Exercise 55 Page 678

We are asked to find the value of x in the given figure. Let's take a look at the given picture. We will label the vertices with consecutive letters.

As we can see, △ ABC is isosceles, so BC is also 6. Since we have all side lengths of △ ABC we can find the measure of ∠ C by using the Law of Cosines. AB^2=BC^2+AC^2-2(BC)(AC)cos CBy substituting the appropriate side lengths, we will solve the above equation.

AB^2=BC^2+AC^2-2(BC)(AC)cos C

SubstituteValues

Substitute values

6^2= 6^2+ 5.9^2-2( 6)( 5.9)cos C

▼

Simplify

CalcPow

Calculate power

36=36+34.81-2(6)(5.9)cos C

Multiply

Multiply

36=36+34.81-70.8cos C

SubEqn

LHS-36=RHS-36

0=34.81-70.8cos C

AddEqn

LHS+70.8cos C=RHS+70.8cos C

70.8cos C=34.81

DivEqn

.LHS /70.8.=.RHS /70.8.

cos C=34.81/70.8

Now we can evaluate the measure of ∠ C using the inverse cosine. cos C=34.81/70.8 ⇓ C=cos ^(-1)34.81/70.8≈ 60.55^(∘) The measure of ∠ C is approximately 60.55^(∘). Let's add this information to our picture.

Now, we will use the Law of Sines to write a proportion for ∠ BCD. According to this law, in a triangle the ratio of the sine of an angle to the opposite side length is constant. sin 68^(∘)/6=sin 60.55^(∘)/x Let's solve this proportion using cross multiplication.

sin 68^(∘)/6=sin 60.55^(∘)/x

CrossMult

Cross multiply

xsin 68^(∘)=6sin 60.55^(∘)

DivEqn

.LHS /sin 68^(∘).=.RHS /sin 68^(∘).

x=6sin 60.55^(∘)/sin 68^(∘)

UseCalc

Use a calculator

x=5.635...

RoundDec

Round to 1 decimal place(s)

x≈ 5.6

The value of x is approximately 5.6.