Question

1. A ladder rests against a vertical wall such that the top of the ladder reaches the top of the wall. The
ladder is inclined at 60° with the ground, and the bottom of the ladder is 1.5m away
from the foot
of the wall. Find (i) the length of the ladder, and (ii) the height of the wall.​

Answer 1

$\large \bf \clubs \:  Given :-$

• A ladder rests against a vertical wall such that the top of the ladder reaches the top of the wall.

• The ladder is inclined at 60° with the ground, and the bottom of the ladder is 1.5m awayfrom the foot of the wall.

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$\large \bf \clubs \:  To \: Find :-$

• the length of the ladder.

• the height of the wall.

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$\large \bf \clubs \:  Solution :-$

☆ In the Attached Figure :

• AB = Wall
• ( height of wall = H meter )

• AC = Ladder
• ( length of ladder = L meter )

》 In △ ABC :-

$\sf \tan60 \degree = \dfrac{P}{B} \\ \\ \sf:\longmapsto \sqrt{3} = \frac{H}{1.5} \\ \\ :\longmapsto \sf1.73 = \frac{H}{1.5} \\ \\ :\longmapsto \sf H = 1.73 \times 1.5 \\ \\ \purple{ \Large :\longmapsto  \underline {\boxed{{\bf H=2.595 \: m } }}}$

Also ,

$\sf \sf \cos60 \degree = \dfrac{B}{H} \\ \\ :\longmapsto \sf\frac{1}{2} = \frac{1.5}{L } \\ \\ :\longmapsto \sf L = 1.5 \times 2 \\ \\ \purple{ \Large :\longmapsto  \underline {\boxed{{\bf L = 3 \: m} }}}$

Hence ,

• Length of Ladder = 3 meter.

• Height of Wall = 2.595 meter.

$\Large\red{\mathfrak{  \text{W}hich \:\:is\:\: the\:\: required} }\\ \LARGE \red{\mathfrak{ \text{ A}nswer.}}$

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Answer 2

$\huge\boxed{\textsf{\textbf{\pink{Given\::-}}}}$

$\:$

• A ladder rests against a vertical wall such that the top of the ladder reaches the top of the wall.

• Ladder is inclined at 60° with the ground, and the bottom of the ladder is 1.5 m away from the foot of wall.

$\:$

$\huge\boxed{\textsf{\textbf{\green{To\:Find\::-}}}}$

$\:$

• (i) Length of the ladder?

• (ii) Height of the wall?

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$\huge\boxed{\textsf{\textbf{\blue{Solution\::-}}}}$

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• Let height of the wall be PQ, distance between foot of the wall and bottom of the ladder be QR and length of the ladder be PR.

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$\small\frak {We\:know\:that} \begin{cases} & \bf{\red{tan\:\theta = \dfrac{Perpendicular}{Base}}} \\ \\ & \bf{\purple{cos\:\theta = \dfrac{Base}{Hypotenuse}}}\end{cases}$

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⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━

$\:$

$\qquad\footnotesize\bf{\dag}\:\underline{\sf{In\:\triangle{ABC}\::-}}$

$\\ :\implies \:\sf cos\:60^{\circ} = \dfrac{Base}{Hypotenuse}$

$\\ :\implies \:\sf \dfrac{1}{2} = \dfrac{QR}{PR}$

$\\ :\implies \:\sf \dfrac{1}{2} = \dfrac{1.5}{PR}$

$\\ :\implies \:\sf PR = 1.5\:\times\:2$

$\\ :\implies \:\underline{\boxed{\bf{\purple{ PR = 3}}}}\:\bigstar$

$\:$

Also,

$\\ :\implies \:\sf tan\:60^{\circ} = \dfrac{Perpendicular}{Base}$

$\\ :\implies \:\sf \sqrt{3} = \dfrac{PQ}{QR}$

$\\ :\implies \:\sf \sqrt{3} = \dfrac{PQ}{1.5}$

$\\ :\implies \:\sf PQ = \sqrt{3}\:\times\:1.5$

$\\ :\implies \:\sf PQ = 1.73\:\times\:1.5$

$\\ :\implies \:\underline{\boxed{\bf{\red{ PQ = 2.595}}}}\:\bigstar$

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• Therefore, length of the ladder is 3 m and height of the wall is 2.595 m.

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$\huge\boxed{\textsf{\textbf{\orange{Know\:More\::-}}}}$

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$\underline{\sf{\bigstar\:Trigonometric\:table\::-}}$

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$\quad$$\large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$

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$\huge\boxed{\textsf{\textbf{\gray{Note\::-}}}}$

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• Diagram is in the attachment!

• If you are not able to see the trigonometric table from app, kindly view it from web (brainly.in), Question link :- https://brainly.in/question/43126511

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