the sides of a right angle triangle are a,a+d,a+2d with a and d both positive.the ratio of a to d

a)1:2 b)1:3 c)3:1 d)5:2

given that tshe sides are a,a+d,a+2d

let a+2d be the side opposite to the hypotenuse

let the angles be p,90-p,90

we apply sine rule

a/sin(p)= a+d/sin(90-p)=a+2d/sin90

a/sin(p)=a+2d      ,        a+d/sin(90-p)=a+2d   {sin(90-p)=cosp}       

sin(p)=a/a+2d         ,     cos(p)=a+d/a+2d

we  know that sin^2(p)+cos^2(p)=1

(a/a+2d)^2+(a+d/a+2d)^2=1

solving this we get

(a+d)^2+(a)62=(a+2d)^2

a^2+d^2+2.a.d=a^2+4.d^2+4.a.d

0=a^2-2.a.d-3.d^2

its quadratic equation in a

here the roots fot a is 3d

a=3d

a/d=1:3

c is the answer .3:1

answer is (c) i.e. 3:1

Solution: Applying Pythagoras Theorem,

we get, (a+2d)²=a²+(a+d)²

=> a²+4d²+4ad=a²+a²+d²+2ad

=> 3d²+2ad-a²=0

=> (d+a)(3d-a)=0

neglecting (d+a)=0, by solving for (3d-a)=0, we get a:d=3:1

Let a+2d be the side opposite to the hypotenuselet the angles be p,90-p,90we apply sine rulea/sin(p)= a+d/sin(90-p)=a+2d/sin90a/sin(p)=a+2d , a+d/sin(90-p)=a+2d {sin(90-p)=cosp} sin(p)=a/a+2d , cos(p)=a+d/a+2dwe know that sin^2(p)+cos^2(p)=1(a/a+2d)^2+(a+d/a+2d)^2=1solving this we get(a+d)^2+(a)62=(a+2d)^2a^2+d^2+2.a.d=a^2+4.d^2+4.a.d0=a^2-2.a.d-3.d^2its quadratic equation in ahere the roots fot a is 3da=3da/d=1:3