Question

a man standing on a tower observes the stone on the ground at an angle of depression of 45°,which is 25m away from the foot of the tower, find the height of the tower ​

Answer 1

Given :-

• A man is standing on a tower observes the stone on the ground at an angle of depression of 45°

• The distance between stone and tower = 25m

Solution :-

According to the question,

Angle of depression ΔBCA = 45°

The distance between tower and stone

( base AC ) = 25m

Let the height AB be h

Now,

Tan45° = AB / AC

Put the required values,

1 = h / 25

25 * 1 = h

h = 25

Hence, The height of the tower is 25 m

[ Note :- Refer the attachment $.$ ]

Explore more :-

• Sin 0° = 0

• Sin 30° = 1/2

• Sin 45° = 1/√2

• Sin 60° = √3/2

• Sin 90° = 1

• Cos 0° = 1

• Cos 30° = √3/2

• Cos 45° = 1/√2

• Cos 60° = 1/2

• Cos 90° = 0

• Tan 0° = 0

• Tan 30° = 1/√3

• Tan 45° = 1

• Tan 60° = √3

• Tan 90° = Not defined

• Cosec 0° = Not defined

• Cosec 30° = 2

• Cosec 45° = √2

• Cosec 60° = 2/√3

• Cosec 90° = 1

• Sec 0° = 1

• Sec 30° = 2/√3

• Sec 45° = √2

• Sec 60° = 2

• Sec 90° = Not defined

• Cot 0° = Not defined

• Cot 30° = √3

• Cot 45° = 1

• Cot 60° = 1/√3

• Cot 90° = 0

Answer 2

Given :-

• Angle of depression (∠CBD) = 45°
• Distance between tower and stone (AC) = 25 m

To Find :-

• Height of the tower (AB) = ?

Concept Implemented :-

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1& \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}$

Diagram :-

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(0,0)(0,5)\qbezier(0,5)(0,5)(5,0)\qbezier(5,0)(5,0)(0,0)\put(.6,4.6){ {\bf{45^\circ}} } \put(2,-.4){\bf{25 m}}\qbezier(0,5)(0,5)(4.5,5)\put(3.9,0.2){ {\bf{45^\circ}} } \put(4.9,0.1){\circle*{.3}}\put(0,5.1){\sf{Man}}\put(4.9,-.4){\bf{C}}\put(-.1,-.4){\bf{A}}\put(-.3,4.8){\bf{B}}\put(5.2,0){\sf{Stone}}\put(4.6,4.8){\bf{D}}\end{picture}$

❍ NOTE : If diagram isn't visible then refer to the attached image.

Solution :-

In ∆ABC

$\sf{tan\:C = \dfrac{AB}{AC}}$

$\sf{\implies tan\:45^\circ = \dfrac{AB}{25\, m}}$

$\sf{\implies 1 = \dfrac{AB}{25\,m}\; \; \; \bigg\lgroup{Putting \; the \; value \; of\; tan\,45^\circ = 1} \bigg\rgroup}$

$\sf{\implies 1 \times 25\, m = AB}$

$\sf{\implies 25\, m = AB}$

$\sf{\implies AB = 25\,m}$

$\sf{\therefore AB = 25\,m}$

∴ The height of the tower is 25 m.