G.SRT.C.8 STUDENT NOTES WS #4 – geometrycommoncore.com 1 Trigonometry problems can escalate in difficulty in a number of ways. One of those ways is to use two triangles and their measurements to come up with a single answer. Really this doesn’t increase the difficulty too much you just need to keep track of the relationship between the two triangles involved. it is mostly an increase of work and not so much difficulty.

Example Problem

A flagpole is at the top of a building. 300 ft from the base of the building, the angle of elevation of the top of the pole is 32 and the angle of elevation of the bottom of the pole is 30. Determine the length of the flagpole (to the nearest foot).

This problem looks quite complex because of the overlapping angles of elevation but really if we think through the problem we can 32° 30° simplify it by solving two different triangles. 300 ft

The way to get the height of just the flagpole would be to calculate the total height from ground to top x of flag (x) and then calculate the y distance from ground to the 32° 30° bottom of the flagpole and then 300 ft 300 ft subtract the two values.

x – y will give us the flagpole height, 187.46 – 173.21 = 14.25 ft

A SYSTEM OF EQUATIONS

In algebra 1 we learned how to solve two variables if we had two equations. This process was called solving a system of equations. We learned a few different ways to solve them; (1) by graphing, (2) by substitution, and finally (3) by elimination. We will look a trigonometry problem that will give us two equations and two variables and we will use substitution to solve it.

Example Problem

Sally and Jonathan are on either side of the tree and 40 ft apart. Sally sees the top of the tree at 42 and Jill sees the top of the steeple at HEIGHT 36. How high is the tree (to the nearest foot)? 42° 36°

The first step is to break up the 40 ft, most students break the 40 ft into two groups of 20 but this can’t be G.SRT.C.8 STUDENT NOTES WS #4 – geometrycommoncore.com 2 because if the distances were equal then so would there angles (the tree would be the perpendicular bisector of an isosceles triangle) but the angles aren’t equal so neither will the distances.

HEIGHT

42° 36° x 40 - x

The tree is 16 ft tall.

This is just one way to solve this problem, as mentioned earlier there are a number of algebraic techniques that could be applied here.