Exercise 72 Page 237

Practice makes perfect

a Since we know the hypotenuse and opposite leg of the reference angle, we can use the sine ratio.

sin θ=Opposite/Hypotenuse

With this information we can write an equation.

Let's solve the equation for x.

sin 22^(∘) = x/17

MultEqn

LHS * 17=RHS* 17

17sin 22^(∘) = x

RearrangeEqn

Rearrange equation

x=17sin 22^(∘)

UseCalc

Use a calculator

x=6.36831...

RoundDec

Round to 3 decimal place(s)

x≈ 6.368

b Here we have been given the opposite and adjacent leg to the reference angle, which means we have to use the tangent ratio.

tan θ=Opposite/Adjacent

Now we can write an equation.

Let's solve the equation for x.

tan 49^(∘) = 7/x

MultEqn

LHS * x=RHS* x

tan 49^(∘)* x = 7

DivEqn

.LHS /tan 49^(∘).=.RHS /tan 49^(∘).

x = 7/tan 49^(∘)

UseCalc

Use a calculator

x=6.08500...

RoundDec

Round to 3 decimal place(s)

x≈ 6.085

c Here we have been given the hypotenuse and adjacent leg to the reference angle, which means we have to use the cosine ratio.

cos θ=Adjacent/Hypotenuse

We can write an equation.

Let's solve the equation for x.

cos 60^(∘) = x/6

MultEqn

LHS * 6=RHS* 6

6cos 60^(∘) = x

RearrangeEqn

Rearrange equation

x=6cos 60^(∘)

UseCalc

Use a calculator

x=3

d In a right triangle the two acute angles are complementary.

Therefore, in Part C the measure of the other acute angle is 30^(∘). 60 ^(∘) + 30 ^(∘) = 90 ^(∘)

Let's look at the diagram with this angle included.

As we can see, this time x is on the opposite side. Therefore, to calculate its value we have to use the sine ratio. sin θ = opposite/hypotenuse Substituting 6 for the length of the hypotenuse, x for the length of the opposite side, and 30^(∘) for θ, we will get an equation which we can solve for x. sin 30^(∘) = x/6 Let's do it!

sin 30^(∘) = x/6

MultEqn

LHS * 6=RHS* 6

6 sin 30 ^(∘) = x

RearrangeEqn

Rearrange equation

x = 6 sin 30 ^(∘)

UseCalc

Use a calculator

x=3