McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
5. Angles of Elevation and Depression
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Exercise 11 Page 666

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

≈ 1309 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 river.

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 45^(∘) and tan 18^(∘). tan 45^(∘)=630/y & (I) tan 18^(∘)=630/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.
tan45^(∘)=630/y & (I) tan18^(∘)=630/y+x & (II)
ytan45^(∘)=630 & (I) tan18^(∘)=630/y+x & (II)
y=630/tan45^(∘) & (I) tan18^(∘)=630/y+x & (II)
y=630 & (I) tan18^(∘)=630/y+x & (II)
The value of y is 630 feet. By substituting this value into the second equation, we can find the value of x.
y=630 & (I) tan18^(∘)=630/y+x & (II)
y=630 & (I) tan18^(∘)=630/630+x & (II)
Solve for x
y=630 & (I) (630+x)tan18^(∘)=630 & (II)
y=630 & (I) (630+x)=630/tan18^(∘) & (II)
y=630 & (I) 630+x=630/tan18^(∘) & (II)
y=630 & (I) x=630/tan18^(∘)-630 & (II)
y=630 x=1308.940...
y=630 x≈1309
The width of the river is approximately 1309 feet.