Question

A ladder leaning against a wall makes an angle of 60° with the horizontal . If the foot of the ladder is 2.5 m away from the wall, find the length of the wall​

Answer 1

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$\large{\red{\underline{\tt{Diagram\::-}}}}$

$\setlength{\unitlength}{1.6cm}\begin{picture}(6,2)\linethickness{0.5mm}\put(7.7,2.9){\large\sf{A}}\put(7.7,1){\large\sf{C}}\put(10.6,1){\large\sf{B}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(8.4,0.7){\sf{\large{2.5m}}}\put(7.4,2){\sf{\large{wall}}}\put(9.3,2){\sf{\large{ladder}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\qbezier(9.8,1)(9.7,1.25)(10,1.4)\put(9.4,1.2){\sf\large{60^{\circ} }}\end{picture}$

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$\large{\red{\underline{\tt{Given\::-}}}}$

• BC = 2.5 m

• $\tt \angle ABC = 60^{\circ}$

$\large{\red{\underline{\tt{To\:find\::-}}}}$

Length of the wall.

$\large{\red{\underline{\tt{Solution\::-}}}}$

$\dagger\:\:{\underline{\tt According\:to\: question\::-}}$

$\dashrightarrow$ $\sf cos \:60^{\circ} = \dfrac{2.5}{l}$

$\dashrightarrow$ $\sf \dfrac{1}{2} = \dfrac{2.5}{l}$

$\dashrightarrow$ $\sf l = 2.5 \times 2$

$\pink\dashrightarrow$ $\large{\underline{\pink{\boxed{\sf l = 5 m}}}}$ $\orange\bigstar$

Hence, length of wall = 5 meter.

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$\bigstar\:\sf Trigonometric\:Values :\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0& \dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D \hat{e} fined\end{tabular}}$

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Answer 2

$\large\bf\underline{Given:-}$

• A ladder leaning against a wall makes an angle of 60° with the horizontal.
• The foot of the ladder is 2.5 m away from the wall.

$\large\bf\underline {To \: find:-}$length of the wall( let it be L )

$\huge\bf\underline{Solution:-}$

Let,

AB be the ladder and CA be the wall (Refer to attachment )

A/q

The ladder make an angle of 60° with horizontal.

∴ ΔABC is a 30°,60° and 90° triangle.

By using Cosine formula,

Cos60° = $\frac{base}{perpendicular}$

Cos60° = $\frac{2.5}{L}$

$\frac{1}{2}$ = $\frac{2.5}{L}$

∴ L = 5 meter

Hence, length of wall is 5 meters.

$\bigstar\:\sf Trigonometric\:Values :\\\boxed{\begin{tabular}{c|c|c|c|c|c}Radians/Angle & 0 & 30 & 45 & 60 & 90\\\cline{1-6}Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1\\\cline{1-6}Cos \theta & 1 & \dfrac{\sqrt{3}}{2}& \dfrac{1}{\sqrt{2}}& \dfrac{1}{2}&0\\\cline{1-6}Tan \theta&0& \dfrac{1}{\sqrt{3}}&1&\sqrt{3}&Not D \hat{e} fined\end{tabular}}$