# Thread: Find the length of the triangle XZ

1. ## Find the length of the triangle XZ

In a triangle XYZ , A is a point on XZ such that $(XA/AZ)= 3$ and angle XAY = 70 degrees. Find the length of XZ if angle XYZ = 110 degrees and YZ = 6cm

Please help me. Will Pythagoras work or do I need to use trigonometry? Or am I taking this too seriously and there is an easier method?

2. Originally Posted by BG5965
In a triangle XYZ , A is a point on XZ such that $(XA/AX)= 3$ and angle XAY = 70 degrees. Find the length of XZ if angle XYZ = 110 degrees and YZ = 6cm

Please help me. Will Pythagoras work or do I need to use trigonometry? Or am I taking this seriously and there is an easier method?
Your question has a typo ..... $(XA/A{\color{red}Z})= 3$, perhaps?

3. Yes, you are right. Sorry for the typo.

4. Did you mean XA/AZ=3? Because XA/AX=1.

HA!
I think this is even better than Pythagoras!
Similar triangles are everyone's friend!

Alright, I want you to find a piece of paper and a pencil or pen and draw the triangle.
I don't care if your book has it drawn already, just draw it.
Remember that angle XYZ=110, so make it obtuse.
Then, make point A so that the distance from A to X is about three times the distance from Z to A.
Now, label everything we know: angle XYZ, angle XAY, and side YZ.
But, we know something else! If angle XAY=70 and angle XAY+angle ZAY=180 (it's on a straight line), then angle ZAY=110.
Now, for the finale. Draw a second triangle. It should look that same as the first. The bigger angle is A.
This is a copy of the smaller triangle AYZ. Why? Well, we have similar triangles! angle ZAY=angle XYZ, and angle AZY=angle XZY.
From there, all you need to do is relate similar sides, and you're done.
Post back with what you think the answer is.

5. Originally Posted by bleesdan
Did you mean XA/AZ=3? Because XA/AX=1.

HA!
I think this is even better than Pythagoras!
Similar triangles are everyone's friend!

Alright, I want you to find a piece of paper and a pencil or pen and draw the triangle.
I don't care if your book has it drawn already, just draw it.
Remember that angle XYZ=110, so make it obtuse.
Then, make point A so that the distance from A to X is about three times the distance from Z to A.
Now, label everything we know: angle XYZ, angle XAY, and side YZ.
But, we know something else! If angle XAY=70 and angle XAY+angle ZAY=180 (it's on a straight line), then angle ZAY=110.
Now, for the finale. Draw a second triangle. It should look that same as the first. The bigger angle is A.
This is a copy of the smaller triangle AYZ. Why? Well, we have similar triangles! angle ZAY=angle XYZ, and angle AZY=angle XZY.
From there, all you need to do is relate similar sides, and you're done.
Post back with what you think the answer is.
I'll just add that if you let angle $XYA = \theta$ this will:

1. Let you see the similarity of triangles XYZ and AYZ

2. Help you set up the appropriate side ratios.

6. Ok, well I have done what you have said and the fact they are similar triangles has shocked me! (Bleesdan) I have also added the angle theta (mr fantastic) But how do I proceed after this? Do I split the 110 degree angle into 90 degree and 20 degree and use trigonometry?

7. Originally Posted by BG5965
Ok, well I have done what you have said and the fact they are similar triangles has shocked me! (Bleesdan) I have also added the angle theta (mr fantastic) But how do I proceed after this? Do I split the 110 degree angle into 90 degree and 20 degree and use trigonometry?
Set up a useful ratio-of-sides between the two similar triangles: $\frac{ZX}{ZY} = \frac{ZY}{ZA}$ ....

8. I'll try:
6/(3x + x)= x + 6
4x^2 = 36
x^2 + 9
x = + {SQRT}9
x = 3
AC = 3x + x
3 x 3 + 3
= 12

9. Originally Posted by BG5965
I'll try:
6/(3x + x)= x + 6
4x^2 = 36
x^2 + 9
x = + {SQRT}9
x = 3
AC = 3x + x
3 x 3 + 3
= 12
And the unit ....?