Question

$\mathfrak {\text{MATHS - CLASS X}}$

$\mathcal{\text{SOME APPLICATIONS OF TRIGONOMETRY}}$

The angles of depression of the top and the bottom of an 8m tall building from the top of a multi-storey building are 30° and 45° respectively. Find the height of the multi-storeyed building and the distance between the two buildings.

$\mathbb{\text{NO SPAMMING}}$

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Answer 1

The angles of depression of the top and the bottom of an 8m tall building from the top of a multi-storey building are 30° and 45° respectively. Find the height of the multi-storeyed building and the distance between the two buildings.

I have created a diagram for this case.

Here, Red line indicates the building.
Orange line indicates the multi storeyed building.

SOLUTION :

Given that, Angle of depression of top of build = α = 30°

Angle of depression of bottom of the building = β = 45°

And height of the building = 8m

Now,

DF || EA,
So Angle DAE = FDA = 30°

Also, DF || BC
Angle FDB = DBC = 45°

Now,
AB = EC = 8m .

Let the height of the multi storeyed building = x
and distance between the two is p.

So ,
DE = x - 8 .

In ΔDBC,
tan45 = DC / CB
x / p = 1
x = p --------> equation 1 .

In ΔDAE,
tan30 = DE/ AE
1/√3 = x - 8 / p
p = √3 ( x - 8 )
x = √3x - 8√3
8√3 = √3x - x

$x = \frac{8 \sqrt{3} }{ \sqrt{3} - 1}$

Substituting √3 = 1.73 we get

x = 8( 1.73 ) / 1.73 - 1

x = 13.84 / 0.73

x = 18.95m approximately.

Therefore, The height of the multistoreyed building is 18.95m and distance between the two buildings is also 18.95m .

HOPE THIS HELPS!

Answer 2

$\bold{\large{\underline{Solution}}}$

$\bold{\large{\underline{Given\: \colon}}}$

The angles of depression of the top and the bottom of an 8m tall building from the top of a multi-storey building are 30° and 45° respectively.

$\bold{\large{\underline{To \: Find\: \colon}}}$

The height of multistorey building = " h " m

The distance between multistorey building and small building = " x " m

$\bold{\large{\underline{Solution\: \colon}}}$

Consider all the statement from the diagram in above statement

According to diagram :

The height of multi - storey building = AB = " h " m

The height of small building = CD = 8 m

DC = EB = 8 m

AB = AE + EB

h = AE + 8

AE = ( h - 8 ) m

Now In ∆ DAE

$\bold{\large{ tan\: 30^{\degree}\: = \: \dfrac{AE}{DE} }}$

$\bold{\large{ \dfrac{1}{\sqrt{3}}\: =\: \dfrac{AE}{DE}}}$

$\bold{\large{ \dfrac{1}{\sqrt{3}}\: =\: \dfrac{(h\: - \: 8)}{DE}}}$

$\bold{\large{ DE \: = \: \sqrt{3}(h\: - \: 8) }}$

$\bold{\large{ x \: = \: \sqrt{3}(h\: - \: 8) }}$

we get equation (1)

In ∆ ABC

$\bold{\large{ tan\: 45^{\degree} \: = \: \dfrac{P}{B} \: = \dfrac{AB}{CB}}}$

$\bold{\large{ 1 \: =\: \dfrac{h}{CB} }}$

we have DE = CB

$\bold{\large{ 1\: = \: \dfrac{h}{x} }}$

$\bold{\large{ x \: = \: h }}$

we get equation (2)

Now equate the equation 1 & 2

h = √3( h - 8 ) m

h = √3h - 8√3

√3h - h = 8√3

h( √3 - 1 ) = 8√3

$\bold{\large{ h \: = \: \dfrac{8\sqrt{3}}{(\sqrt{3}\: - \: 1})}}$

Rationalising the denominator and numerator

$\bold{\large{ h \: = \: \dfrac{8\sqrt{3}}{(\sqrt{3}\: - \: 1})\:\times\: \dfrac{(\sqrt{3}\: + \: 1)}{(\sqrt{3}\: + \: 1)}}}$

$\bold{\large{ h \: = \: \dfrac{8\sqrt{3}(\sqrt{3}\: -\: 1)}{[(\sqrt{3})^{2}\: - \: 1^{2}]}}}$

$\bold{\large{ h \: = \: \dfrac{8\sqrt{3}(\sqrt{3}\: -\: 1)}{[3\: - \: 1]}}}$

$\bold{\large{ h \: = \: \dfrac{8\sqrt{3}(\sqrt{3}\: +\: 1)}{[2]}}}$

$\bold{\large{ h \: = \: 4\sqrt{3}(\sqrt{3}\: +\: 1)}}$

$\bold{\large{ h \: = \: 4(3\: + \: \sqrt{3})}}$

AB = 4( 3 + 1.732)

AB = 4(4.732)

AB = 18.928 m

The height of multistorey building is 18.928 m

and according to equation second x = h = 18.928 m

$\bold{\large{\underline{Answer}}}$

$\bold{\large{\therefore\: Height\: of \: multistorey \: building\: = \: 18.928 \: m }}$

$\bold{\large{\therefore \:Distance \: between \: building \: = 18.928 \: m }}$

$\bold{\large{Thanks}}$