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Grade: 12


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

8 years ago

Answers : (4)

veeravalli mohan
33 Points

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


solving this we get




its quadratic equation in a

here the roots fot a is 3d



8 years ago
Rohit Yadav
34 Points

c is the answer .3:1

8 years ago
Deepali Singh
32 Points

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

8 years ago
Neel Shobhan Dutta
11 Points
							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
3 years ago
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