[tex]\sf\huge\underline{\star Solution:-}[/tex]
[tex]\rightarrow[/tex]Let the women's height be AE and distance between the women and tower be AC.
Also let the height of tower be BC.
[tex]\rightarrow[/tex]Now, clearly it is forming a triangle.
So, in triangle ABC,
[tex]\rightarrow[/tex][tex]\sf{Tan50° \:= \: \frac{BC}{AC}}[/tex]
[tex]\rightarrow[/tex][tex]\sf{1.19 \:= \: \frac{BC}{50}}[/tex]
[tex]\rightarrow[/tex][tex]\sf{1.19 ×50\:= \: BC}[/tex]
[tex]\rightarrow[/tex][tex]\sf{59.5m\:= \: BC}[/tex]
[tex]\rightarrow[/tex]Hence, BC = 59.5m
So, BD(total height of the tower)[tex]\sf{ = \:BC+CD}[/tex]
[tex]\sf{= \:59.5+1.65}[/tex]
[tex]\sf{=\: 61.15m}[/tex]
Therefore, total height of the tower =[tex]\sf\purple{ 61.15m.}[/tex]
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Hope it helps you:)
