Question

If the sides of a right triangle are in A.P., then the sum of the sines of the two acute angles is:

• A. $\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$5$}\right.$
• B. $\raisebox{1ex}{$8$}\!\left/ \!\raisebox{-1ex}{$5$}\right.$
• C. $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.$
• D. $\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$5$}\right.$

Answer 1

The correct option is A

$\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$5$}\right.$

Explanation for the correct option:

Step 1: Find the sides of the triangle

It is given that the sides of the right-angled triangle are in A.P.

So, let the sides be $\left(a-d,a,a+d\right)$

The hypotenuse is the longest side of a triangle

So, Hypotenuse$=a+d$ and other sides are $a$ and $a-d$

Applying Pythagoras theorem, ${\left(\mathrm{Hypotenus}e\right)}^2={\left(\mathrm{Perpendicular}\right)}^2+{\left(\mathrm{Base}\right)}^2$, we get,

$$\begin{gathered} \Rightarrow {(a+d)}^2=a^2+{(a-d)}^2 \\ \Rightarrow \cancel{a^2}+\cancel{d^2}+2ad=\cancel{a^2}+a^2+\cancel{d^2}-2ad \\ \Rightarrow a^2-4ad=0 \\ \Rightarrow a(a-4d)=0 \end{gathered}$$

Either $a=0$ or $a-4d=0$

So, $a=4d$

Replace this in $\left(a-d,a,a+d\right)$, we get, $(3d,4d,5d)$

Step 2: Evaluate the trigonometric value of the acute angles

We need to evaluate the $\sin$ of the acute angles,

So, as $\sin \theta =\frac{\mathrm{Perpendicular}\left(\mathrm{P}\right)}{\mathrm{Hypotenuse}\left(\mathrm{H}\right)}$

When $4d$ is the base,

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

When $3d$ is the base,

$\Rightarrow \sin {\theta }_2=\frac{4d}{5d}=\frac{4}{5}$

Therefore, the sine of the acute angles are $\frac{3}{5},\frac{4}{5}$

The sum of the sine of two acute angles$=\left(\frac{3}{5}+\frac{4}{5}\right)=\frac{7}{5}$

Thus, option (A) is the correct option.