Answer my question

Question

You have given an equal-sided triangle with side length a. A straight line connects the center of the bottom side to the border of the triangle with an angle of α. Derive an expression for the enclosed area A(α) with respect to the angle

Answer 1

Here is your solution

Comments

hey, r u sure about ur answer. Don't misunderstand me, please.

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BECAUSE I WASNT ABLE TO SEND THE FIGURE

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and I am not able to send the figure so

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Suppose there are 2 triangle.In 1st triangle PQR is an equal sided triangle.MN is a straight line connects the center of the bottom side to the border of the triangle with an angle which is angle MNR is α1. In 2nd all are same.just angle MNQ is α2.

Answer 2

Let SQ = h and PQ = b

from \(\triangle\)PQS

tan\(\alpha\) = \(\frac{h}{b}\) \(\Rightarrow\) h = b tan\(\alpha\) ...........(i)

from \(\triangle\)SQR

tan60° =\(\frac{h}{\frac{a}{2}-b}\) = h = \(\big(\frac{a}{2}-b\big)\)tan60°

\(\Rightarrow\) h = \(\big(\frac{a}{2}-b\big)\sqrt3\) ............(ii)

From equation (i) and (ii)

b tan\(\alpha\) = \(\big(\frac{a}{2}-b\big)\sqrt3\)

\(\Rightarrow\) b(tan\(\alpha\) + \(\sqrt3\)) = \(\frac{a}{2}\sqrt3\)

\(\Rightarrow\) b = \(\frac{\sqrt3a}{2(tan\alpha+\sqrt3)}\) .............(iii)

Area of triangle PRQ = Enclosed Area by \(\alpha\) = P(\(\alpha\)) (Let)

P(\(\alpha\)) = \(\frac{1}{2}\)x \(\frac{a}{2}\) x h  (\(\because\) Area of triangle = \(\frac{1}{2}\)x base x height)

\(\Rightarrow\) P(\(\alpha\)) = \(\frac{1}{2}\)x \(\frac{a}{2}\) x b tan\(\alpha\)  (from(i))

= \(\frac{a}{4}\) \(\frac{\sqrt3\,a}{2(tan\alpha+\sqrt3)}\)tan\(\alpha\)  (from (iii))

= \(\frac{\sqrt3\,a^2tan\alpha}{8(tan\alpha+\sqrt3)}\)

Special case:-  If \(\alpha\) = 60°

then P(\(\alpha\)) = \(\frac{\sqrt3a^2\sqrt3}{8(\sqrt3+\sqrt3)}\)

= \(\frac{\sqrt3a^2}{16}\)

Comments

I know you are a participate of this contest..i saw you give some problem of IYMC you asked..

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hi, Rafsan which test r u talking abt I really don't know. These questions have been given to me by my teacher who is from IIT . so his level of questions is different. I don't understand what is this IYMC? i m not a participant. submission for what???

yes, but idk that they are the problems of iymc . sorry if u r partcipating

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i even read abt the contest and even its rules and you know what it is written that u cant copy from sites brainly, chegg ,sarthaks etc. So no one is so dumb to copy and i would have never copied.

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please leave me alone i am already stressed abt my exam tomorrow..... sorry to even reply to your rude and rubbish comments. have a good day, great year ahead. I hope u survive this pandemic. :)

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And also try to improve ur reading,writing (especially english)and seeing skills, coz u need otherwise u cant survive in this world. Hope corona doesn't harm you much. have a great evening:):).....