Guys, pls help me? trigonometry?

Question

A man stands on a scaffold at B, which is directly above D. Object A lies on the ground, due north of D. The angle of depression of object A from the man is 14 degrees. the man walks 140m due east and the angle of depression of A changes to 10 degrees. Find the height of the scaffold.

Answer 1

Consider first the triangle formed by the lines B-A and B-C (C is the point 140m due east of A). That triangle's angle at B is 4º (14º - 10º), and the side opposite that is 140m. You can find the length of line B-A from the law of sines if you know the angle opposite that segment. That angle is 10º, since the due east line A-C is parallel to the line against which the depression angle is measured (horizontal). Therefore the lenght of the B-A segment is given by

140/sin(4º) = B-A/sin(10º)

B-A = 140*sin(10º)/sin(4º) = 348.5m

The B-A line is the hypotenuse of right triangle B-D and D-A. B-D is the height; the angle opposite B-D is 14º therefore sin(14º) = B-D/B-A

B-D = 348.5*sin(14º) = 84.3m

Answer 2

A 3D diagram is required:-
Draw right angled triangle BDA right angle at D:-
B is above D
A is to right of D
BD = h, the height of the scaffold.
Horizontal line DA = x m
Angle DAB = 14°
Draw right angled triangle DAC where C is directly below A and AC = 140m
Draw right angled triangle BDC by joining C to B.
BDC is right angle.
Now have a 3 dimensional drawing consisting of triangles BDA , DAC and DBC.
In triangle BDA:-
tan 14° = h / DA
DA = h / tan 14°
In triangle DAC:-
DC² = DA² + AC²
DC² = h² / (tan 14°)² + 140²
DC² = 16.1 h² + 19600
DC = √(16.1 h² + 19600)
In triangle BDC:-
tan 10° = h / √(16.1h² + 19600)
0.18.(√(16.1h² + 19600)) = h
0.03.(16.1 h² + 19600) = h²
0.5 h² + 588 = h²
0.5 h² = 588
h = 34.3
Height of scaffold = 34.3 m

Answer 3

Draw the picture.

Three right triangles. One is on the ground.

Use trigonometric relations and the Pythagorean formula.

Answer 4

Yucky