IB maths Sin, Cos, angle of elevation, midpointWatch
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
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????
So what I did was say that CM is 65m (yes, AC is 130m 😃) 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 29.1° (3sf)