Question

In the right triangle shown, m\angle V = 60\degreem∠V=60°m, angle, V, equals, 60, degree and UV= 18UV=18U, V, equals, 18.

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

Problem 1)

$m\angle T=30^o$

$TU=18\sqrt{3}\ units$

$TV=36\ units$

Problem 2)

$m\angle U=30^o$

$TU=9\sqrt{3}\ units$

$TV=9\ units$

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

$m\angle U=90^o$

$m\angle V=60^o$

step 1

Find the measure of angle T

we know that

$m\angle V+m\angle T=90^o$ ---> by complementary angles in a right triangle

substitute the given value

$60^o+m\angle T=90^o$

$m\angle T=90^o-60^o=30^o$

step 2

Find the length side TU

we know that

In the right triangle TUV

$tan(60^o)=\frac{TU}{UV}$ ---> by TOA (opposite side divided by adjacent side)

we have

$tan(60^o)=\sqrt{3}$

$UV=18\ units$

substitute the given values

$\sqrt{3}=\frac{TU}{18}$

$TU=18\sqrt{3}\ units$

step 3

Find the length side TV

we know that

In the right triangle TUV

$cos(60^o)=\frac{UV}{TV}$ ---> by CAH (adjacent side divided by the hypotenuse)

we have

$cos(60^o)=\frac{1}{2}$

$UV=18\ units$

substitute the given values

$\frac{1}{2}=\frac{18}{TV}$

$TV=36\ units$

Problem 2

Solve the right triangle TUV

The right angle is ∠T

so

$m\angle T=90^o$

$m\angle V=60^o$

step 1

Find the measure of angle U

we know that

$m\angle V+m\angle U=90^o$ ---> by complementary angles in a right triangle

substitute the given value

$60^o+m\angle U=90^o$

$m\angle U=90^o-60^o=30^o$

step 2

Find the length side TU

we know that

In the right triangle TUV

$sin(60^o)=\frac{TU}{UV}$ ---> by SOH (opposite side divided by hypotenuse)

we have

$sin(60^o)=\frac{\sqrt{3}}{2}$

$UV=18\ units$

substitute the given values

$\frac{\sqrt{3}}{2}=\frac{TU}{18}$

$TU=9\sqrt{3}\ units$

step 3

Find the length side TV

we know that

In the right triangle TUV

$cos(60^o)=\frac{TV}{UV}$ ---> by CAH (adjacent side divided by the hypotenuse)

we have

$cos(60^o)=\frac{1}{2}$

$UV=18\ units$

substitute the given values

$\frac{1}{2}=\frac{TV}{18}$

$TV=9\ units$