Question

6) The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From
another point 10 m vertically above the first, its angle of elevation is 45°. Find the height of the
tower.
<​

Answer 1

Answer:

What would you like to ask?

10th

Maths

Some Applications of Trigonometry

Heights and Distances

The angle of elevation of t...

MATHS

The angle of elevation of the top of a vertical tower from a point on the ground is 60

0

. from another point 10 m vertically above the first, its angle of elevation is45

0

.Find the height of the tower.

MEDIUM

Help best friend

Study later

ANSWER

R.E.F. Image.

Height of the tower, H = (h+10)m

In △ABC,

tan60

∘

=

AB

h+10

AB=

3

h+10

⋯(i)

In △DCO,

tan45

∘

=

AB

h

∴AB=h⋯(ii)

comparing eq.(i) and (ii), we get :

h=

3

h+10

∴

3

h=h+10

⇒

3

h−h=10

⇒h(

3

−1)=10

∴h=

3

−1

10

Height of tower = h + 10

=

3

−1

10

+10

=

3

−1

10+10

3

−10

H=

3

−1

10

3

m

Answer 2

Step-by-step explanation:

Let the height be EC

Let the the angle EAD be 45° and the angle EBC be 60°

Let EB be x m and in reactangle ABCD we know that opposite sides are equal: AB = CD = 10 m and BC = AD

Now,

$\sf In \: \Delta \: AED,$

$:\implies \sf tan \: 45^{\circ} = \dfrac{ED}{AD}$

$:\implies \sf tan \: 45^{\circ} = \dfrac{ED}{BC}$

$:\implies \sf tan \: 45^{\circ} = \dfrac{x}{BC}$

$:\implies \sf 1= \dfrac{x}{BC}$

$:\implies \sf x = BC \:\: \: \: \:\:\:\Bigg\lgroup \bf Equation\:(i)\Bigg\rgroup$

_______________________

$\sf In \: \Delta \: EBC,$

$:\implies \sf tan \: 60^{\circ} = \dfrac{EC}{BC}$

$:\implies \sf \sqrt{3} = \dfrac{x + 10}{BC}$

$:\implies \sf BC (\sqrt{3} ) = x + 10 \:\: \: \: \:\:\:\Bigg\lgroup \bf taking\:BC \: towards \:LHS \Bigg\rgroup$

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

$:\implies \sf x \: \left(\sqrt{3} - 1 \right) = 10$

$:\implies \sf x = \dfrac{10}{( \sqrt{3} - 1) } \:\: \: \: \:\:\:\Bigg\lgroup \bf Equation\:(ii)\Bigg\rgroup$

_______________________

★ Height of the tower will be :

$\dashrightarrow\:\: \sf Height= x+ 10$

$\dashrightarrow\:\: \sf Height= \frac{10}{ \sqrt{3 } - 1} + 10$

$\dashrightarrow\:\: \sf Height= \dfrac{10 \sqrt{3}}{ \sqrt{3} - 1 }$

$\dashrightarrow\:\: \sf Height=15 + 5 \sqrt{3}$

$\dashrightarrow\:\: \sf Height=15 + 5 \times 1.73 \: \: \: \Bigg\lgroup \bf Putting \: \sqrt{3} = 1.73 \Bigg\rgroup$

$\dashrightarrow\:\: \sf Height=15 + 8.65$

$\dashrightarrow\:\: \underline{ \boxed{ \sf Height=23.65 \: m}}$

⠀

$\therefore \: \underline{\textsf{The height of the tower is \textbf{23.65 m}}}.$