Answer
Ans. $h=d(tanβ-tanα)$
Work Step by Step
Given:-
In ∆ABD
$tanα=BD/AB$
$tanα=x/d$
$x=dtanα $ eq. ~1
In∆ABC
$tanβ=BC/AD$
$tanβ=\frac{x+h}{d}$
$dtanβ=dtanα+h$
$h=d(tanβ-tanα)$
Ans.
Calculus, 10th Edition (Anton)
by Anton, Howard
Published by
Wiley
ISBN 10:
0-47064-772-8
ISBN 13:
978-0-47064-772-1
Appendix B - Trigonometry Review - Exercise Set B - Page A25: 43
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