OT is a vertical mast of height h. From A, a m due south of O, the angle of elevation of T is 30 degree. If B is a point 2a m from A, on a bearing of 060 degree, find the angle of elevation of T from B.

Answer 1

tan 30 = h/a

tan T = h/2a

tan T = (1/2) tan 30

T = 8.21 degrees

Answer 2

note: AB=2a m, AO=a m,OB=? And angle OAB=60 degree

Answer 3

So we are looking at a 3-D diagram?

Triangle AOT is a 30-60-90° triangle with sides in the ratio of 1:√3:2
so h:a:AT = 1:√3:2
h/1 = a/√3
h = a/√3
AT/2 = a/√3
AT = 2a/√3 -- may not need this

in triangle AB , the ground plane ...
we can use the cosine law to find OB
OB^2 = a^2 + 4a^2 - 2(a)(2a)cos60
= 5a^2 - 4a^2(1/2)
= 3a^2
OB = √3a

now in triangle BOT
tan (OBT) = (a/√3) / (√3a)
= 1/3
angle B)t = appr 18.4°

check my arithmetic

Answer 4

that's great Reiny, you are perfectly correct. Thank you somuch