Exercise 28 Page 675

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 the angle measure between Trail 1 and Trail 2. We will call it A.

Using the Law of Cosines, we can write an equation for A. 4^2= 2^2+ 3^2-2( 2)( 3)cos A Let's solve the equation. We will start with isolating cos A.

4^2=2^2+3^2-2(2)(3)cos A

▼

Simplify

CalcPow

Calculate power

16=4+9-2(2)(3)cos A

Multiply

Multiply

16=4+9-12cos A

AddTerms

Add terms

16=13-12cos A

SubEqn

LHS-13=RHS-13

3=-12cos A

DivEqn

.LHS /(-12).=.RHS /(-12).

3/-12=cos A

MoveNegDenomToFrac

Put minus sign in front of fraction

-3/12=cos A

ReduceFrac

a/b=.a /3./.b /3.

-1/4=cos A

RearrangeEqn

Rearrange equation

cos A=-1/4

Next let's use the inverse cosine to find the value of A. cos A=-1/4 ⇓ A=cos^(-1)-1/4≈104.5^(∘) The angle between Trail 1 and Trail 2 is approximately 104.5^(∘).