Question

A ladder 15 metres long just reaches the top of a vertical
wall. If the ladder makes an angle of 60° with the wall, then the
height of the wall will be​

Answer 1

QUESTION :

A ladder 15 metres long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, then the height of the wall will be ?

SOLUTION :

It is given that ladder is 15 metres long and makes 60° with the wall.

In the diagram ( refer attachment ) :-

• AB = height of wall
• AC = ladder of 15m
• angle CAB = 60°

Let height of wall be x.

We know that :- cos t = base/ hypotenuse

therefore :- cos A = AB / AC

⟹cos A = AB / AC

⟹cos 60° = x / 15 [ AC = 15]

we know that Cos 60° = 1/2

⟹1/2 = x / 15

now cross multiply :-

⟹15= 2x

or

⟹2x = 15

⟹x = 15/2

Therefore height of wall = 15/2 m or 7.5 m

Answer : 15/2 m or 7.5 m

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LEARN MORE :-

Sin 30° = 1/2

cos 30° = √3/2

tan 30° = 1/√3

Sin 45° = 1/√2

cos 45° = 1/√2

tan 45° = 1

Sin 60° = √3/2

cos 60° = 1/2

tan 60° = √3

Sin 90° = 1

cos 90° = 0

tan 90° = infinite

cosec x = 1/ sin x

sec x = 1/ cosx

cot x = 1/tan x

____________________

Answer 2

Answer :

➥ The Height of the wall = 7.5 m

Given :

➤ Length of the ladder (AC) = 15 m

➤ Angle made by the ladder with the wall = 60°

To Find :

➤ Height of the wall (h) = ?

Required Solution :

Let ,

The height of the wall be "h"

In ∆ABC

$\tt{:\implies \dfrac{AB}{AC}=\dfrac{p}{h} = cos \theta}$

$\tt{:\implies \dfrac{h}{15} = cos \: {60}^{ \circ} }$

$\tt{:\implies \dfrac{h}{15} = \dfrac{1}{2} }$

$\tt{:\implies 15 \times 1 = 2 \times h }$

$\tt{:\implies 15 = 2h}$

$\tt{:\implies \dfrac{15}{2} = h }$

$\tt{:\implies 7.5 = h }$

$\bf{:\implies \underline{ \: \: \underline{ \red{ \: \: h = 7.5 \: m \: \: }} \: \: }}$

║Hence, the height of the wall is 7.5 m.║

$\:$

Additional Information :

$\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \rm Not \: De fined \\ \\ \rm cosec A & \rm Not \: De fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: De fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}$

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