Math, asked by bangtanboys95, 17 days ago

Charmi chowdary is observing an object from a building of height 6mts with 60° of angle of depression, then the horizontal distance between building and object is

A) 6mts
B) 6√3 mts
C) \sf\frac{3}{ \sqrt{3}} mts
D) 2√3 mts​

Answers

Answered by XxLUCYxX
1

\mathcal{\fcolorbox{aqua}{azure}{\red{❖NONE\:OF\:THESE\:AND\:THE\:CORRECT\: ANSWER \:IS\:50METRE}}} </p><p>

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 \huge\mathcal \fcolorbox{gold}{orange}{MORE\:TO\:KNOW}

\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf\red{ Trigonometry\: Table} \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A &amp; \bf{0}^{ \circ} &amp; \bf{30}^{ \circ} &amp; \bf{45}^{ \circ} &amp; \bf{60}^{ \circ} &amp; \bf{90}^{ \circ} \\ \\ \rm sin A &amp; 0 &amp; \dfrac{1}{2}&amp; \dfrac{1}{ \sqrt{2} } &amp; \dfrac{ \sqrt{3}}{2} &amp;1 \\ \\ \rm cos \: A &amp; 1 &amp; \dfrac{ \sqrt{3} }{2}&amp; \dfrac{1}{ \sqrt{2} } &amp; \dfrac{1}{2} &amp;0 \\ \\ \rm tan A &amp; 0 &amp; \dfrac{1}{ \sqrt{3} }&amp;1 &amp; \sqrt{3} &amp; \rm \infty \\ \\ \rm cosec A &amp; \rm \infty &amp; 2&amp; \sqrt{2} &amp; \dfrac{2}{ \sqrt{3} } &amp;1 \\ \\ \rm sec A &amp; 1 &amp; \dfrac{2}{ \sqrt{3} }&amp; \sqrt{2} &amp; 2 &amp; \rm \infty \\ \\ \rm cot A &amp; \rm \infty &amp; \sqrt{3} &amp; 1 &amp; \dfrac{1}{ \sqrt{3} } &amp; 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}

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