Exercise 57 Page 576

Consider the given triangle.

Note that the two sides of the triangle are congruent. Therefore, both of them have the length of 32 and the triangle is isosceles. Also, both angles near the base of the triangle have equal measures. Let's add this information to our diagram.

We want to find the values of x and y. To do that we will consider the smaller triangle, which is a right triangle. Let's find the variables one at a time.

Finding x

Let's take a look at the right triangle.

Note that the given side is the hypotenuse of the triangle, and the side we want to find is adjacent to the given angle. Therefore, we will use the cosine ratio. cos θ = Length of leg adjacent toθ/Length of hypotenuse In our triangle, we have that θ =54^(∘) and the length of the hypotenuse is 32. We want to find the length of the leg adjacent to the angle.

cos θ = adjacent/hypotenuse

SubstituteValues

Substitute values

cos 54^(∘) = x/32

▼

Solve for x

MultEqn

LHS * 32=RHS* 32

cos 54^(∘) * 32 = x

UseCalc

Use a calculator

18.80912...=x

RoundDec

Round to 1 decimal place(s)

18.8≈ x

RearrangeEqn

Rearrange equation

x≈ 18.8

Finding y

Let's add the obtained information to the diagram.

We know the length of the hypotenuse and this time we want to find the length of the side opposite to the given angle. Therefore, we will use the sine ratio. sin θ = Length of leg opposite toθ/Length of hypotenuse In our triangle, we have that θ =54^(∘) and the length of the hypotenuse is 32. We want to find the length of the leg opposite to the angle.

sin θ = opposite/hypotenuse

SubstituteValues

Substitute values

sin 54^(∘) = y/32

▼

Solve for y

MultEqn

LHS * 32=RHS* 32

sin 54^(∘) * 32 = y

UseCalc

Use a calculator

25.88854...=y

RoundDec

Round to 1 decimal place(s)

25.9≈ y

RearrangeEqn

Rearrange equation

y≈ 25.9