Question #8107e

1 Answer
Feb 15, 2017

see explanation.

Explanation:

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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)