McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
5. Angles of Elevation and Depression
Continue to next subchapter

Exercise 18 Page 667

Recall that the angle of depression and the angle of elevation are congruent.

≈ 290 ft

Practice makes perfect

Let's begin with redrawing the given picture. We will name the missing sides. Let x represents the width of the field.

Let's recall that the angle of depression and the angle of elevation are congruent by the Alternate Interior Angles Theorem.
To evaluate the lengths of the missing sides, we can use one of the trigonometric ratios. Let's recall that the tangent of ∠ A is the ratio of the leg opposite ∠ A to the leg adjacent ∠ A. Using this definition, we can create the equations for tan 32^(∘) and tan 27^(∘). tan 32^(∘)=800/y & (I) tan 27^(∘)=800/y+ x & (II) As we can see, we need to solve the system of equations. To do this, we can use the Substitution Method. Our first step will be to isolate y in the first equation.
tan32^(∘)=800/y & (I) tan27^(∘)=800/y+x & (II)
ytan32^(∘)=800 tan27^(∘)=800/y+x
y=800/tan32^(∘) tan27^(∘)=800/y+x
y=1280.267... tan27^(∘)=800/y+x
y≈1280 tan27^(∘)=800/y+x
The value of y is approximately 1280 feet. By substituting this value into the second equation, we can find the value of x.
y≈1280 tan27^(∘)=800/y+x
y≈1280 tan27^(∘)=800/1280+x
Solve for x
y≈1280 (1280+x)tan27^(∘)=800
y≈1280 (1280+x)=800/tan27^(∘)
y≈1280 1280+x=800/tan27^(∘)
y≈1280 x=800/tan27^(∘)-1280
y≈1280 x=290.0884...
y≈1280 x≈290
The width of the field is approximately 290 feet.