Question

If the length of sides of a right triangle are in A.P., then the sines of the acute angle are

• A. $\frac{3}{5},\frac{4}{5}$
• B. $\frac{2}{3}$ , $\frac{1}{3}$
• C. $\frac{5}{2}$ , $\frac{5}{3}$
• D. none of these

Answer 1

The correct option is A $\frac{3}{5},\frac{4}{5}$

Let the sides of the triangle be $a,b,c$.

Since $a,b,c$ are in A.P, therefore

$2b=a+c$.

And

$c^2=a^2+b^2$ [By Pythagoras's theorem]

Hence

$c^2=a^2+(\frac{a+c}{2}{)}^2$

$4c^2=4a^2+(a+c{)}^2$

$4c^2=5a^2+c^2+2ac$

$3c^2=5a^2+2ac$

$5a^2+2ac-3c^2=0$

$a=\frac{-2c\pm \sqrt{4c^2+60c^2}}{10}$

$a=\frac{-2c\pm 8c}{10}$

$a=\frac{-c\pm 4c}{5}$

Since sides cannot be negative.

Hence

$a=\frac{-c+4c}{5}$
$a=\frac{3c}{5}$ ..(i)

Since

$b=\frac{a+c}{2}$

$=\frac{\frac{3c}{5}+c}{2}$

$=\frac{8c}{10}$

$b=\frac{4c}{5}$

Hence

$a=\frac{3c}{5},b=\frac{4c}{5}$

Thus

$sin\theta =\frac{a}{c}=\frac{3}{5}$

And

$sin(\pi /2-\theta )=\frac{b}{c}=\frac{4}{5}$.