Asked by somay
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.
Answers
Answered by
Damon
tan 30 = h/a
tan T = h/2a
tan T = (1/2) tan 30
T = 8.21 degrees
tan T = h/2a
tan T = (1/2) tan 30
T = 8.21 degrees
Answered by
somay
note: AB=2a m, AO=a m,OB=? And angle OAB=60 degree
Answered by
Reiny
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
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
Answered by
somay
that's great Reiny, you are perfectly correct. Thank you somuch
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