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
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).
