Exercise 34 Page 521

We are given that a firefighter sees that fire breaks through the window near the top of a building, and the angles of elevation to the windowsill and to the top of the building are 28^(∘) and 42^(∘). We also know that the firefighter is 75 ft from the building and her eyes are 5 ft above the ground. Let's sketch a diagram.

Since we are asked to find the roof-to-windowsill distance, let x represent this distance. We will also label the distance between the windowsill and the firefighter's line of sight as y.

To find the values of x and y, we can use one of the trigonometric ratios. Let's recall that the tangent of an angle is the ratio between the leg opposite this angle and the leg adjacent to this angle. Using this definition, we can write equations for tan 28^(∘) and tan42^(∘). tan28^(∘)=y/75 & (I) tan42^(∘)=x+y/75 & (II) Next we can solve the above system of equations using the Substitution Method. Let's start by solving the first equation for y.

tan28^(∘)=y/75 & (I) tan42^(∘)=x+y/75 & (II)

MultEqn

(I):LHS * 75=RHS* 75

75tan28^(∘)=y tan42^(∘)=x+y/75

RearrangeEqn

(I):Rearrange equation

y=75tan28^(∘) tan42^(∘)=x+y/75

UseCalc

(I):Use a calculator

y=39.8782... tan42^(∘)=x+y/75

RoundDec

(I):Round to 1 decimal place(s)

y≈39.9 tan42^(∘)=x+y/75

Now, by substituting the value of y into the second equation we will find x.

y≈39.9 tan42^(∘)=x+y/75

Substitute

(II):y= 39.9

y≈39.9 tan42^(∘)=x+ 39.9/75

MultEqn

(II):LHS * 75=RHS* 75

y≈39.9 75tan42^(∘)=x+39.9

SubEqn

(II):LHS-39.9=RHS-39.9

y≈39.9 75tan42^(∘)-39.9=x

RearrangeEqn

(II):Rearrange equation

y≈39.9 x=75tan42^(∘)-39.9

UseCalc

(II):Use a calculator

y≈39.9 x=27.6303...

RoundDec

(II):Round to 2 decimal place(s)

y≈39.9 x≈27.6

The values of x and y are approximately 27.6 and 39.9. Let's add this information to our diagram.

The firefighter can report by radio that the roof-to-windowsill distance is approximately 27.6 feet.