Equal area!

In general, the area of a sector of a circle of radius $r$ subtended by a central angle of $\theta,$ (in radian measure), is $\dfrac{1}{2}r^{2}\theta.$ So with $|AM| = r$ the area of the shaded sector with "base" $AM$ is $\dfrac{1}{2}r^{2}(\pi - \alpha).$

Next, triangle $\Delta MBC$ has a base length of $|MB| = r$ and height $|BC| = |MB|\tan(\alpha) = r\tan(\alpha).$ The area of the shaded region of the triangle is then the area of $\Delta MBC$ minus that of the circular sector subtended by the angle $\alpha.$ This area is then

$\dfrac{1}{2}|MB||BC| - \dfrac{1}{2}|MB|^{2}\alpha = \dfrac{1}{2}r^{2}\tan(\alpha) - \dfrac{1}{2}r^{2}\alpha = \dfrac{1}{2}r^{2}(\tan(\alpha) - \alpha).$

For the two shaded regions to then have equal area, we then require that

$\dfrac{1}{2}r^{2}(\pi - \alpha) = \dfrac{1}{2}r^{2}(\tan(\alpha) - \alpha) \Longrightarrow \pi = \tan(\alpha) \Longrightarrow \alpha = \boxed{\arctan(\pi)}.$

$\begin{aligned} \mathbb{S}_1 & = \mathbb{S}_2 \\ \mathbb{S}_1 + \mathbb{S}_3 & = \mathbb{S}_2 +\mathbb{S}_3 \\ \frac{\pi \cdot \textrm{(MB)}^2}{2} & = \frac{\textrm{MB} \cdot \textrm{BC}}{2}\\ \frac{\pi \cdot \textrm{(MB)}^2}{2} & = \frac{\textrm{MB} \cdot \textrm{MB} \cdot \tan \alpha}{2}\\ \tan \alpha = \pi & \Rightarrow \alpha = \boxed{\arctan \pi} \end{aligned}$

Given the three options, the answer can only be $arctan(\pi)$ since the domain of both $arcsin(x)$ and $arccos(x)$ is $-1\le x \le 1$ , of which $\pi$ is not an element.

Among the 3 options, only $\arctan \pi$ is real value. Not only to guess, we can find $\alpha$ as follows:

Area of sector = $\frac12 r^2 \theta$

Let height of triangle be h and with $r$ = 1 as we only need to find an angle,

$\frac12 (\pi - \alpha) = \frac12 h - \frac12 \alpha.$

$\implies h = \pi = tan \alpha$

$\implies \alpha = \arctan \pi$

Answer: $\boxed{\arctan \pi}$

If the two shaded regions have same area also the triangle and the semicircle have same area.

• Area of the semicircle = pi * r *r * 1/2
• Area of the triangle = r * (r * tan alfa) * 1/2

tan (alfa) = pi ,

alfa = arctan (pi)

Simple analysis, since the range of sine and cosine functions are from -1 to 1, and that pi takes the value greater than 1, then it would be obvious that the angle formed would be the inverse tangent of pi (based on the choices :D).. :D

How do you all "type" those equations?

tana=(BC)/r if r is the circle radius.then(BC)=r(tana) The area of the triangle MBC is:(1/2)r(BC)=(1/2)r^2(tana). The area of the semicircle is:(1/2)πr^2 Therefore:(1/2)r^2(tana)=(1/2)πr^2 and tana=π therefore a=arctanπ.