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