A triangle has sides 65, 65 and 10.75 and no known angles. Find all three angles.

Step-by-step solution

1. 1

   Choose the right tool for SSS. With three sides and no angles, the sine rule is unusable - it always needs one matched side-angle pair. The law of cosines is the only starting point:

   $c^2=a^2+b^2-2ab\cos C\;\Longrightarrow\;\cos C=\frac{a^2+b^2-c^2}{2ab}$

2. 2

   Exploit the isosceles symmetry. Two sides are equal ($65=65$), so the two angles opposite them are equal. That means one law-of-cosines computation - for the angle $C$ opposite the odd side $10.75$ - determines everything.

3. 3

   Substitute the numbers.

   $\cos C=\frac{65^2+65^2-10.75^2}{2(65)(65)}=\frac{4225+4225-115.5625}{8450}=\frac{8334.4375}{8450}=0.98632396$

4. 4

   Take the inverse cosine.

   $C=\cos^{-1}(0.98632396)=9.4867^{\circ}$

   That the cosine is so close to $1$ is the arithmetic signature of a very thin sliver of a triangle, which is exactly what sides $65,65,10.75$ describe.

5. 5

   Get the two base angles from the angle sum.

   $A=B=\frac{180^{\circ}-9.4867^{\circ}}{2}=85.2567^{\circ}$

6. 6

   Check independently with the area. Heron's formula with $s=70.375$ gives area $=348.178$. The same area as $\tfrac12 ab\sin C=\tfrac12(65)(65)\sin(9.4867^{\circ})=348.178$ ✓, and $9.4867+85.2567+85.2567=180.0001^{\circ}$ to rounding.