Geometry and TrigonometryEasy–MediumExterior angle theorem / angles in a triangle
Geometry and Trigonometry practice question
In triangle ABC, side BC is extended past C to point D. The exterior angle ACD measures 130°, and angle A measures 55°.
What is the measure, in degrees, of angle B?
- A. 50
- B. 65
- C. 75Correct
- D. 125
Answer: C. 75
Angles ACB and ACD form a linear pair, so angle ACB = 180° − 130° = 50°. The angles of a triangle sum to 180°, so angle B = 180° − 55° − 50° = 75°. (Equivalently, the exterior angle equals the sum of the two remote interior angles: 130° = 55° + B, so B = 75°.) Choice A is the measure of interior angle ACB, the intermediate step, mistaken for the final answer. Choice B comes from halving the exterior angle (130 ÷ 2), treating the triangle as if it were isosceles. Choice D comes from computing 180° − 55°, using only the given interior angle and ignoring angle ACB entirely.
Why the other answers are wrong
- A. 50
- 50° is angle ACB, the intermediate step from 180° − 130°. Now subtract both known angles from 180°: angle B = 180° − 55° − 50° = 75°.
- B. 65
- Halving the exterior angle (130 ÷ 2) assumes the triangle is isosceles, which isn't given. Use 130° = 55° + B to get B = 75°.
- D. 125
- You computed 180° − 55° and ignored angle ACB. All three interior angles must sum to 180°: 180° − 55° − 50° = 75°.
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