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April 24, 2014

April 24, 2014

Posted by **lin** on Friday, July 5, 2013 at 8:33am.

D be a point on line segment AB such that

AD=BC. What is the measure (in degrees) of ∠ADC?

- heeeeeeeelp math -
**MathMate**, Friday, July 5, 2013 at 10:51amGiven ΔABC,

A=20°

B=80° =>

C=80°

=> isosceles triangle with base BC=x.

Let E=mid-point of BC, then

ΔAEC is a right triangle right-angled at E

EC=x/2

By definition of cosine,

AC=(x/2)/cos(80°)

=x/(2cos(80°))

Consider ΔADC,

AD=x (given)

AC=x/(2cos(80°)) (from above)

∠DAC=20° (given),

we find DC by cosine rule

DC=sqrt(AD²+AC²-2*AD*AC*cos(20°) )

=sqrt(x²+x/(2cos(80°))-x²cos(80°))

=1.879x (approx.)

∠ ADC can be found by the sine rule:

sin(ADC)=(x/(2cos(80°))*sin(20°)/DC

=sin(20°)/(1-2cos(80°))

∠ADC=asin(sin(20°)/(1-2cos(80°)))

=31.6° approx.

Please check my arithmetic.

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