Question

In ΔOPQ, p = 75 cm, o = 32 cm and ∠O=5°. Find all possible values of ∠P, to the nearest degree.

Answer 1

Using the Law of Sines and trigonometric functions, we find that angle P can be approximately 11.81° or 168.19° in triangle OPQ.

To find all possible values of angle P in triangle OPQ, we'll use the Law of Sines:

sin(P)/75 = sin(5°)/32

First, isolate sin(P):

sin(P) = (75 * sin(5°))/32

Now, take the arcsin of both sides to find the possible values for angle P:

P = arcsin((75 * sin(5°))/32)

Using a calculator:

P ≈ arcsin((75 * 0.08716)/32)

P ≈ arcsin(6.516/32)

P ≈ arcsin(0.203625)

P ≈ 11.81°

Now, since sin(P) has two solutions, the second possible value for angle P is found by subtracting the first result from 180°:

P' = 180° - 11.81°

P' ≈ 168.19°

So, the two possible values for angle P are approximately 11.81° and 168.19°.