1. ## elevation

From a point A south of the vertical tower , the angle of elevation of the top of the tower is observed to be 34 degree . From A , an observer walked a distance of 270 m on a bearing 070 degree and found that his bearing from the tower is 135 degree . Cal culate the height of the tower to the nearest metre .

2. Originally Posted by thereddevils
From a point A south of the vertical tower , the angle of elevation of the top of the tower is observed to be 34 degree . From A , an observer walked a distance of 270 m on a bearing 070 degree and found that his bearing from the tower is 135 degree . Cal culate the height of the tower to the nearest metre .
make a sketch ...

let point T be the tower's position and point B be the position the observer walks to from point A.

triangle TAB is formed ...

angle A = 70

angle B = 65

angle T = 45

AB = 270 m

use the law of sines to find AT, then use the tangent ratio to find the tower's height ...

tan(34) = h/(AT)

3. This is a 3D problem. I did one similar to it in this thread: http://www.mathhelpforum.com/math-he...g-problem.html . But there are some differences.

First you'll need to imagine a bird's eye view of the ground. See attachment #1. You've got a triangle ABT. Remember that bearings are measured with 0° = N. Angle T is 45°, angle A is 70°, and angle B is 65°. Use the Law of sines to find side AT:
$\frac{\sin 45^{\circ}}{270} = \frac{\sin 65^{\circ}}{AT}$

Why AT? Because you need that length to find the height of the tower at T. See attachment #2. (Actually, the placement of T here is wrong -- it should be at the base of the tower, not the top. Sorry about that.) Use the tangent ratio to find AT.
$\tan 34^{\circ} = \frac{h}{AT}$

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EDIT: Beaten to it by skeeter...

4. thanks .. but i don seem to be able to see from my diagram to get the 3 angles . Maybe my sketch is wrong and i am pretty confused with the diagram . Do u mind posting up ur sketch? Its ok if it's too much of work . Thank you once again .

5. here is the basic bird's eye view ... you should be able to find the value of every angle from this diagram.

6. Originally Posted by yeongil
This is a 3D problem. I did one similar to it in this thread: http://www.mathhelpforum.com/math-he...g-problem.html . But there are some differences.

First you'll need to imagine a bird's eye view of the ground. See attachment #1. You've got a triangle ABT. Remember that bearings are measured with 0° = N. Angle T is 45°, angle A is 70°, and angle B is 65°. Use the Law of sines to find side AT:
$\frac{\sin 45^{\circ}}{270} = \frac{\sin 65^{\circ}}{AT}$

Why AT? Because you need that length to find the height of the tower at T. See attachment #2. (Actually, the placement of T here is wrong -- it should be at the base of the tower, not the top. Sorry about that.) Use the tangent ratio to find AT.
$\tan 34^{\circ} = \frac{h}{AT}$

01

EDIT: Beaten to it by skeeter...

I can follow up still $\tan 34^{\circ} = \frac{h}{AT}$

why not $\sin 34^{\circ} = \frac{h}{AT}$

How come the placement of T is at the bottom ?

7. Originally Posted by thereddevils
I can follow up still $\tan 34^{\circ} = \frac{h}{AT}$

why not $\sin 34^{\circ} = \frac{h}{AT}$

How come the placement of T is at the bottom ?
Use the definition of Sine:

$\sin(angle)=\dfrac{leg}{hypotenuse}$

Since AT is the hypotenuse this distance must be the denominator. (By the way: The confusion is caused by yeongil's labelling of the first sketch: In a bird eye's view the points T and F are of course on the same place, but ...)

Maybe it's easier to label the footpoint of the tower by F. Then you'll get:

$\tan 34^{\circ} = \frac{h}{AF}$

8. Yes, apologies again for the confusion. I need to stress that T is supposed to be at the base of the tower, not the top. The reason T is not at the top is because when I found AT from the 1st diagram, AT was the distance from A to the tower along the ground, so that corresponds to T being the base of the tower in the 2nd diagram.

Therefore, we cannot use $\sin 34^{\circ} = \frac{h}{AT}$. We don't even know the distance from point A to the top of the tower (hypotenuse) - just the distance along the ground (adjacent).

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