f.5 trigonometry (urgent)

11.
In ΔTYZ :
tan27° = (3m)/YZ
YZ = 3/tan27°

In ΔXYZ :
sinÐYXZ / YZ = sin ÐXZY / XY (sine law)
sin12° / (3/tan27°) =sinÐXZY / 24
sinÐXZY = 24 sin12° tan27° /3
ÐXZY = 57.9°
57.9° +12° =69.9°
Bearing of T (Y) from Z = S 69.9° W


14.
(a)
ÐBXY = 90° - 60° =30°
ÐBYX = 90° - 30° =60°

ÐXBY + ÐBXY + ÐBYX = 180° (Ð sum of Δ)
ÐXBY + 30° + 60° =180°
ÐXBY = 90°

(b)
In rt.Ð ΔBXY :
sinÐBXY = BY / XY
sin30° =BY / (24 km)
BY = 24 sin30° km
BY = 12 km

In rt.Ð ΔABY :
tanÐAYB = AB / BY
tan10° = AB/ (12 km)
AB = 12 tan10° km
AB = 2.16 km
The height of the aeroplane A above B = 2.16km


(c)
In rt.Ð ΔBXY :
cosÐBXY = BX / XY
cos30° =BX / (24 km)
BX = 24 cos 30° km

In rt.Ð ΔABX :
tanÐAXB = AB / BX
tanÐAXB = (12 tan10°) / (24 cos30°)
ÐAXB = 5.81°
The angle of elevation of the aeroplane A from X = 5.81°