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

Exercise 9 Page 666

Recall the definition of tangent.

≈ 9.23 ft

Practice makes perfect

We are given that Austin is standing on the high dive at the local pool and that two of his friends are in the water. We also know that one friend is 5 feet beyond the other one. Let y represent the horizontal distance from the friend that is nearer to Austin to the edge of the pool.

We are also given the angles of depression to each of Austin's friends. Since the angle of depression is always congruent to the corresponding angle of elevation, we can say that the angles of elevations are 30^(∘) and 40^(∘) respectively.

As we are asked to evaluate how tall the platform is, let x represent this height. Notice that we have two right triangles in our diagram.

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 40^(∘) and tan 30^(∘). tan 40^(∘)=x/y & (I) tan 30^(∘)=x/5+ y & (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 x in the first equation.
tan40^(∘)=x/y & (I) tan30^(∘)=x/5+y & (II)
ytan40^(∘)=x tan30^(∘)=x/5+y
x=ytan40^(∘) tan30^(∘)=x/5+y
Next we will substitute the value of x into the second equation.
x=ytan40^(∘) tan30^(∘)=x/5+y
x=ytan40^(∘) tan30^(∘)=ytan40^(∘)/5+y
(II):Solve for y
x=ytan40^(∘) (5+y)tan30^(∘)=ytan40^(∘)
x=ytan40^(∘) 5tan30^(∘)+ytan30^(∘)=ytan40^(∘)
x=ytan40^(∘) 5tan30^(∘)=ytan40^(∘)-ytan30^(∘)
x=ytan40^(∘) 5tan30^(∘)=y(tan40^(∘)-tan30^(∘))
x=ytan40^(∘) 5tan30^(∘)/(tan40^(∘)-tan30^(∘))=y
x=ytan40^(∘) y=5tan30^(∘)/(tan40^(∘)-tan30^(∘))
x=ytan40^(∘) y=11.0286...
x=ytan40^(∘) y≈11
The value of y is approximately 11 feet. By substituting this value into the first equation, we can find the value of x.
x=ytan40^(∘) y≈11
x= 11tan40^(∘) y≈11
x=9.23009... y≈11
x≈9.23 y≈11
The platform is approximately 9.23 feet tall.