Find the height of the tower.

Question

David walks along a straight road. at one point he notices a tower on a bearing of 053 degrees, with an angle of elevation of 21 degrees. After walking 230 metres, the tower is on a bearing of 342 degrees, with an angle of elevation of 26 degrees. Find the height of the tower.

Answer 1

This question really needs a good diagram which is hard to do here so I hope you can follow the basic diagram at the bottom.

 

Point T is the the Tower, A is where the first observation is made and B is where the second observation is made.

 

The length of AB=230 m.

 

Let the distance from A to T be d1 and the distance from B to T be d2, and let the tower height be h. Points C and D are points due north of the observation locations A and B respectively.  D is chosen so that it also lies on the ray extended from A through T.

 

First, from the elevation angles we have

 

h=d1 tan(21°) = d2 tan(26°) ....(1)

 

Over relatively short distances we can assume that AC is parallel to BD (both are lines pointing north).

 

So from the bearing angle at A

<CAD=53° = <BDA (alternate angles)

From the bearing angle at B

<DBT=360-342=18°

Then <BTD=180-53-18=109°

and <BTA=71°

 

Now in triangle ABT you can apply the Law of Cosines to get

 

230² = d1² +d2² -2d1.d2cos(71°).....(2)

 

Now d2=d1 tan(21°)/tan(26°)  from equation (1)

 

Substitute this for d2 in equation (2) and factor out d1².

 

230²=d1²(1+(tan(21°)/tan(26°))²-2(tan(21°)/tan(26°))cos71°)

 

I got d1=218.6059 m

 

Then h=d1 tan(21°)=83.915m

 

 

 

                                           C  •                                    •  D

 

 

                                                                        T  

                                                                        •

 

 

 

 

                                                                                     • B

 

 

                                          A  •