Answer:
Problem 1)
[tex]m\angle T=30^o[/tex]
[tex]TU=18\sqrt{3}\ units[/tex]
[tex]TV=36\ units[/tex]
Problem 2)
[tex]m\angle U=30^o[/tex]
[tex]TU=9\sqrt{3}\ units[/tex]
[tex]TV=9\ units[/tex]
Step-by-step explanation:
I will analyze two cases
see the attached figure to better understand the problem
Problem 1
Solve the right triangle TUV
The right angle is ∠U
so
[tex]m\angle U=90^o[/tex]
[tex]m\angle V=60^o[/tex]
step 1
Find the measure of angle T
we know that
[tex]m\angle V+m\angle T=90^o[/tex] ---> by complementary angles in a right triangle
substitute the given value
[tex]60^o+m\angle T=90^o[/tex]
[tex]m\angle T=90^o-60^o=30^o[/tex]
step 2
Find the length side TU
we know that
In the right triangle TUV
[tex]tan(60^o)=\frac{TU}{UV}[/tex] ---> by TOA (opposite side divided by adjacent side)
we have
[tex]tan(60^o)=\sqrt{3}[/tex]
[tex]UV=18\ units[/tex]
substitute the given values
[tex]\sqrt{3}=\frac{TU}{18}[/tex]
[tex]TU=18\sqrt{3}\ units[/tex]
step 3
Find the length side TV
we know that
In the right triangle TUV
[tex]cos(60^o)=\frac{UV}{TV}[/tex] ---> by CAH (adjacent side divided by the hypotenuse)
we have
[tex]cos(60^o)=\frac{1}{2}[/tex]
[tex]UV=18\ units[/tex]
substitute the given values
[tex]\frac{1}{2}=\frac{18}{TV}[/tex]
[tex]TV=36\ units[/tex]
Problem 2
Solve the right triangle TUV
The right angle is ∠T
so
[tex]m\angle T=90^o[/tex]
[tex]m\angle V=60^o[/tex]
step 1
Find the measure of angle U
we know that
[tex]m\angle V+m\angle U=90^o[/tex] ---> by complementary angles in a right triangle
substitute the given value
[tex]60^o+m\angle U=90^o[/tex]
[tex]m\angle U=90^o-60^o=30^o[/tex]
step 2
Find the length side TU
we know that
In the right triangle TUV
[tex]sin(60^o)=\frac{TU}{UV}[/tex] ---> by SOH (opposite side divided by hypotenuse)
we have
[tex]sin(60^o)=\frac{\sqrt{3}}{2}[/tex]
[tex]UV=18\ units[/tex]
substitute the given values
[tex]\frac{\sqrt{3}}{2}=\frac{TU}{18}[/tex]
[tex]TU=9\sqrt{3}\ units[/tex]
step 3
Find the length side TV
we know that
In the right triangle TUV
[tex]cos(60^o)=\frac{TV}{UV}[/tex] ---> by CAH (adjacent side divided by the hypotenuse)
we have
[tex]cos(60^o)=\frac{1}{2}[/tex]
[tex]UV=18\ units[/tex]
substitute the given values
[tex]\frac{1}{2}=\frac{TV}{18}[/tex]
[tex]TV=9\ units[/tex]
