In a triangle ABC, D is a point on BC such that AD is the internal bisector of `angle A`. Let `angle B = 2 angle C` and CD = AB. Then `angle A` is

Options

A. `18^(@)`
B. `36^(@)`
C. `54^(@)`
D. `72^(@)`

Answer

D

Explanation

Since `angle B=2 angle C`
sin B = sin 2C
`rArr sin B = 2 sin C cos C`
`rarr b(2ab)=2c(a^(2)+b^(2)-c^(2))`
`rarr ab^(2)=ca^(2)+cb^(2)-c^(3)`
`rArr ab^(2)-cb^(2)=ca^(2)-c^(3)`
`rArr b^(2)(a-c)=c(a^(2)-c^(2))`
`rArr b^(2)=c(a+c)`
Now CD = c and BD = a-c
Also D is angle bisector
`therefore (a-c)/(c )=(c )/(b)`
`rArr c^(2)=ab-bc`
`rArr b^(2)-ac=ab-bc` (using `b^(2)=c(a+c)`)
`rArr b^(2)+bc=ac+ab`
`rArr b(b+c)=a(b+c)`
`rArr a=b`
`rArr angle A = 2 angle C`
Now `angle A+ angle B + angle C = pi`
`therefore 2angle C +2 angle C + angle C = pi`
`therefore angle C = (pi)/(5)=(180^(@))/(5)=36^(@)`
`therefore angle A = 72^(@)`