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.

  ∠AKB = 22°

Step-by-step 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°

68°

Step-by-step explanation:

BC is parallel to BK

Angles BCA and BKA are equal

180 - 59 - 53 = 68°

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step-by-step explanation:

the answer is y < -3 or   y > 3

-3y > 7 + 2

-3 y > 9

y < -3