Question

ABC is a right triangle and right angled at B such that angle BCA is equal to 2 angle BAC.
Show that hypotenuse AC is equal to 2BC.

Answer 1

In Triangle ABC
angle BAC + angle BCA = 90°
=> angle BAC + 2×angle BAC = 90° [Given]
=> 3×angle BAC = 90°
=> angle BAC = 90°/3
=> angle BAC = 30°
then, angle BCA = 2×30° = 60°

Now,
Cos BCA = BC/AC
=> Cos 60° = BC/AC
=> 1/2 = BC/AC
=> 2BC = AC
=> AC = 2BC
Hence proved

Answer 2

Answer:

Given: <BCA = 2<BAC

To Prove: AC = 2BC

∴ We know that, <BCA = <BAC

=> As <B + <C + <A = 180 (Angle Sum Property)

= 90 + 2A + A = 180 (<C=2<A)

= 3A = 90

= <A = 30 - - (1)

∵ Now, If we take a cosec of <CAB

=> Cosec A = AC/BC (Cosec 30 = 2)

= Cosec 30 = AC/BC (From (1))

= 2 = AC/BC

= AC = 2BC