Question

i will mark as brainlist

From the top of a building 100 m high, the angles of depression of the top and bottom of a tower are observed to be 45° and 60° respectively. Find the height of the tower. Also
find the distance between the foot of the building and bottom of the tower.

pls ans fast
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Answer 1

ANSWER:-

Given:

From the top of a building 100m high, the angles of depression of the top and bottom of a tower are observed to be 45° & 60° respectively.

To find:

Find the height of the tower. Also find the distance between the foot of the building & bottom of the tower.

Solution:

Above attachment a figure.

Let CD be the building = 100m

Let AB be the tower of height h m.

⚫∠EAC = 45°

⚫∠DBC= 60°

Let BD= AE= x

Therefore,

In right ∆AEC,

$tan45 \degree = \frac{CE}{AE} \\ \\ = > 1 = \frac{100 - h}{x} \\ \\ = > x = 100 - h.........(1)$

And,

In right ∆BDC,

$tan60 \degree = \frac{CD}{BD} \\ \\ = > \sqrt{3} = \frac{100}{x} \\ \\ = > x = \frac{100}{ \sqrt{3} } ...............(2)$

So,

Comparing equation (1) & (2),we get;

$100 - h = \frac{100}{ \sqrt{3} } \\ \\ = > \sqrt{3} (100 - h) = 100 \\ \\ = > 100 \sqrt{3} - \sqrt{3h} = 100 \\ \\ = > \sqrt{3h} = 100 \sqrt{3} - 100 \\ \\ = > h = \frac{100 \sqrt{3} - 100}{ \sqrt{3} } \\ \\ = > h = \frac{100( \sqrt{3} - 1)}{ \sqrt{3} } \\ \\ = > h = \frac{100(1.732 - 1)}{1.732} \\ \\ = > h = \frac{100 \times 0.732}{1.732} \\ \\ = >h = \frac{73.2}{1.732} = \frac{73.2 \times 1000}{1.732 \times 1000} \\ \\ = > h = \frac{73200}{1732} \\ \\ = > h = 42.26m$

Hence,

Height of the tower(AB) is 42.26m

Answer 2

To find

Height of the tower and distance between the foot of the tower and building

Method

AB=building of height 100m

CD=height of tower h m

BC=distance between foot of the building and the tower

$\bullet \bigtriangleup ABC$

$it \: is \: a \: right \: angled \: triangle$

$tan60\degree=\dfrac{AB}{BC}$

$\sqrt{3}=\dfrac{100}{a}$

$\huge\boxed {a=\dfrac{100}{\sqrt{3}}}$

$\bullet from \bigtriangleup AED$

$tan45\degree=\dfrac{AE}{ED}$

$1=\dfrac{100-h}{\dfrac{100}{\sqrt{3}}}$

since AE=XD and ED=BC

$100=100\sqrt{3}-h\sqrt{3}$

$-h\sqrt{3}=-100(\sqrt{3}-1)$

$h=\dfrac{100(0.7)}{1.7}=\dfrac{700}{17}$

$\huge\boxed {h=42}$

Therefore

distance between the foots =100/√3 m

height of the tower =42 m