HELP needed please!!!! (1 Viewer)

[grasshopper1]

I just cant get this question :burn:

Pete walks along a straight road. At one point he notices a tower on a bearing of 053 degrees with an angle of elevation of 21degrees. After walking 230m, the tower is on a bearing of 342degrees, with an angle of elevation of 26degrees. What is the height of the tower?

please explain cause i just cant get it:mad1:

 

[Drongoski]

I just cant get this question :burn:

Pete walks along a straight road. At one point he notices a tower on a bearing of 053 degrees with an angle of elevation of 21degrees. After walking 230m, the tower is on a bearing of 342degrees, with an angle of elevation of 26degrees. What is the height of the tower?

please explain cause i just cant get it:mad1:

Is the height about 83.9 m ; If so I'll get back and explain although without the diagramming this is going to be a challenge.

 

[Drongoski]

I'll do my best without the diagrams. 'degrees' will be understood where applicable to save time.

Let Pete be at A when he first observes tower CD, base C and top D.

Let Pete be at B when he makes 2nd observation.

Now there are 3 triangles of interest: ABC (flat on ground), right-angled ACD and BCD.

Let height of tower be 'h' m. Then AB = 230 m and angle C = 53 + 18 = 71 deg.

AC = h/tan 21 and BC = h/tan 26

Re triangle ABC, using cosine rule:

230² = AC^2 + BC^2 - 2 AC.BC.cos 71

= (h/tan 21)^2 + (h/tan 26)^2 - 2 x (h/tan 21)(h/tan 26) cos 71

= h^2{(1/tan 21)^2 + (1/(tan26)^2 - 2 x (1/tan 21) (1/tan 26) cos 71 }

Therefore h² = 230^2/ { . . . }

Hence h = (you work this out)