Find k when a sector is divided into two equal areas

Key concept: This is the same area setup as the general case, but now the angle is fixed at $\pi/6$. Once you write the triangle area and sector area correctly, the value of $k$ drops out neatly.

Step by step

For the sector of radius $a$ and angle

$\theta=\frac{\pi}{6},$

the equal-area condition gives

$\frac12k^2a^2\sin\theta=\frac14a^2\theta$

So

$k^2=\frac{\theta}{2\sin\theta}$

Substitute $\theta=\frac\pi6$:

$k^2=\frac{\pi/6}{2\cdot\sin(\pi/6)}$

Since

$\sin\left(\frac\pi6\right)=\frac12,$

we get

$k^2=\frac{\pi/6}{2\cdot 1/2}=\frac\pi6$

Therefore

$k=\sqrt{\frac\pi6}$

Now evaluate numerically:

$k\approx 0.7236$

Answer

$\boxed{0.7236}$

Pitfall alert

A small but costly error is forgetting that the triangle area uses $\sin\theta$, not $\sin(\theta/2)$. Also, keep the angle in radians when using the sector area formula $\frac12r^2\theta$.

Try different conditions

If the angle were $\pi/3$ instead, the same method would give

$k=\sqrt{\frac{\pi/3}{2\sin(\pi/3)}}$

So the structure of the solution never changes; only the trig value inside the square root changes.

Further reading

sector area, chord, radians