Question

In triangle ABC, it is given that angle A is 59 degrees and angle B is 53 degrees. The altitude from B to line AC is extended until it intersects the line through A that is parallel to segment BC; they meet at K. Calculate the size of angle AKB.

Answer 1

68°

Explanation:

BC is parallel to BK

Angles BCA and BKA are equal

180 - 59 - 53 = 68°

Answer 2

∠AKB = 22°

Explanation:

Call the point of intersection of BK and AC point P. Then line BK is a transversal to parallel lines AK and BC. Alternate interior angles AKB and PBC will be congruent.

We know that angle PCB is 68°, the difference between 180° and the sum of the two given angles in ΔABC:

∠PCB = 180° -∠PAB -∠ABC = 180° -59° -53° = 68°

Since PB is an altitude, ∠CPB is a right angle. Then ∠PBC is the complement of ∠PCB, so is ...

∠PBC = 90° -68° = 22°

As we said, this is congruent to ∠AKB, so ...

∠AKB = 22°