Describing a Triangle in a Circle

Anything

As shown in the diagram let the radius of the circle be $r$

Also it is easy to notice that the height of the triangle is $rsin(\theta)$

Hence area of the triangle ABC = $\frac{1}{2}(2r)(rsin(\theta))={r}^{2}sin(\theta)$

Given ratio of area of triangle and the area of the circle is $2\pi$ .

So we have $2\pi=\frac{\pi {r}^{2}}{{r}^{2}sin(\theta)}$

Solving we have :

$sin(\theta)=\frac{1}{2}$

$\Rightarrow \theta={30}^{0}$

Hence angle CAB= $90-\frac{\theta}{2}=90-\frac{30}{2}={75}^{0}$

And angle CBA= $\frac{\theta}{2}={15}^{0}$

Sum of digital product of the non-right angles is :

$\boxed { 7 \times 5 + 1 \times 5 = 40 }$

In a right angle triangle , the median drawn to the hypotenuse is equal to half the hypotenuse. This can be easily proved using coordinate system.Here, median is equal to the radius. So we get two right isosceles triangles. as asked angles are 45 each. So answer is 20 +20=40 Notice that the area evaluation is not required.

Let the radius of the circle be $r$ , and $\angle ABC = \theta$ . Then the area $A$ of $\triangle ABC$ is:

$A = \frac {1}{2} (2r\sin{\theta})(2r\cos{\theta}) = r^2\sin{2\theta}$

It is given that: $A=\dfrac {\pi r^2}{2 \pi} = \dfrac {r^2}{2}$

$\Rightarrow A = r^2\sin{2\theta} = \dfrac {r^2}{2} \quad \Rightarrow \sin {2\theta} = \frac {1}{2} \quad \Rightarrow 2\theta = 30^o \quad \Rightarrow \theta = 15^o$

$\angle CAB = 90^o - \angle ABC = 90^o - \theta = 90^o - 15^o = 75^o$

The required answer $= 1\times 5 + 7\times 5 = 5 + 35 = \boxed {40}$

If angle AOC = 90 then the area of ACB is r squared and the ratio of the area to the area of the triangle is pi. To get a ratio of 2 pi the area of ACB must be 1/2 of that mentioned above. Therefore the height has to be 1/2 r. Therefore angle AOC = 30 degrees. Thus angle ABC = 15 and angle CAB = 75.

1 5 + 7 5 = 40