# Thread: Finding the angle measure of a triangle

1. ## Finding the angle measure of a triangle

In triangle AXZ, angle XAZ measures 80, segment XY bisects angle AXZ, and ZY bisects AZX. What is the measure of angle XYZ?

The answer should be 130 but I don't see how to obtain it.

2. ## Re: Finding the angle measure of a triangle

Originally Posted by dannyc
In triangle AXZ, angle XAZ measures 80, segment XY bisects angle AXZ, and ZY bisects AZX. What is the measure of angle XYZ? The answer should be 130 but I don't see how to obtain it.
Well $m\left( {\angle YXZ} \right) = 40^o \;\& \,m\left( {\angle YZX} \right) = 10^o$. How is that true?

3. ## Re: Finding the angle measure of a triangle

Hmmm to be honest, I'm not sure Plato--must be a theorem I'm forgetting?

4. ## Re: Finding the angle measure of a triangle

Originally Posted by dannyc
Hmmm to be honest, I'm not sure Plato--must be a theorem I'm forgetting?
You are given $m\left( {\angle AXZ} \right) = 80^o \;\& \,m\left( {\angle XAZ} \right) = 80^o$

What does that tell you about $m\left( {\angle AZX} \right) = ?^o$

5. ## Re: Finding the angle measure of a triangle

But why is AXZ 80? How can we conclude that with only knowing that XAZ is 80?

6. ## Re: Finding the angle measure of a triangle

Originally Posted by dannyc
But why is AXZ 80? How can we conclude that with only knowing that XAZ is 80?
Maybe I am misreading the given. I thought that is what is meant.

7. ## Re: Finding the angle measure of a triangle

From the given, we're only given the one angle of 80 at angle A and the two bisectors (I, too, thought about using the properties of isosceles triangles but it seems there is not enough info to do so which is why I'm stuck).

8. ## Re: Finding the angle measure of a triangle

You know (angle XAZ=80°) and the triangle is isosceles, so that means (angle AXZ) =50° =(angle AZX). Because XY bisects angle AXZ that means the bisector divides the (angle AXZ = 50°) in two equal angles of 25°. So if you make the addition: 180°-(25°+25°)=130°=(angle XYZ).

9. ## Re: Finding the angle measure of a triangle

The calculations certainly make sense, but how can you deduce that either triangle would be isosceles? That is the segments that are given that bisect the base angles of triangle AXZ are necessarily congruent base angles..

10. ## Re: Finding the angle measure of a triangle

The triangle needs not be isosceles; the 2 base angles total 100 (180 - 80)
and these are bisected, so 50: 180 - 50 = 130

11. ## Re: Finding the angle measure of a triangle

Hello, dannyc!

$\text{In }\Delta AXZ,\;\angle XAZ\,=\,80^o.$

$\text{Segment }XY\text{ bisects }\angle AXZ\text{, and }ZY\text{ bisects }\angle AZX.$

$\text{What is the measure of }\angle XYZ\,?$

Code:
                A
*
* *
*80d*
*     *
*   Y   *
*    *    *
*   * @ *   *
*  *       *  *
* *           * *
** 1           2**
X *  *  *  *  *  *  * Z

Let: . $\begin{Bmatrix}\angle X \:=\:\angle AXZ \\ \angle Z \:=\:\angle AZX \end{Bmatrix}\qquad \begin{Bmatrix}\angle 1 &=^& \angle YXZ &=& \frac{1}{2}\angle X\\ \angle 2 &=& \angle YZX &=& \frac{1}{2}\angle Z\\ \theta &=& \angle XYZ \end{Bmatrix}$

$\text{In }\Delta AXZ\!:\;\angle X + \angle Z + 80^o \:=\:180^o$

. . . . . . . . . . . . . $\angle X + \angle Z \:=\:100^o$

$\text{Divide by 2:}$ . . $\tfrac{1}{2}\angle X + \tfrac{1}{2}\angle Z \:=\:50^o$ .[1]

$\text{In }\Delta XYZ\!:\;\angle 1 + \angle 2 + \theta \:=\:180^o$

. . . . . . $\tfrac{1}{2}\angle X + \tfrac{1}{2}\angle Z + \theta \:=\:180^o$

Substitute [1]: . $50^o + \theta \:=\:180^o$

. . . Therefore: . . = . . $\theta \:=\:130^o$

12. ## Re: Finding the angle measure of a triangle

Thought of just to add on a point as well.
The point of intersection y is called in-centre. If you draw a circle taking it as the centre and perpendicular distance from it to any of the side the circle will pass through the other two points of the triangle

13. ## Re: Finding the angle measure of a triangle

Let's not confuse Danny more than he is...

14. ## Re: Finding the angle measure of a triangle

you only get it as 130 if you consider the whole triangle as an isosceles, have you consider that the one who wrote this question forgot to provide that, unless there is a fact i'm still not aware of ?