If the shadow of a tower when suns altitude is 45

The height of a tower is 10 m. What is the length of its shadow when Sun's altitude is 45°?

Let BC be the length of shadow is x m 

Given that: Height of tower is 10 meters and altitude of sun is 45°

Here we have to find length of shadow.

So we use trigonometric ratios. 

In a triangle ABC,

`⇒ tan = (AB)/(BC)`   

`⇒ tan 45°=(AB)/(AC)` 

`⇒1=10/x`

`⇒x=10`

Hence the length of shadow is 10 m.

Concept: Heights and Distances

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If the ratio of the height of a tower and the length of its shadow is `sqrt3:1`, what is the angle of elevation of the Sun?

Let C be the angle of elevation of sun is θ. 

Given that: Height of tower is `sqrt3` meters and length of shadow is 1.

Here we have to find angle of elevation of sun.

In a triangle ABC, 

`⇒ tanθ =(AB)/(BC)` 

`⇒ tan θ=sqrt3/1`        ` [∵ tan 60°=sqrt3]`

`⇒ tan θ=sqrt3`

`⇒ θ=60 °`

Hence the angle of elevation of sun is 60°.

Concept: Heights and Distances

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The shadow of a tower, when the angle of elevation of the sun is 45∘, is found to be 10 m. longer than when it was 60∘. Find the height of the tower.

Solution

The given information can be represented as,

Here, AB is the tower. BD is the shadow when $\angle ADB={45}^{\circ }$, CB is the shadow when $\angle ACB={60}^{\circ }$. By the given condition, CD = 10 m.

To find: Length of AB

In right triangle ADB, $\tan {45}^{\circ }=\frac{\text{Opposite side}}{\text{Adjacent side}}$

$\therefore 1=\frac{AB}{BD}$ [$\tan {45}^{\circ }=1$]

$\Rightarrow 1=\frac{AB}{BC+CD}$

$AB=BC+10\dots \dots \left(i\right)$

In right $△ACB,\tan {60}^{\circ }=\frac{\text{Opposite side}}{\text{Adjacent side}}$

$\sqrt{3}=\frac{AB}{BC}$ [$\tan {60}^{\circ }=\sqrt{3}$]

$AB=\sqrt{3}BC\dots \dots \left(ii\right)$

From $\left(i\right)$ and $\left(ii\right),$ we have

$BC+10=\sqrt{3}BC$

$\sqrt{3}BC-BC=10$

$BC(\sqrt{3}-1)=10$

$BC=\frac{10}{(\sqrt{3}-1)}=\frac{10}{(\sqrt{3}-1)}\times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}$

$=\frac{10(\sqrt{3}+1)}{3-1}=5(\sqrt{3}+1)m$

$\therefore AB=BC+10=\left[5\right(\sqrt{3}+1)+10]m=5\left[\right(\sqrt{3}+3\left)\right]m$

Thus, the height of the tower is $5\left[\right(\sqrt{3}+3\left)\right]=5[1.73+3]=5\times 4.73=23.65m$

Mathematics

RD Sharma

Standard X