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Due to short circuit, a fire has broken out in New Home Complex. Two buildings, namely X and Y have mainly been affected. The fire engine has arrived and it has been stationed at a point which is in between the two buildings. A ladder at point O is fixed in front of the fire engine.
The ladder inclined at an angle $60^\circ$ to the horizontal is leaning against the wall of the terrace (top) of the building Y. The foot of the ladder is kept fixed and after some time it is made to lean against the terrace (top) of the opposite building X at an angle of $45^\circ$ with the ground. Both the buildings along with the foot of the ladder, fixed at 'O' are in a straight line.
Based on the above given information, answer the following questions :
(i) Find the length of the ladder.
(ii) Find the distance of the building Y from point 'O', i.e. OA.
(iii) (a) Find the horizontal distance between the two buildings.
OR
(b) Find the height of the building X.
The ladder inclined at an angle $60^\circ$ to the horizontal is leaning against the wall of the terrace (top) of the building Y. The foot of the ladder is kept fixed and after some time it is made to lean against the terrace (top) of the opposite building X at an angle of $45^\circ$ with the ground. Both the buildings along with the foot of the ladder, fixed at 'O' are in a straight line.
Based on the above given information, answer the following questions :
(i) Find the length of the ladder.
(ii) Find the distance of the building Y from point 'O', i.e. OA.
(iii) (a) Find the horizontal distance between the two buildings.
OR
(b) Find the height of the building X.
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(i) In $\triangle OAP$,
$\frac{OP}{12\sqrt{3}} = \cosec 60^\circ = \frac{2}{\sqrt{3}}$
$\Rightarrow OP = 24$ m
$\therefore$ Length of ladder is $24$ m
(ii) In $\triangle OAP$,
$\frac{OA}{12\sqrt{3}} = \cot 60^\circ = \frac{1}{\sqrt{3}}$
$\Rightarrow OA = 12$ m
$\therefore$ the distance of the building Y from point O ie.,OA is $12$ m
(iii) (a) OP = OR = $24$ m
$\therefore$ In $\triangle OCR$,
$\frac{OC}{24} = \cos 45^\circ = \frac{1}{\sqrt{2}}$
$\Rightarrow OC = 12\sqrt{2}$ m
$\therefore$ distance between two buildings $=$ OA + OC
$= (12 + 12\sqrt{2})$ m or $12(1 + \sqrt{2})$ m
OR
(iii) (b) OP = OR = $24$ m
$\therefore$ In $\triangle OCR$,
$\frac{RC}{24} = \sin 45^\circ = \frac{1}{\sqrt{2}}$
$\Rightarrow RC = 12\sqrt{2}$ m
$\therefore$ height of building X is $12\sqrt{2}$ m
$\frac{OP}{12\sqrt{3}} = \cosec 60^\circ = \frac{2}{\sqrt{3}}$
$\Rightarrow OP = 24$ m
$\therefore$ Length of ladder is $24$ m
(ii) In $\triangle OAP$,
$\frac{OA}{12\sqrt{3}} = \cot 60^\circ = \frac{1}{\sqrt{3}}$
$\Rightarrow OA = 12$ m
$\therefore$ the distance of the building Y from point O ie.,OA is $12$ m
(iii) (a) OP = OR = $24$ m
$\therefore$ In $\triangle OCR$,
$\frac{OC}{24} = \cos 45^\circ = \frac{1}{\sqrt{2}}$
$\Rightarrow OC = 12\sqrt{2}$ m
$\therefore$ distance between two buildings $=$ OA + OC
$= (12 + 12\sqrt{2})$ m or $12(1 + \sqrt{2})$ m
OR
(iii) (b) OP = OR = $24$ m
$\therefore$ In $\triangle OCR$,
$\frac{RC}{24} = \sin 45^\circ = \frac{1}{\sqrt{2}}$
$\Rightarrow RC = 12\sqrt{2}$ m
$\therefore$ height of building X is $12\sqrt{2}$ m