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

  • 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°))

    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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