Geometry: Finding The Representation For An Angle

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Homework Help Overview

The discussion revolves around a geometry problem involving angle bisectors. The original poster presents expressions for two angles formed by a bisector and seeks to determine the measure of the larger angle.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the angles and question the setup, particularly the notation used for the bisector. There is discussion about whether to add the angle representations and how to properly account for the bisection in calculations.

Discussion Status

There is an ongoing exploration of the problem, with participants providing guidance on how to approach the calculations. Some participants suggest setting the angle expressions equal to each other, while others clarify the implications of the angle bisector concept.

Contextual Notes

Participants note confusion regarding the notation and the correct interpretation of the problem setup, which may affect their understanding of how to proceed with the calculations.

Bashyboy
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[itex]\stackrel{\rightarrow}{PQ}[/itex] bisects [itex]\angle{ABC}[/itex].
If [itex]m\angle{ABD}[/itex] can be represented by [itex]3a + 10[/itex] and [itex]m\angle{DBC}[/itex] can be represented by [itex]5a - 6[/itex], what is [itex]m\angle{ABC}[/itex]?

What I did was add the two representations together and multiplied by two, which would give me [itex]16a + 8[/itex]. Does that sound correct?
 
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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:
[itex]\stackrel{\rightarrow}{PQ}[/itex] bisects [itex]\angle{ABC}[/itex].
If [itex]m\angle{ABD}[/itex] can be represented by [itex]3a + 10[/itex] and [itex]m\angle{DBC}[/itex] can be represented by [itex]5a - 6[/itex], what is [itex]m\angle{ABC}[/itex]?

What I did was add the two representations together and multiplied by two, which would give me [itex]16a + 8[/itex]. Does that sound correct?
What does [itex]\stackrel{\rightarrow}{PQ}[/itex] 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 [itex](1/2 + 1/2)2[/itex], 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 [itex](1/2 + 1/2)2[/itex], 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.
 

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