IB maths Sin, Cos, angle of elevation, midpoint

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mikaelalrc
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Hello people! (@JackMoseley and others)

A flat horizontal area, ABC, is such that AB = 100m, BC = 50m, angle <ACB = 43.7degrees.
<BAC = 20.2 degrees (3s.f.)
Area of triangle ABC = 2245.24
AC = 130 (3sf) I think

So,
A vertical pole, TB, is constructed at point B and has height 25m.
Calculate the angle of elevation of T from, M, the midpoint of the side AC.

What's an angle of elevation????
Last edited by mikaelalrc; 1 year ago
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Squirmz
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The angle between TB and and plane ABC
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mikaelalrc
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oh, thank you!!!
but how do i calculate it?
(Original post by mazenod)
The angle between TB and and plane ABC
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Squirmz
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(Original post by mikaelalrc)
oh, thank you!!!
but how do i calculate it?
Okay give me a second :-)
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Squirmz
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So what I did was say that CM is 65m (yes, AC is 130m πŸ˜ƒ) and use triangle CMB and the cosine rule (a^2=b^2 c^2 -2bc x cos(A)) to find side MB, which I got as 45m (2sf). Then you have 2 sides of a right angled triangle BMT, the opposite and adjacent to the angle of elevation, so you do tan(x)=opposite/adjacent= 25/45 = 5/9. Then x = tan^-1(5/9). I got 29.1Β° (3sf)
Last edited by Squirmz; 1 year ago
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Squirmz
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It wont let me put a plus sign in for b^2 plus c^2 :/
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mikaelalrc
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hey, how did you get 45m? I got 12.8 (2sf) lol

(Original post by mazenod)
So what I did was say that CM is 65m (yes, AC is 130m πŸ˜ƒ) and use triangle CMB and the cosine rule (a^2=b^2 c^2 -2bc x cos(A)) to find side MB, which I got as 45m (2sf). Then you have 2 sides of a right angled triangle BMT, the opposite and adjacent to the angle of elevation, so you do tan(x)=opposite/adjacent= 25/45 = 5/9. Then x = tan^-1(5/9). I got 29.1Β° (3sf)
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Squirmz
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Square root (65^2 plus 50^2 - 2 x 65 x 50 x cos (43.7)) :-)
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