Question

If the lengths of sides of right angled triangle are in A.P then the sines of the acute angles are

• A. $3$$/5\cdot 4/5$
• B. $\sqrt{2/3}\cdot \sqrt{1/3}$
• C. $\sqrt{\frac{\sqrt{5}-1}{2}},\sqrt{\frac{\sqrt{5}+1}{2}}$
• D. $\sqrt{\frac{\sqrt{3}-1}{2}},\sqrt{\frac{\sqrt{3}+1}{2}}$

Answer 1

The correct option is A $3$$/5\cdot 4/5$

Let the sides be $(a-d)$, & $(a+d)$

${(a-d)}^2+a^2={(a+d)}^2$

$\Rightarrow a^2-2ad+d^2+a^2=a^2+2ad+d^2$

$\Rightarrow a^2=4ad\Rightarrow a=4d$

$\therefore$ sides are $3d$, $4d$ & $5d$

$\therefore$ $\sin {\theta }_1=\frac{3d}{5d}=\frac{3}{5}$

$\sin {\theta }_1=\frac{3d}{5d}=\frac{4}{5}$ [A]