Trigonometry Problem - Confused over multiple possible angle sizes.

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Homework Help Overview

The discussion revolves around a trigonometry problem involving triangle PQR, where the lengths of two sides and one angle are known, and participants are exploring the possible sizes of another angle. The context includes the application of the Sine Rule and the implications of having multiple possible angle values.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of the Sine Rule to find angle sizes and express confusion about the conditions that lead to multiple possible angles. Some share similar problems to illustrate their understanding and seek clarification on detecting scenarios that yield multiple solutions.

Discussion Status

The discussion is ongoing, with participants actively questioning the conditions under which multiple angle solutions arise. Some have provided insights into the geometric implications of the triangle's configuration, while others are seeking further explanation on how to visualize and calculate these angles.

Contextual Notes

Participants note the importance of the sum of interior angles in a triangle and the constraints imposed by the known angle and side lengths. There is also mention of the need for sketches to better understand the problem setup.

CallMeSwayze
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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?