Exercise 38 Page 705

Let's begin with recalling the Law of Cosines. If △ ABC has lengths of a, b, and c and angle measures of A, B, and C, then we can write equations that relate the side lengths of this triangle and the cosine of one of its angles.

Now let's take a look at the given picture. We are asked to evaluate how much cable Ed needs if he needs twice as much as the length of SB. Let x represent the length of this segment.

Using the Law of Cosines, we can write an equation for x and then solve it. Notice that since x represents a length, we will consider only a positive case when taking a square root of x^2.

x^2= 1000^2+ 864^2-2( 1000)( 864)cos 150^(∘)

▼

Solve for x

CalcPow

Calculate power

x^2=1 000 000+746 496-2(1000)(864)cos 150^(∘)

Multiply

Multiply

x^2=1 000 000+746 496-1 728 000cos 150^(∘)

AddTerms

Add terms

x^2=1 746 496-1 728 000cos 150^(∘)

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(x^2)=sqrt(1 746 496-1 728 000cos 150^(∘))

SqrtPowToNumber

sqrt(a^2)=a

x=sqrt(1 746 496-1 728 000cos 150^(∘))

UseCalc

Use a calculator

x=1800.8297...

RoundInt

Round to nearest integer

x≈ 1801

The length of SB is approximately 1801 feet. Since we know that Ed needs twice as much cable as the length of SB, we can evaluate the appropriate length of the cable by multiplying 1801 by 2. 1801*2=3602 Ed needs approximately 3602 ft of the cable.