A tower subtends angle A, 2A and 3A at three points A, B, C repectively lying on a horizontal line through the foot of the tower. Find the ratio of AB/BC

consider XY the tower by tanA.in triangle XYA we can find the value of AY .then in triangle XYB we can get the value of AB. in triangle XYC we can the value of bc then find the ratio

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consider XY the tower by tanA.in triangle XYA we can find the value of AY .then in triangle XYB we can get the value of AB. in triangle XYC we can the value of bc then find the ratio

  • -9
Let the height of tower be 'h' and tower be ED  (E at top and D at bottom)
Then,


A+BEA=2A     (EXTERIOR ANGLE PROPERTY)
BEA=A
Similarly CEB= A

So EB is angle bisector of CEA  

Therefore, EA/EC = AB/BC  1

SinA = h/EA   and   Sin3A = h/EC

Sin3A/SinA = h/EC * EA/h = EA/EC  2

From 1 and 2 

Sin3A/SinA = AB/BC
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Etawah substance and angle alpha 2 alpha 3 Alpha at ABC all lying on the horizontal line through the foot of the tower then AB/BC
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