In square ABCD, P is on BC such that BP = 4 and PC = 1, and Q is on CD such that DQ= 4 and QC = 1. Find sin angle PAQ

Guest Jun 19, 2023

#1**0 **

We are given that square ABCD, P is on BC such that BP = 4 and PC = 1, and Q is on CD such that DQ= 4 and QC = 1. This means that triangle BPQ and triangle CQP are both right triangles.

We are asked to find sin angle PAQ. We can do this by using the following steps:

Find the length of side PQ.

Use the Pythagorean Theorem to find the length of side PA.

Use the ratio of opposite to hypotenuse to find sin angle PAQ.

Step 1:

The length of side PQ can be found by using the Pythagorean Theorem on triangle BPQ.

PQ^2 = BP^2 + PC^2 PQ^2 = 4^2 + 1^2 PQ^2 = 17 PQ = sqrt(17)

Step 2:

The length of side PA can be found by using the Pythagorean Theorem on triangle CQP.

PA^2 = CQ^2 + QP^2 PA^2 = 4^2 + sqrt(17)^2 PA^2 = 33 PA = sqrt(33)

Step 3:

The sine of angle PAQ can be found by using the ratio of opposite to hypotenuse.

sin angle PAQ = PQ / PA sin angle PAQ = sqrt(17) / sqrt(33) sin angle PAQ = sqrt(17/33)

**Therefore, the sin angle PAQ is sqrt(17/33).**

Guest Jun 19, 2023