Question #8107e

1 Answer
CW
Feb 15, 2017

see explanation.

Explanation:

enter image source here

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)

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