Answer:
∠DCA = 10°
Step-by-step explanation:
i know of no simple way to show the angle is 10° and that ΔAFC is isosceles (see the attached figure). Using the Law of Sines with respect to ΔADC, you can show that ∠DCA = 10°.
The measure of side AC is ...
AC·cos(80°) = EC
The measure of side AD is ...
AD = 2·EC
AD = 2·AC·cos(80°) . . . . . . . substituting for EC
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Another relation involving these sides comes from the law of sines.
sin(∠DCA)/AD = sin(∠ADC)/AC
The sum of angles in a triangle is 180°, so ...
∠DCA + ∠ADC + ∠A = 180°
Solving for ∠ADC, we have ...
∠ADC = 180° - (∠A + ∠DCA)
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By the symmetry of the sine function, we can rewrite the above law of sines expression as ...
sin(∠DCA)/(AC·2·cos(80°)) = sin(∠A + ∠DCA)/AC . . . . . . [eq1]
Multiplying by the denominator on the left and expanding the sine of the sum of angles, we have ...
sin(∠DCA) = 2·cos(80°)·(sin(20°)cos(∠DCA) +cos(20°)sin(∠DCA))
Separating sine and cosine terms, we have ...
sin(∠DCA)·(1 -cos(20°)·2·cos(80°)) = cos(∠DCA)·sin(20°)·2·cos(80°)
And this gives us an expression for the tangent of ∠DCA:
tan(∠DCA) = 2·sin(20°)·cos(80°) / (1 -2·cos(20°)·cos(80°))
arctan(2·sin(20°)·cos(80°) / (1 -2·cos(20°)·cos(80°))) = ∠DCA = 10° (exactly)
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You can simplify the tangent further using various trig identities, but that's more work than we like. You can show the answer is exactly 10° by substituting that value into the rewritten law of sines equation [eq1] above. This gives ...
sin(10°)/(2·cos(80°)) = sin(30°)
Since cos(80°) = sin(10°), this is a tautology:
1/2 = sin(30°)
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Possible alternate solution approach
You may be able to do something with the fact that AE is an angle bisector, so DF/DA = CF/CA.
