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Find alpha and triange abc is isosceles........................ 
3 months ago

Answers : (1)

Aditya Gupta
1819 Points
							
hello pushkar, this is a direct application of sine rule
note that ADE will also be isosceles since angle ADE= angle AED= 80deg (since DE parallel to BC so that angle ADE= angle ABC and similarly)
so, DE= 2xcos80, where x= AE= BC
now, let BE= p and DB= EC= y. call alpha= q
apply sine rule in BCE, we have
sin(100 – q)/x= sin80/p.....(1)
apply sine rule in BDE we have,
sin(80 – q)/2xcos80= sin100/p.......(2)
noting that sin100= sin80, we have (1)= (2)
so sin(100 – q)/x= sin(80 – q)/2xcos80
or sin(80+q)/sin(80 – q)= 1/2cos80
this is clearly an eqn in one variable q, which can be solved with ease to find that q= 70 deg
kindly approve :=)
3 months ago
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