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I am having a hard time with this problem, So I know triangle WXZ is isosceles because the sides are the same which means the opposite angles are the same as well.

Now i'm trying to determine similar triangles or congruent triangles and I can tell that triangle WXY and WYZ are congruent but I can't seem to find WY.

I decided to use law of cosines and find angle X and then from there, I decided to use law of sines however, I got WY to be negative which is not possible but the absolute value of that length was very close to choice A.

However, I'm not sure if that is a right procedure. Any ideas on solving this problem?

If $WX=YZ=13$ and $WZ=10$, what is the length of $WY$ ? Figure

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  • $\begingroup$ Use the area formula --- $[\triangle] = \dfrac {ab \sin C}{2}$. $\endgroup$ – Mick Aug 18 '16 at 3:43
  • $\begingroup$ Actually WXY is not congruent to WYZ. WZ is 10 while W X =13 so they are not conguent. $\endgroup$ – fleablood Aug 18 '16 at 7:24
  • $\begingroup$ Note $wy^2 + yx^2 =169$. $wy^2 + yz^2 = 100$ . $yx+ yz= 13$ three equations, three unknown (albeit not linear -- but still worth attempting to solve). $\endgroup$ – fleablood Aug 18 '16 at 7:28
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Hint: If we can express the area in two ways then maybe we can find the length of that height!

Solution Drop a perpendicular from $X$ to $WZ$ and let it intersect at $P$. By Pythagorean Theorem, $XP^2=13^2-5^2=144$ so $XP=12$. This means the area of triangle $XWZ$ is just $\frac{bh}{2}=\frac{WZ\cdot XP}{2}=\frac{10\cdot 12}{2}=60$. Now, note that $WY$ is the height from $W$ to $XZ$. So the area of triangle $XWZ$ can also be expressed as $\frac{bh}{2}=\frac{WY\cdot XZ}{2}=\frac{WY\cdot 13}{2}$. Since we know the area is $60$, we get $\frac{WY\cdot 13}{2}=60$ which means $WY=\frac{120}{13}\approx \boxed{9.23}$. So the answer is A :)

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