Question

Question #8107e

Answer 1

see explanation.

Explanation:

Given `AB=AC, and angleCAB=20^@`,
`=> angleABC=angleACB=(180-20)/2=80^@`
Given `angleBDC=30^@, => angleBCD=70^@`
`=> angleACD=80-70=10^@`

In `DeltaADC, (AD)/sin10=(CD)/sin20`
`=> AD=CD*sin10/sin20=0.50771xxCD`

In `DeltaBDC, (BC)/sin30=(CD)/sin80`
`=> BC=CD*sin30/sin80=0.50771xxCD`

Hence, `AD=BC`

proof of `sin10/sin20=sin30/sin80`

cross multiplying :
`sin10sin80=sin20sin30`
`LHS = sin10cos(90-80)=sin10cos10`
`=1/2*sin20` ....(as `sinxcosx=1/2sin2x`)
`RHS=sin20sin30=sin20*1/2=1/2*sin20`

`LHS=RHS` (proved)