Problem:

You must find the horizontal distance between two towers (points A and B) at the same elevation on opposite sides of a wide canyon running east and west. The towers lie directly north and south of each other. You mark off an east/west line running perpendicular to AB. (Drawing not to scale!)

For three points each answer the following questions showing all your work. You may need to add to the sketch above, or make your own, in order to fully understand the problem.

A. From C you measure the angle between the two towers (angle ACB). Given the distance from C to B, which trig function helps you find the distance AB? (note: AB is perpendicular to CD.

B. Write an equation and solve it to find an expression for the distance AB.

C. If the angle at C is 87.3° and the distance CB is 217 feet, what is the distance between the towers to the nearest whole foot?

D. You want to check your work to make sure it’s right.You should be able to both measure and compute the angle at D. Knowing the distance between the two towers and the distance BD, what inverse trig function would allow you to compute the angle at D?

E. Given the distance you found in part C and the distance BD is 450 feet, what is the angle at D to the nearest one-tenth degree?

A. Working with triangle ABC: Like AB is perpendicular to CD, the angle CBA is a right angle (90°), and the triangle ABC is a right triangle. In this triangle the side opposite to the right angle (side AC) is the hypotenuse, and the other two sides (AB and CB) are the legs.

If we measure from C the angle between the two towers (angle ACB), and gicen the distance from C to B (adjacent leg to angle ACB), to find the distance AB (opposite leg to angle ACB) we can use a trigonometric function that relates the opposite leg with the adjacent leg to an angle. This trigonometric function is tangent.

B.

tan(<ACB) = (Opposite leg to <ACB) / (Adjacent leg to <ACB)

tan(<ACB)=AB/CB

Solving for AB: Multiplying both sides of the equation by CB:

CB*tan(<ACB) = CB * AB/CB

CB*tan(<ACB)=AB

AB=CB*tan(<ACB) (eq. 1)

C. Angle at C = <ACB=87.3°

CB=217 feet

The distance between the towers: AB=?

Replacing the known values in the eq. 1

AB=(217 feet)*tan(87.3°)

tan(87.3°)=21.20494878. Replacing in the formula above:

AB=(217 feet)*21.20494878

AB=4,601.473885 feet

Rounding to the nearest whole foot:

AB=4,601 feet

D. Working with triangle ABD: Like AB is perpendicular to CD, the angle DBA is a right angle (90°), and the triangle ABD is a right triangle. In this triangle the side opposite to the right angle (side AD) is the hypotenuse, and the other two sides (AB and BD) are the legs.

Knowing the distance between the two towers (AB, opposite leg to angle ADB); and the distance BD (adjacent leg to angle ADB), the inverse trig function would allow us to compute the angle at D (<ADB) is tan^(-1)

E. AB=4,601.473885 feet

BD=450 feet

Angle at D: <ADB=?

tan(<ADB)=(Opposite leg to <ADB)/(Adjacent leg to <ADB)

tan(<ADB)=AB/BD

Replacing the known values:

tan(<ADB)=(4,601.473885 feet)/(450 feet)

Dividing:

tan(<ADB)=10.22549752

Solving for <ADB

<ADB=tan^(-1) (10.22549752)

<ADB=84.41453435°

Rounding to the nearest one-tenth degree:

<ADB=84.4°