Question

In a right angled triangle if the square of the hypotenuse is twice the product of the other two sides, find one of the angles of the triangle​

Answer 1

Answer:

In a triangle ABC, Angle B is 90

ThenAB, BC, CA are let x, y, z respectively

z^2=x^2+y^2

Given,

z^2=2xy

Csc B=z/x, Cos B=y/z

z x z=2xy

z/x=2y/z

Csc B=2Cos B

1=2Cos B x Sin B-------(1)

(Sin B + Cos B)^2=(Sin B) ^2+(Cos B) ^2 +2Sin B CosB

From (1),

(Sin B+ Cos B) ^2=1+1

Sin B+ Cos B=√2

If B = 45

Sin 45+ Cos 45=√2

1/√2+1√2=√2

2/√2 x √2/√2=√2

2√2/2=√2

√2=√2

Therefore Angle B = 45

Answer 2

Let the three sides of triangle be,

Base=B, hypotenus=h and perpendicular=P

According to the given questions,

h²=2(b.p)

⇒b²+p²=2(b×p)

⇒(b−p)² =0⇒b=p

∴ triangle is isosceles.