Answer:
Problem 1)
![m\angle T=30^o](https://tex.z-dn.net/?f=m%5Cangle%20T%3D30%5Eo)
![TU=18\sqrt{3}\ units](https://tex.z-dn.net/?f=TU%3D18%5Csqrt%7B3%7D%5C%20units)
![TV=36\ units](https://tex.z-dn.net/?f=TV%3D36%5C%20units)
Problem 2)
![m\angle U=30^o](https://tex.z-dn.net/?f=m%5Cangle%20U%3D30%5Eo)
![TU=9\sqrt{3}\ units](https://tex.z-dn.net/?f=TU%3D9%5Csqrt%7B3%7D%5C%20units)
![TV=9\ units](https://tex.z-dn.net/?f=TV%3D9%5C%20units)
Step-by-step explanation:
I will analyze two cases
see the attached figure to better understand the problem
Problem 1
<em>Solve the right triangle TUV</em>
The right angle is ∠U
so
![m\angle U=90^o](https://tex.z-dn.net/?f=m%5Cangle%20U%3D90%5Eo)
![m\angle V=60^o](https://tex.z-dn.net/?f=m%5Cangle%20V%3D60%5Eo)
step 1
Find the measure of angle T
we know that
---> by complementary angles in a right triangle
substitute the given value
![60^o+m\angle T=90^o](https://tex.z-dn.net/?f=60%5Eo%2Bm%5Cangle%20T%3D90%5Eo)
![m\angle T=90^o-60^o=30^o](https://tex.z-dn.net/?f=m%5Cangle%20T%3D90%5Eo-60%5Eo%3D30%5Eo)
step 2
Find the length side TU
we know that
In the right triangle TUV
---> by TOA (opposite side divided by adjacent side)
we have
![tan(60^o)=\sqrt{3}](https://tex.z-dn.net/?f=tan%2860%5Eo%29%3D%5Csqrt%7B3%7D)
![UV=18\ units](https://tex.z-dn.net/?f=UV%3D18%5C%20units)
substitute the given values
![\sqrt{3}=\frac{TU}{18}](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D%3D%5Cfrac%7BTU%7D%7B18%7D)
![TU=18\sqrt{3}\ units](https://tex.z-dn.net/?f=TU%3D18%5Csqrt%7B3%7D%5C%20units)
step 3
Find the length side TV
we know that
In the right triangle TUV
---> by CAH (adjacent side divided by the hypotenuse)
we have
![cos(60^o)=\frac{1}{2}](https://tex.z-dn.net/?f=cos%2860%5Eo%29%3D%5Cfrac%7B1%7D%7B2%7D)
![UV=18\ units](https://tex.z-dn.net/?f=UV%3D18%5C%20units)
substitute the given values
![\frac{1}{2}=\frac{18}{TV}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%3D%5Cfrac%7B18%7D%7BTV%7D)
![TV=36\ units](https://tex.z-dn.net/?f=TV%3D36%5C%20units)
Problem 2
<em>Solve the right triangle TUV</em>
The right angle is ∠T
so
![m\angle T=90^o](https://tex.z-dn.net/?f=m%5Cangle%20T%3D90%5Eo)
![m\angle V=60^o](https://tex.z-dn.net/?f=m%5Cangle%20V%3D60%5Eo)
step 1
Find the measure of angle U
we know that
---> by complementary angles in a right triangle
substitute the given value
![60^o+m\angle U=90^o](https://tex.z-dn.net/?f=60%5Eo%2Bm%5Cangle%20U%3D90%5Eo)
![m\angle U=90^o-60^o=30^o](https://tex.z-dn.net/?f=m%5Cangle%20U%3D90%5Eo-60%5Eo%3D30%5Eo)
step 2
Find the length side TU
we know that
In the right triangle TUV
---> by SOH (opposite side divided by hypotenuse)
we have
![sin(60^o)=\frac{\sqrt{3}}{2}](https://tex.z-dn.net/?f=sin%2860%5Eo%29%3D%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D)
![UV=18\ units](https://tex.z-dn.net/?f=UV%3D18%5C%20units)
substitute the given values
![\frac{\sqrt{3}}{2}=\frac{TU}{18}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%3D%5Cfrac%7BTU%7D%7B18%7D)
![TU=9\sqrt{3}\ units](https://tex.z-dn.net/?f=TU%3D9%5Csqrt%7B3%7D%5C%20units)
step 3
Find the length side TV
we know that
In the right triangle TUV
---> by CAH (adjacent side divided by the hypotenuse)
we have
![cos(60^o)=\frac{1}{2}](https://tex.z-dn.net/?f=cos%2860%5Eo%29%3D%5Cfrac%7B1%7D%7B2%7D)
![UV=18\ units](https://tex.z-dn.net/?f=UV%3D18%5C%20units)
substitute the given values
![\frac{1}{2}=\frac{TV}{18}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%3D%5Cfrac%7BTV%7D%7B18%7D)
![TV=9\ units](https://tex.z-dn.net/?f=TV%3D9%5C%20units)