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?

See:

http://www.jiskha.com/display.cgi?id=1373027638

150

That's true, it should be the supplement of asin()=180-31.6=148 (nearest degree).

To find the measure of ∠ADC, we first need to determine the measure of ∠ACD.

Let's start by drawing triangle ABC with the given angle measures ∠A = 20° and ∠B = 80°.

Next, we know that AD = BC. Since ∠BAC and ∠BCA are equal angles (opposite sides are equal in an isosceles triangle), this implies that triangle ABC is an isosceles triangle.

Since the sum of the angles in any triangle is always 180°, we can use this property to find the measure of ∠C. We know that ∠C = 180° - ∠A - ∠B. Plugging in the given values, we have ∠C = 180° - 20° - 80° = 80°.

Since triangle ADC is also an isosceles triangle (AD = BC), we can conclude that ∠ACD = ∠ADC.

Therefore, ∠ADC = ∠ACD = 80°.