# Area of triangle $OAB$

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 $BAQ=\theta$

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.

• $$OA\perp PQ,$$ right? Nov 26 '13 at 7:55
• The tangent is perpendicular to the radius. Nov 26 '13 at 8:41
• Hmm...this question teaches me that intelligent people could sometimes also fail on simple questions :) Mar 2 '14 at 12:43
• @Sawarnik : Intelligent? Whom are you referring to :P
– user87543
Mar 2 '14 at 14:23
• @Sawarnik : Thank you :P
– user87543
Mar 3 '14 at 9:41

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 $(\angle AOB)=2\theta$

• I am sorry, I could not understand...
– user87543
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...
– user87543
Nov 26 '13 at 9:18
• Yes, you are welcome.
– Ömer
Nov 26 '13 at 9:24

Isosceles triangles have equal legs opposite equal base angles. Tangents to a circle at a point intersect at right angles to the radius at that point. It follows that the base angles are equal and complementary to $$\theta$$.

Area of a triangle = $$\frac{1}{2}Base\cdot Height$$

Suppose $$\alpha + \theta=\frac{\pi}{2}$$ and $$\alpha$$ is the measure of the base angle if this Isosceles triangle.

Area = $$\frac{1}{2}(r\sin{\alpha})(2r\cos{\alpha})$$

$$r=1$$ and the product $$(\sin{\alpha}\cos{\alpha}) =(\sin{\theta}\cos{\theta})$$ holds for complementary angles.