Question

find the length of side a.
b 40°
c
a 120° 8 c
c = 20°
a = ?
c =

Explanation:

Step1: Identify the Law of Sines

We use the Law of Sines, which states that in any triangle, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$. Here, we know angle $A = 120^\circ$, angle $C = 20^\circ$, and side $b$ (wait, actually side opposite angle $B$? Wait, no, side $AC$ is 8, which is side $b$? Wait, let's label the triangle properly. In triangle $ABC$, angle at $A$ is $120^\circ$, angle at $B$ is $40^\circ$, angle at $C$ is $20^\circ$. Side $AC$ is 8, which is side $b$ (opposite angle $B$)? Wait, no, side $a$ is opposite angle $A$? Wait, no, standard notation: side $a$ is opposite angle $A$, side $b$ opposite angle $B$, side $c$ opposite angle $C$. Wait, in the diagram, side $a$ is $BC$, side $c$ is $AB$, and side $AC$ is 8, which is side $b$ (opposite angle $B$). Wait, angle at $B$ is $40^\circ$, angle at $A$ is $120^\circ$, angle at $C$ is $20^\circ$. So side $AC$ (length 8) is opposite angle $B$ (40°), so side $b = 8$, angle $B = 40^\circ$, angle $A = 120^\circ$, and we need to find side $a$ (opposite angle $A$).

So by Law of Sines: $\frac{a}{\sin A}=\frac{b}{\sin B}$

Step2: Plug in the values

We have $A = 120^\circ$, $B = 40^\circ$, $b = 8$. So:

$\frac{a}{\sin 120^\circ}=\frac{8}{\sin 40^\circ}$

Solve for $a$:

$a = \frac{8 \times \sin 120^\circ}{\sin 40^\circ}$

Calculate $\sin 120^\circ = \sin (60^\circ) = \frac{\sqrt{3}}{2} \approx 0.8660$

$\sin 40^\circ \approx 0.6428$

So $a = \frac{8 \times 0.8660}{0.6428} \approx \frac{6.928}{0.6428} \approx 10.78$ (rounded to two decimal places)

Answer:

$\approx 10.78$ (or more precisely, using exact values: $\frac{8 \times \frac{\sqrt{3}}{2}}{\sin 40^\circ} = \frac{4\sqrt{3}}{\sin 40^\circ} \approx 10.78$)