Question

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)​

Answer 1

Given

➝The Angle of the elevation of the top of the tower from point A and B is 30⁰ and 45⁰

To Find

➝Distance Between Point A and B

According to Question

➝We form a  ΔACD

➝DC = h , BC = x , AB =x' and AC = x'+x

Now Take  ΔBCD , where

➝∠B = 45⁰ and ∠C = 90⁰

Now Using

➝TanB = Perpendicular(p)/Base(b)

➝Tan45⁰ = DC/BC [ Tan45⁰ = 1]

➝1 = h/x

➝h = x (i)

Now Take  ΔACD ,where

➝∠A = 30⁰ and ∠C = 90⁰

Now Using

➝TanA = Perpendicular(p)/Base(b)

➝Tan30⁰ = h/(x'+x) [Tan30⁰ = 1/√3]

➝1/√3 = h/(x' +h) [Putting x = h from (i)eq]

➝x' + h = (√3)h

➝x' = (√3)h - h

➝x' = (√3-1)h

Answer

➝Distance Between A and B is (√3-1)h Units

Answer 2

Answer:

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

$\begin{cases} \sf Height =h \\ \sf AO = \frak{30}\\ \sf BO = 45 \end{cases}$

Now,

$\sf \tan(45)^{\circ} = \dfrac{OC}{BC}$

• Tan 45⁰ = 1

$\sf \: 1 = \dfrac{h}{x}$

$\sf \: 1 \times x = h$

$\sf \: x = h...(1)$

Again

In ∆ACO

Since,

ACO is a right angled triangle so,

$\sf \angle A = 30^{\circ}$

$\sf \angle C = 90^{\circ}$

$\sf \angle O = 60^{\circ}$

Again

$\sf \tan({30}^{ \circ} ) = \dfrac{h}{x + h}$

• tan 30⁰ = 1/√3

$\sf \: \dfrac{1}{ \sqrt{3} } = \dfrac{h}{h + x}$

$\sf \: 1(h + x) = \sqrt{3} (h)$

$\sf \: h + x = \sqrt{3}h$

$\sf \: x = \sqrt{3} - 1(h)$

$\sf \: x \: = \sqrt{3} - 1h$

Or

$\bf \: x \: = 0.73 \: \: m$