Two pillars of equal heights are at the point A and B on the opposite side of the road which is 120 m wide. From a point C on the line joining the foots of the pillars, the angle of elevation of the top of the pillar at A and B are 60° and 30° respectively. Find AC.
WBBSE
Class X
Mathematics
Application of Trigonometric Ratios
5
2024
Let the common height of the two pillars be h.
Let AC = x. Then CB = 120−x.
In Δ ACD
tan 30° = h\over x
⇒ h = x tan 30° = x\over √3
In Δ BCE
tan 60° = h\over 120 - x
⇒ √3 = {x\over √3}\over 120 - x
⇒ √3 = x\over √3(120 - x)
⇒ √3 × √3(120 – x) = x
⇒ 3 (120 – x) = x
⇒ 360 – 3x = x
⇒ 360 = 4x
⇒ x = 360\over 4 = 90 m
∴ AC = 90 m
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