Let's begin noting that a triangle is isosceles if and only if two of its angles are congruent. We can thus find the angle <ABP, recalling that the sum of the interior angles of a triangle is equal to 180°.

$\angle \: ABP = \frac{180 - 80}{2} = 50 \degree$

Finally, let point K be the intersection between segments BC and PQ, and let's note that the triangle PQB is a right isosceles triangle, since all the angles in a square are equal to 90°, and the two triangles APB and BQC are congruent.

Therefore, the angle BKQ is equal to 180-50-45=85°.

Of course angle BKP=180-85=95°.

Hope this helps :)

6 0

2 years ago

Given the circle with the equation (x - 3)2 + y2 = 49, determine the location of each point with respect to the graph of the cir

Svetllana [295]

B....................................................

3 0

3 years ago