Question 1112321: A tower T is observed from two points A and B, which are 240 metres apart. Angle TAB and angle TBA are found to be 57° and 78° respectively. Find the distance of the tower from A
Found 3 solutions by mananth, ikleyn, josgarithmetic:
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website!
tan 78 = TG/(240-x)
tan 78(240-x) =TG
tan 57= TG/x
x *tan 57 = TG
tan 78(240-x)=x *tan 57
4.70(240-x) = 1.54x
1128 -4.70x=1.54x
6.24x=1128
x=180.76
Height of tower =180.77*tan 57=282.28 m
x=
Answer by ikleyn(52898)  (Show Source):
You can put this solution on YOUR website! .
I do not know why the tutor @mananth talks about the height of the tower.
Nobody asked in this problems about the height of the tower.
And the problem is ABSOLUTELY IRRELEVANT to the tower's height.
It is about totally different thing.
The tutor @mananth misread and misinterpreted the problem, and his writing is I R R E L E V A N T.
Answer by josgarithmetic(39630)  (Show Source):
You can put this solution on YOUR website! Description may be not completely precise.
A, B, T, for a triangle. Imagine A and B are on opposite sides of whatever platform the tower sits. Tower is at T on the top of this platform.
AB given as 240 meters.
Angle TAB, 57 degree
Angle TBA, 78 degree
Sum of interior angles is 180 degrees, so
Angle ATB is 45 degrees.
Question asks for length TA.
Law Of Sines:
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