Question

A ladder 7.5 m long is resting against a wall. If the ladder makes an angle of 60°
with the ground, find
(i) the height where the ladder is resting on the wall.
(i) The distance between the feet of the ladder and the wall.​

Answer 1

Answer:

Let AC be a ladder, AB is the wall and BC is the ground as shown in the figure.

We know that

cos60  

o

=  

AC

BC

​  

 

2

1

​  

=  

AC

2.5

​  

 

AC=5 m

Therefore, the length of ladder is 5 m.

it may helped you..

Answer 2

ANSWER:

Given:

• Length of ladder = 7.5m
• Angle made by ladder with ground = 60°

To Find:

• Height where ladder is resting
• Distance between feet of ladder and wall

Solution:

(Refer attachment)

We can see that in the attachment,

⇒ AC = Length of Ladder = 7.5m,

⇒ AB = Distance between foot of Ladder and wall

⇒ BC = Height of wall,

⇒ Angle CAB = 60°

So,

We know that,

$\implies\sf\sin\theta=\dfrac{Perpendicular}{Hypotenuse}$

So,

$\implies\sf\sin60^{\circ}=\dfrac{BC}{AC}$

Now, as sin60° = √3/2,

$\implies\sf\dfrac{\sqrt3}{2}=\dfrac{BC}{7.5}$

So,

$\implies\sf BC=\dfrac{7.5\sqrt3}{2}$

Hence,

$\implies\sf BC=\dfrac{7.5\times1.73}{2}\approx\dfrac{12.975}{2}$

Therefore,

⇒ BC = 6.4875 ≈ 6.5m

Hence, the height where the ladder is resting on wall is 6.5m.

Now,

We know that,

$\implies\sf\cos\theta=\dfrac{Base}{Hypotenuse}$

So,

$\implies\sf\cos60^{\circ}=\dfrac{AB}{AC}$

Now, as cos60° = 1/2,

$\implies\sf\dfrac{1}{2}=\dfrac{AB}{7.5}$

So,

$\implies\sf AB=\dfrac{7.5}{2}$

Hence,

Therefore,

⇒ AB = 3.75m

Hence, the distance between the foot of ladder and the wall is 3.75m.