Question

The foot of the ladder is 9.5m away from the wall and and the ladder is inclined at an angle of 60 degree withthe groud . The length of ladder will be______ solve it.
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Answer 1

now to find out multiply 9.5 with 6 and the second digit has value but not used at sum like these so we will get 57m hope it was helpful

Answer 2

Given:-

• Distance of foot of ladder is 9.5 m away from the wall = 9.5 m
• Ladder is inclined at an angle of 60° from ground.

To Find:-

• The length of the ladder.

Solution:-

In ∆ABC,

We know,

Cos C = $\sf{\dfrac{Base}{Hypotenuse} = \dfrac{AB}{BC}}$

Hence,

Cos60° = $\sf{\dfrac{BC}{AC}}$

From Trigonometric table we have,

• Cos60° = $\sf{\dfrac{1}{2}}$

Putting the value,

$\sf{\dfrac{1}{2} = \dfrac{9.5}{AC}}$

$\sf{:\implies AC = 9.5\times 2}$

$\sf{:\implies AC = 19\:m}$

$\sf{\therefore The\:length\:of\:the\:ladder\:is\:19\:m}$

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Explore More!!

Trigonometric Table is as follows:-

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

Important points!!

• Sinθ = $\sf{\dfrac{Perpendicular}{Hypotenuse}}$

• Cosθ = $\sf{\dfrac{Base}{Hypotenuse}}$

• Tanθ = $\sf{\dfrac{Perpendicular}{Base}}$

• Cosecθ = $\sf{\dfrac{Hypotenuse}{Perpendicular}}$

• Secθ = $\sf{\dfrac{Hypotenuse}{Base}}$

• Cotθ = $\sf{\dfrac{Base}{Perpendicular}}$

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