Question

the angle of elevation of the top of a vertical tower from a point on the ground is 60 degrees. from another point 10m vertically above the first, its angle of elevation is 45 degrees. find the height of the tower

Answer 1

Answer:

23.66 m

Step-by-step explanation:

Refer the attached figure  

Let the point on the ground be B

There i another point A vertically 10 m above the point B

So, AB = 10 m

Height of the tower is EC = ED+DC=ED+10

The angle of elevation of the top of a vertical tower from a point on the ground is 60 degrees i.e.∠EAD = 45°

From another point 10 m vertically above the first, its angle of elevation is 45 degrees.i.e.∠EBD = 60°

Let ED be x

In ΔAED

Using trigonometric ratio

$Tan\theta = \frac{Perpendicular}{Base}$

$Tan45^{\circ} = \frac{ED}{AD}$

$AD= ED =x$ -1

In ΔEBC

Using trigonometric ratio

$Tan\theta = \frac{Perpendicular}{Base}$

$Tan 60^{\circ} = \frac{EC}{BC}$

$Tan 60^{\circ} = \frac{x+10}{BC}$

$BC = \frac{x+10}{\sqrt{3}}$  ---2

Since BC = AD

So, equate 1 and 2

$x= \frac{x+10}{\sqrt{3}}$

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

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

$10=0.732050x$

$\frac{10}{0.732050}=x$

$13.66=x$

Substitute the value of x in 2

$EC=10+13.66$

$EC=23.66$

Hence the height of the tower is 23.66 m

Answer 2

Answer:

$Height \:of \:the \:tower =23.660\:m$

Step-by-step explanation:

Given ,the angle of elevation of the top of a vertical tower from a point on the ground is 60 degrees.

From the figure,

Angle of elevation <ACB = 60°,

Distance from the point on the ground to base of the tower = BC ,

<ADE = 45°,

DC = BE = 10m,

Let AE = x m,

Height of the tower = AB m

$i ) In \triangle ACB ,\\tan\angle ACB = \frac{AB}{BC}$

$\implies tan 60\degree= \frac{x+10}{BC}$

$\implies \sqrt{3}=\frac{x+10}{BC}$

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

$ii) In \triangle ADE ,\\tan\angle ADE = \frac{AE}{DE}$

$\implies tan 45\degree= \frac{x}{BC}$

$\implies 1 = \frac{x}{BC}$

$\implies BC = x ---(2)$

/* From (1) & (2) */

$x = \frac{x+10}{\sqrt{3}}$

$\implies \sqrt{3}x = x + 10$

$\implies \sqrt{3}x - x =10$

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

$\implies x = \frac{10}{\sqrt{3}-1}\\=\frac{10(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}\\=\frac{10(\sqrt{3}+1)}{(3-1)}\\=\frac{10(\sqrt{3}+1)}{2}\\=5(\sqrt{3}+1)\\=5(1.732+1)\\=5\times 2.732\\=13.660$

$Height \:of \:the \:tower(AB)\\=AE+EB\\=x+10\\=13.660+10\\=23.660\:m$

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