Also $5(sinA+sinB)=5\sqrt{2}cos\frac{A-B}{2}=7$