Question

A ladder is placed against a wall such that it just touches the top of the wall. the foot of latter is 10m away from the wall and the ladder is inclined at an angle 60° with the ground. find the height of the wall and length of the ladder. ​

Answer 1

Given : A ladder is placed against a wall such that it just touches the top of the wall. the foot of latter is 10m away from the wall and the ladder is inclined at an angle 60° with the ground

Answer:

$\qquad\qquad\tiny\underline{\frak{ By \: Using \: trignometry \: property \: we \: get :}}$

$: \implies \sf \tan( \theta) = \dfrac{Perpendicular}{Base}$

$: \implies \sf \tan( \theta) = \dfrac{AB}{BC}$

$: \implies \sf \tan(60) = \dfrac{AB}{10}$

$: \implies \sf \dfrac{ \sqrt{3} }{1} = \dfrac{AB}{10}$

$: \implies \sf AB = 10 \sqrt{3} \: m$

$: \implies \sf AB = 10 \times 1.732 \qquad\Bigg\lgroup \textsf{\textbf{Taking \sqrt{3} = 1.732}}\Bigg\rgroup \: m$

$: \implies \underline{ \boxed{\sf AB = 17.32 \: meters}}$

$\therefore\underline{\textsf{The height of the wall is \textbf{17.32 meters}}}.$

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$\begin{array}{| c | c | c | c | c | c |}\cline{1-6} \bf Angles & \bf 0^{o} & \bf 30^{o} & \bf 45^{o} & \bf 60^{o} & \bf 90^{o} \\\cline{1-6} \tt Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}}& \dfrac{\sqrt{3}}{2}& 1\\\cline{1-6} \tt cos \theta & 1 & \dfrac{\sqrt{3}}{2} &\dfrac{1}{\sqrt{2}}&\dfrac{1}{2}&0\\\cline{1-6} \tt tan \theta & 0 & \dfrac{1}{\sqrt{3}} & 1& \sqrt{3} & \infty \\\cline{1-6} \tt cosec \theta & \infty & 2 & \sqrt{2} & \dfrac{2}{\sqrt{3}} &1\\\cline{1-6} \tt sec \theta & 1 & \dfrac{2}{\sqrt{3}} & \sqrt{2} & 2 & \infty \\\cline{1-6} \tt cot \theta & \infty & \sqrt{3} &1& \dfrac{1}{\sqrt{3}} &0\\\cline{1-6}\end{array}$