Geometry: Finding The Representation For An Angle

Click For Summary
The discussion revolves around finding the measure of angle ABC, given that line segment PQ bisects it. The expressions for angles ABD and DBC are provided as 3a + 10 and 5a - 6, respectively. Participants clarify that instead of multiplying by two directly, one should first solve for 'a' using either expression and then double that result to find the full angle. There was confusion regarding the notation, with PQ mistakenly referenced instead of BD. Ultimately, both methods of solving for the angle are validated, emphasizing the importance of understanding angle bisection in the calculations.
Bashyboy
Messages
1,419
Reaction score
5
\stackrel{\rightarrow}{PQ} bisects \angle{ABC}.
If m\angle{ABD} can be represented by 3a + 10 and m\angle{DBC} can be represented by 5a - 6, what is m\angle{ABC}?

What I did was add the two representations together and multiplied by two, which would give me 16a + 8. Does that sound correct?
 
Physics news on Phys.org
Where is the point D? Also, is it supposed to read ##\vec{BD}## instead of ##\vec{PQ}##?

Also, why would you multiply by 2?
 
Bashyboy said:
\stackrel{\rightarrow}{PQ} bisects \angle{ABC}.
If m\angle{ABD} can be represented by 3a + 10 and m\angle{DBC} can be represented by 5a - 6, what is m\angle{ABC}?

What I did was add the two representations together and multiplied by two, which would give me 16a + 8. Does that sound correct?
What does \stackrel{\rightarrow}{PQ} have to do with anything?

Why multiply by 2 ?
 
Sorry to both of you. It is not suppose to be PQ, but BD; I was looking at the wrong problem. Well, I was thinking, since it was a bisection of an angle, it would be the whole angle cut in half. But, know that I really consider it, that would be equivalent to (1/2 + 1/2)2, which not represent the whole angle. Now, since I know that those two expressions represent the same angle measurement, could I set them equal to each other, solve for a, and finally multiply that value by two get get the un-bisected angle?
 
Bashyboy said:
Sorry to both of you. It is not suppose to be PQ, but BD; I was looking at the wrong problem. Well, I was thinking, since it was a bisection of an angle, it would be the whole angle cut in half. But, know that I really consider it, that would be equivalent to (1/2 + 1/2)2, which not represent the whole angle. Now, since I know that those two expressions represent the same angle measurement, could I set them equal to each other, solve for a, and finally multiply that value by two get get the un-bisected angle?

Okay, that makes more sense! You almost have the right idea. You are absolutely correct in your method for solving for a. However, a by itself does not represent either angle, so multiply a by 2 would serve no purpose. However, plugging in a for either of the two equations (##3a+10## or ##5a-6##) will give you the equation for one half of the angle. Then you are correct, you would multiply THAT angle by two to get the entire angle.

Alternatively, you can add ##3a + 10## and ##5a-6## together and plug in a to get the entire angle. Both methods work!
 
Thank you for helping me.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
4K
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
12
Views
4K
  • · Replies 4 ·
Replies
4
Views
663
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K