Solve for a: 120^2 = 150^2 + a^2 - 2 * 150 * a * cos(30 degrees).

Step-by-step solution

1. 1

   Recognise the law of cosines with the unknown on the wrong side. In $c^{2}=b^{2}+a^{2}-2ba\cos C$ the unknown $a$ appears both squared and linearly, so this is a quadratic in $a$, not a one-step evaluation. Expect up to two positive answers — the ambiguous SSA case, where two different triangles share the given data.

2. 2

   Substitute the known values. With $120^{2}=14400$, $150^{2}=22500$ and $\cos 30^{\circ}=\dfrac{\sqrt3}{2}$:

   $14400=22500+a^{2}-2(150)a\cdot\frac{\sqrt3}{2}=22500+a^{2}-150\sqrt3\,a$

   The factor 2 cancels against the denominator of $\cos 30^{\circ}$, leaving the tidy coefficient $150\sqrt3$.

3. 3

   Rearrange into standard quadratic form.

   $a^{2}-150\sqrt3\,a+8100=0$

   since $22500-14400=8100$.

4. 4

   Apply the quadratic formula and simplify the surds.

   $\left(150\sqrt3\right)^{2}=67500,\qquad 67500-4(8100)=35100,\qquad \sqrt{35100}=30\sqrt{39}$

   so

   $a=\frac{150\sqrt3\pm 30\sqrt{39}}{2}=75\sqrt3\pm 15\sqrt{39}$

5. 5

   Convert to decimals carefully. With $\sqrt3=1.7320508$ and $\sqrt{39}=6.2449980$:

   $75\sqrt3=129.9038,\qquad 15\sqrt{39}=93.6750$

   $a=129.9038+93.6750=223.5788\qquad\text{or}\qquad a=129.9038-93.6750=36.2288$

6. 6

   Check both in the original equation. For $a=223.5788$: $22500+49987.5-2(150)(223.5788)(0.866025)=14400$ ✓. For $a=36.2288$: $22500+1312.5-9412.5=14400$ ✓. Both are positive, so both triangles are geometrically valid and both must be reported.