Trigonometry Problem - Confused over multiple possible angle sizes.

AI Thread Summary
In triangle PQR, with QR measuring 4 mm, PR measuring 3.5 mm, and angle Q at 60°, the possible sizes for angle P include 81.79 degrees and others, but some options exceed the total of 180 degrees when combined with angle Q. The Sine Rule was used to find one solution of 81.79 degrees, but confusion arises regarding the existence of multiple angle sizes. To determine when multiple angles occur, it's essential to visualize the triangle and consider the geometric constraints, as certain configurations can lead to two valid angles. A sketch can aid in understanding how to construct the triangle and identify all potential angles. Understanding these principles clarifies the conditions under which multiple angle solutions arise in trigonometric problems.
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Homework Statement


In triangle PQR, the length of QR is 4 mm, the length of PR is 3.5 mm and the size of angle Q is 60°. What are the possible size(s) of angle P (to 2 decimal places)
The possible solutions are 138.81, 81.79, 118.96, 61.04, 98.21

The Attempt at a Solution



I've solved for 81.79 with the Sine Rule.
Sine P/4 = Sine 60/3.5
Sine P/4 = 0.247
Sine P = 0.98
Arcsin P = 81.79.

What has me confused is where I can get multiple possible sizes of the angle. How can I detect when this is going to occur and how can I calculate all the possible angles? Thank you.
 
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It's not clear what situations you think will result in multiple angles.

For triangles, it is important to remember that the sum of the interior angles must equal 180 degrees.

For this problem, angle Q is given as 60 degrees, so that would eliminate the first choice (138.81 degrees) automatically, since 60 + 138.81 > 180.
 
It's a multiple choice that says select all possible angles, so there may be more than one answer and I've heard of this before.
Here's another problem that's similar:
In the triangle XYZ, the length of YZ is 15 cm, the length of XZ is 30 cm. If the size of angle Y is 20°, what are the possible size(s) of angle X (to 2 decimal places)?
Solutions are 170, 10, 9.85, 30, 170.15
I know one is 9.85 as I solved it with the sine rule, I just don't know where I'd get multiple values for the angle, which I'm sure does happen.
 
CallMeSwayze said:

Homework Statement


In triangle PQR, the length of QR is 4 mm, the length of PR is 3.5 mm and the size of angle Q is 60°. What are the possible size(s) of angle P (to 2 decimal places)
The possible solutions are 138.81, 81.79, 118.96, 61.04, 98.21

The Attempt at a Solution



I've solved for 81.79 with the Sine Rule.
Sine P/4 = Sine 60/3.5
Sine P/4 = 0.247
Sine P = 0.98
Arcsin P = 81.79.

What has me confused is where I can get multiple possible sizes of the angle. How can I detect when this is going to occur and how can I calculate all the possible angles? Thank you.
Maybe draw a sketch of the triangle. From the sketch that I drew, I can see that the values given for QR and PQ and angle Q open up a couple of possibilities for the length of QP.
 
I've drawn a sketch already to use the Sine Rule. Would you mind explaining how you can get multiple angles and how you can calculate them?
 
Draw the triangle using (in principle, at least) ruler and compass construction. Start with the line QR. At Q you have an angle of 60 degrees, so you can draw a line from there, P lying somewhere on it. You know the distance of P from R. How would you construct the point P?
 

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