# heeeeeeeelp math

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In triangle ABC, ∠A=20∘ and ∠B=80∘. Let
D be a point on line segment AB such that

• heeeeeeeelp math -

Given 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°))

AC=x/(2cos(80°)) (from above)
∠DAC=20° (given),
we find DC by cosine rule
=sqrt(x²+x/(2cos(80°))-x²cos(80°))
=1.879x (approx.)

∠ ADC can be found by the sine rule:
=sin(20°)/(1-2cos(80°))

=31.6° approx.

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