Answered

Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Experience the convenience of getting reliable answers to your questions from a vast network of knowledgeable experts. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

Will award brainliest to correct answer!
Given square ABCD. Two congruent i s o - Δ's ABP and BCQ are constructed with bases AB and BC. Each of these triangles has vertex angle of 80°. Point P lies in the interior of the square, while point Q lies outside of the square. Find the angle measure between PQ and BC.

Thanks in advance :D

Sagot :

Answer:

85° and 95°

Step-by-step explanation:

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°.

[tex] \angle \: ABP = \frac{180 - 80}{2} = 50 \degree[/tex]

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 :)

We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.