$F r o m a w i n d o w A, 10 m a b o v e t h e g r o u n d t h e a n g l e o f e l e v a t i o n o f t h e t o p C o f a t o w e r i s x^{0}, w h e r e t a n x^{0} = 5 / 2 a n d t h e a n g l e o f d e p r e s s i o n o f t h e f o o t D o f t h e t o w e r i s y^{0}, w h e r e t a n y^{0} = \frac{1}{4} . C a l c u l a t e t h e h e i g h t C D o f t h e t o w e r i n m e t r e s .$

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Solution

$i n △ A E C, w e h a v e t a n x^{0} = \frac{C E}{A E} \Rightarrow \frac{5}{2} = \frac{C E}{A E} . . . . . (i) I n △ A B D, w e h a v e t a n y^{0} = \frac{A B}{B D} \Rightarrow \frac{1}{4} = \frac{10}{A E} [∵ B D = A E] A E = 40 S u b s t i t u t i n g t h e v a l u e o f A E i n (i) w e g e t \frac{5}{2} = \frac{C E}{40} \Rightarrow C E = \frac{5}{2} \times 40 = 100 ∴ C D = C E + D E = C E + A B [∵ B D = A E] C D = (100 + 10) m = 110 m$

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From a window A, 10 m above the ground the angle of elevation of the top C of a tower is x0, where tan x0=5/2 and the angle of depression of the foot D of the tower is y0,where tany0=14.Calculate the height CD of the tower in metres.

in △ AEC,we havetanx0=CEAE⇒52=CEAE.....(i)In △ ABD,we havetany0=ABBD⇒14=10AE [∵BD=AE]AE=40Substituting the value of AE in (i) we get52=CE40⇒CE=52×40=100∴CD=CE+DE=CE+AB [∵BD=AE]CD=(100+10)m=110m