Ac and cb are parts of minar ac and cb makes angles alpha and beta at the point p if tan alpha=1/2 and tan beta =1/3 find height of minar of pb base is 16 meter
Answers
The height of the minar is 16 m.
Step-by-step explanation:
Referring to the figure attached below, let’s make some assumptions
AB = height of the minar of which AC and CB are the parts
PB = 16 m = the distance of point P from the base B of the minar
α = angle APC = angle made by AC at point P
β = angle CPB = angle made by CD at point P
Consider ∆CBP, by applying the trigonometry ratios of a triangle, we have
tan β = perpendicular/base
⇒ tan β = CB/PB
substituting the given values of tanβ = 1/3 & PB = 16
⇒ 1/3 = CB/16
⇒ CB = 16/3 m …….. (i)
Consider ∆ABP, by applying the trigonometry ratios of a triangle, we have
tan (α+β) = perpendicular/base
⇒ tan (α+β) = AB/PB
⇒ tan (α+β) = (AC + CB)/PB
using the formula, tan (α+β) = [tan α + tan β] / [1 – tanα tanβ]
⇒ [tan α + tan β] / [1 – tanα tanβ] = (AC + CB)/PB
substituting tanβ = 1/3 & tanα = ½, PB = 16 and value of CB from (i), we get
⇒ [½ + 1/3] / [1 – {(1/2) * (1/3)}] = {AC + (16/3)}/16
⇒ [5/6] / [1 – (1/6)] = [3AC + 16] / 48
⇒ [5/6] / [5/6] = [3AC + 16] / 48
⇒ 48 = 3AC + 16
⇒ 3AC = 32
⇒ AC = 32/3 m ……. (ii)
Thus, by substituting from (i) & (ii), we get
The height of the minar is,
= AB
= AC + CB
= (32/3) + (16/3)
= 48/3
= 16 m
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