If a chord divides a circle into parts with areas in ratio 5:7, what is the angle between a tangent line to the circle at one of the ends of the chord and the chord? (1) 15° (2) 75° (3) 90° (4) 115°.
I tried using sines theorem and drawing the centered triangles out, but I'm not sure how to figure out the angle out.
If you only need to quickly find a solution without any strict proof.
Intuition. It can't be $>90°$ since the angle between two lines by definition $\leq 90°$. Also it can not be $90°$ since this is the diameter case. $15°$ correspond to very small chord. So the only residual choice is $75°$.
Let $\theta$ the central angle of the circular segment whose area is $\frac{5}{12}$ of the circle. The desired angle is the half of $\theta$.
So find out $\theta$ using: $$\frac{r^2}{2}(\theta- \sin\theta)=\frac{5}{12}\pi r^2$$ You will get: $$\theta=2.878 \text{rad},$$ or $$\theta =164.9^\circ{}.$$ Therefore the angle is $82.45^\circ{}$.
Let $ a$ be the diameter of circle touching y-axis at the origin.
$\psi$ is the required angle and so on basis of $\psi$, all areas are computed.
$ \dfrac{\sin(\psi)} {r} = \dfrac{1}{a} $ ...(1*)
$ \psi^ {'} = \dfrac{1}{a} $...(2*)
Curvature $ k_g = \sin (\psi)/ r + \psi^{'} = 1/a + 1/a =2/a $...(3*)
where primes are with respect to arc length.
$ \dfrac{dA}{d\psi} = \dfrac{r^2}{2} \dfrac{d\theta}{ds} \dfrac{ ds}{d\psi} $ ...(4*)
$ A = \int_\psi ^{\pi/2} \sin^2(\psi)* a^2/2 * d\psi $..(5*)
$ a^2/2 * ( \pi/4 -( \psi/2 - \sin(2 \psi)/4 ) = \pi (a/2)^2 *\frac{1}{12} $...(6*)
Remaining area fraction is $ \frac{1}{12}$ because portions $ \frac{5}{12}$ is situated above and $ \frac{1}{2}$ in semi-circle below.
The equation numerically solves to $ \psi =82.4566 $ degrees, an option not among those given.
