The angle of elevation of the top of the tower from point A and B respectively is 30 and 45 degree. If it is known that point B lies between point A and the base of the tower on the straight line joining point A and the base of the tower, what is the distance between point A and B?(height =h)
Answers
Given :-
The angle of elevation of the top of the tower from point A and B respectively is 30 and 45 degree. If it is known that point B lies between point A and the base of the tower on the straight line joining point A and the base of the tower,
To Find :-
Distance between point A and B
Solution :-
We need to take height as h
So,
We have
\begin{gathered}\begin{cases} \sf Height =h \\ \sf AO = \frak{30}\\ \sf BO = 45 \end{cases}\end{gathered}⎩⎪⎪⎨⎪⎪⎧Height=hAO=30BO=45
Now,
\sf \tan(45)^{\circ} = \dfrac{OC}{BC}tan(45)∘=BCOC
Tan 45⁰ = 1
\sf \: 1 = \dfrac{h}{x}1=xh
\sf \: 1 \times x = h1×x=h
\sf \: x = h...(1)x=h...(1)
Again
In ∆ACO
Since,
ACO is a right angled triangle so,
\sf \angle A = 30^{\circ}∠A=30∘
\sf \angle C = 90^{\circ}∠C=90∘
\sf \angle O = 60^{\circ}∠O=60∘
Again
\sf \tan({30}^{ \circ} ) = \dfrac{h}{x + h}tan(30∘)=x+hh
tan 30⁰ = 1/√3
\sf \: \dfrac{1}{ \sqrt{3} } = \dfrac{h}{h + x}31=h+xh
\sf \: 1(h + x) = \sqrt{3} (h)1(h+x)=3(h)
\sf \: h + x = \sqrt{3}hh+x=3h
\sf \: x = \sqrt{3} - 1(h)x=3−1(h)
\sf \: x \: = \sqrt{3} - 1hx=3−1h
Or
\bf \: x \: = 0.73 \: \: mx=0.73m
Hope tie helps u
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