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Question is :

Consider a circle of unit radius centered at $O$ in the plane. let $AB$ be a chord which makes an angle $\theta$ with the tangent to the circle at $A$ .find the area of triangle $OAB$

What all i could do was is just draw the picture and even in that i am not sure if he mean angle $BAP=\theta$ or $BAQ=\theta$. I am assuming for some time that $BAQ=\theta$ enter image description here

Now, I would have some hope if angle $OAB$ can be calculated from given data in terms of theta then i would use the area formula as

$\text{Area of triangle $OAB= \frac{1}{2}.OA.OB.\sin (\angle AOB)$}$ and as $OA=OB=1$ we would then get

Area of triangle $OAB=\frac{1}{2}.\sin (\angle AOB)$

But i am not sure how to relate $\theta$ with $\angle AOB$.

I am not even sure if there is any way to relate this.

I would be thankful if some one can help me with this.

Thank you.

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2  
$$OA\perp PQ,$$ right? –  lab bhattacharjee Nov 26 '13 at 7:55
2  
The tangent is perpendicular to the radius. –  Lucian Nov 26 '13 at 8:41
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Hmm...this question teaches me that intelligent people could sometimes also fail on simple questions :) –  Sawarnik Mar 2 at 12:43
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@Sawarnik : Intelligent? Whom are you referring to :P –  Praphulla Koushik Mar 2 at 14:23
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@Sawarnik : Thank you :P –  Praphulla Koushik Mar 3 at 9:41
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1 Answer

up vote 1 down vote accepted

The angle $(\angle BAQ)$ is a tangent (chord) angle.

Theorem An Angle formed by a chord and a tangent that intersect on a circle is half the measure of the intercepted arc. So $$\theta=\frac{1}{2} m \overset{\frown}{(AB)}$$ from there $(\angle AOB)=2\theta$

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I am sorry, I could not understand... –  Praphulla Koushik Nov 26 '13 at 8:50
    
Did you see theorem? In your case the angle $(\angle BAQ)$ is formed by a chord ($|AB|$) and a tangent ($|PQ|$) that intersect on a circle. So the measure of $\overset{\frown}{(AB)}$ is $2\theta$ and because of $(\angle AOB)$ is a center angle its measure is equal to $m \overset{\frown}{(AB)}$. Think $m \overset{\frown}{(AB)}$ as small arc on the circle. –  Ömer Nov 26 '13 at 9:01
    
OK.. OK.. now i got it.. SO, area of triangle $OAB$ is $\frac{1}{2}. \sin (\angle AOB)=\frac{1}{2}. \sin (2\theta)=\frac{1}{2}. 2 sin(\theta) \cos(\theta)= sin(\theta) \cos(\theta)$.. Thank you so much for your support... –  Praphulla Koushik Nov 26 '13 at 9:18
    
Yes, you are welcome. –  Ömer Nov 26 '13 at 9:24
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