Question

A ladder rests against a vertical wall such thay the top of the ladder reach the top of the wall the ladeer is inclined at 60degeree with the ground and a bottom of ladder is 15m away from the foot of the wall find (a) length of the ladder. (b) find the height of the wall​

Answer 1

Given :

• Angle of inclination = 60°

• Distance from the foot of the wall = 15 m.

To find :

• The height of the wall.

• The length of the ladder.

Solution :

To find the height of the wall :

Let the height of the wall be x m.

According to the diagram, AB is the height of the triangle and CB is the base of the triangle.

And we know that :-

$\bf{\tan\:\theta = \dfrac{P}{B}}$

Where :-

• P = Height
• B = Base

So using tan θ and substituting in it , we get :

$:\implies \bf{\tan\:\theta = \dfrac{P}{B}}$

$:\implies \bf{\tan\:60^{\circ} = \dfrac{x}{15}}$

$:\implies \bf{\sqrt{3} = \dfrac{x}{15}}\quad[\because \bf{\tan\:60^{\circ} = \sqrt{3}}]$

$:\implies \bf{15\sqrt{3} = x}$

$\boxed{\therefore \bf{Height\:(x) = 15\sqrt{3}\:m}}$

Hence the height of the wall is 15√3 m.

To find the length of the ladder :

Method (i) !!

Given :-

• Height (P) = 15√3 m

• Base (B) = 15 m

Let the Hypotenuse be h m.

By using the Pythagoras theorem and substituting the values in it , we get :

$\underline{:\implies \bf{H^{2} = P^{2} + B^{2}}}$

Where :-

• H = Hypotenuse
• B = Base
• H = Height

$:\implies \bf{h^{2} = (15\sqrt{3})^{2} + 15^{2}}$

$:\implies \bf{h^{2} = 675 + 225}$

$:\implies \bf{h^{2} = 900}$

$:\implies \bf{h = \sqrt{900}}$

$:\implies \bf{h = 30}$

$\boxed{\therefore \bf{Hypotenuse\:(h) = 30\:m}}$

Hence the length of the ladder is 30 m.

Method (ii) !!

According to the diagram, CB is base of the triangle and AC is the Hypotenuse of the triangle.

And we know that :-

$\bf{\cos\:\theta = \dfrac{B}{H}}$

Where :-

• H = Hypotenuse
• B = Base

So using cos θ and substituting in it , we get :

$:\implies \bf{\cos\:\theta = \dfrac{B}{H}}$

$:\implies \bf{\cos\:60^{\circ} = \dfrac{15}{H}}$

$:\implies \bf{\dfrac{1}{2} = \dfrac{15}{H}}\quad[\because \bf{\cos\:60^{\circ} = \dfrac{1}{2}}]$

$:\implies \bf{\dfrac{1}{2 \times 15} = \dfrac{1}{H}}$

$:\implies \bf{\dfrac{1}{30} = \dfrac{1}{H}}$

$:\implies \bf{H = 30}$

$\boxed{\therefore \bf{Hypotenuse\:(h) = 30\:m}}$

Hence the length of the ladder is 30 m.