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Difficulty: MediumTriangle Properties and Angle Theorems

In △ABC\triangle ABC, point DD lies on side BCBC such that AD=BDAD = BD. If the measure of ∠ADC\angle ADC is 112∘112^\circ and the measure of ∠BAC\angle BAC is 85∘85^\circ, what is the measure of ∠C\angle C, in degrees?

Answer: 39 degrees

Answer

The measure of ∠C\angle C is 39∘39^\circ.
The measure of ∠C\angle C is found by first calculating the interior angle ∠ADB=180∘−112∘=68∘\angle ADB = 180^\circ - 112^\circ = 68^\circ since BDCBDC forms a straight line. Because AD=BDAD = BD, △ABD\triangle ABD is isosceles with ∠B=∠BAD\angle B = \angle BAD. Using the angle sum of △ABD\triangle ABD, we have 2(∠B)+68∘=180∘2(\angle B) + 68^\circ = 180^\circ, which yields ∠B=56∘\angle B = 56^\circ. Finally, using the angle sum of △ABC\triangle ABC, we calculate ∠C=180∘−(85∘+56∘)=39∘\angle C = 180^\circ - (85^\circ + 56^\circ) = 39^\circ.

Step-by-Step Solution

1
Find the measure of ∠ADB\angle ADB using the linear pair relationship with ∠ADC\angle ADC.
∠ADB=68∘\angle ADB = 68^\circ
Angles on a straight line add up to 180∘180^\circ. Since DD lies on BCBC, ∠ADB+∠ADC=180∘\angle ADB + \angle ADC = 180^\circ.
2
Calculate the measure of ∠B\angle B using the properties of the isosceles triangle ABDABD.
∠B=56∘\angle B = 56^\circ
Since AD=BDAD = BD, the base angles opposite to these sides are equal: ∠BAD=∠B\angle BAD = \angle B. The sum of angles in △ABD\triangle ABD is 180∘180^\circ, so 2(∠B)+68∘=180∘2(\angle B) + 68^\circ = 180^\circ.
3
Find the measure of ∠C\angle C using the triangle angle sum theorem on the large triangle ABCABC.
∠C=39∘\angle C = 39^\circ
The sum of the angles in △ABC\triangle ABC is 180∘180^\circ, meaning ∠BAC+∠B+∠C=180∘\angle BAC + \angle B + \angle C = 180^\circ. Substituting the known values gives 85∘+56∘+∠C=180∘85^\circ + 56^\circ + \angle C = 180^\circ.

Key Concept

Using the Isosceles Triangle Theorem, the Triangle Angle Sum Theorem, and linear pairs to trace unknown angles in a geometric figure.

Practice More

Try finding the missing angles when a transversal cuts two parallel lines that form a triangle with a third intersecting line.

Alternative Method

Instead of finding ∠B\angle B first and then solving for ∠C\angle C in △ABC\triangle ABC, we can find the angle ∠DAC\angle DAC first. Since ∠ADC=112∘\angle ADC = 112^\circ is an exterior angle to △ABD\triangle ABD, we have ∠ADC=∠B+∠BAD\angle ADC = \angle B + \angle BAD. Since ∠B=∠BAD\angle B = \angle BAD, we get 2(∠BAD)=112∘  ⟹  ∠BAD=56∘2(\angle BAD) = 112^\circ \implies \angle BAD = 56^\circ. Because ∠BAC=85∘\angle BAC = 85^\circ, we have ∠DAC=85∘−56∘=29∘\angle DAC = 85^\circ - 56^\circ = 29^\circ. Now looking at △ADC\triangle ADC, we can solve for ∠C\angle C directly: ∠C=180∘−(112∘+29∘)=39∘\angle C = 180^\circ - (112^\circ + 29^\circ) = 39^\circ.
Estimated Time:1m 30s
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Topics

Triangle Properties and Angle Theorems

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