How can we calculate the side $AC$?

Question

In the triangle $ABC$ it holds that $|AB| = 10$, for the midpoint $C'$ of the side $AB$ it holds that $|CC'| = 8$ and the interior angle at $C$ at the triangle $BCC'$ is $\frac{\pi}{6}$.

(a) What can we say about the distance between $A$ and $C$?

(b) Let $M$ be the midpoint ofthe circumcenter of the triangle. Determine all possibilities for the angle $\angle AMB$.

Using the cosine law we could calculate the side $BC$, right?

But how can we calculate the side $AC$?

Answer 1

a) Apply the sine rule to $BCC’$

$$\sin B= \frac85\sin\frac\pi6=\frac45,\>\>\>\>\>\cos B = \pm\frac35 $$ Then, apply the cosine rule to $ACC’$

$$AC^2 = 8^2+ 5^2 - 2\cdot8\cdot 5\cos(B+\frac\pi6)=121\pm24\sqrt3 $$ Thus, two possibilities for the distance $AC = \sqrt{ 121\pm24\sqrt3} $.

b) Note $\angle AMB =2C$ and apply the sine rule to $ABC$ $$ {\sin C}=\frac {AB}{AC}\sin B = \frac{10}{ \sqrt{ 121\pm24\sqrt3} }\cdot \frac45= \frac8{ \sqrt{ 121\pm24\sqrt3} }$$ Thus, two possible anglular values $$\angle AMB = 2C = 2\arcsin \frac8{\sqrt{ 121\pm24\sqrt3} }$$

Answer 2

$\sin (\pi/6)/5=\sin \beta /8$, then third angle in the triangle is $\tau=180-\beta-30$, then angle $\rho=180-\tau$. We have now the second triangle, there is the angle $\rho$, and $AC'=5,CC'=8$, we can use cosine formula $(AC)^2=8^2+5^2-80*\cos \rho$.