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

Options

A. `3/5,4/5`
B. `1/sqrt(3),sqrt(2/3)`
C. `1/2,sqrt(3)/(2)`
D. none of these

Answer

A

Explanation

Let `angleC=90^(@)` being greatest and B=`90^(@)`-A
The sides are a-d,a, and a+d
We have `(a+d)^(2)=(a+d)^(2)+a^(2)` (using Pythagoras therom)
`therefore 4ad-a^(2)=0` or a=4d
Hence, the sides are 3d, 4d , 5d
Clearly,
sinA=`(BC)/(AB)=(a-d)/(a+d)=(3d)/(5d)=3/5`
sinB=`(AC)/(AB)=(a)/(a+d)=(4d)/(5d)`
`=4/5`