A 120° isosceles construction and the cosine law

Problem

Right triangle $△ABC$ has $AB=\sqrt{7}$, $AC=\sqrt{3}$, and $BC=2$. On bases $AB$ and $AC$, construct isosceles triangles $△MAB$ and $△NAC$ outside $△ABC$, each with vertex angle $120^{\circ }$. Find $M{N}^2$. The original wording and the answer format are on the CEEC 112 AST Mathematics I paper.

Where it goes wrong

The CEEC item statistics give this item a correct rate of $29\%$ and a discrimination of $0.59$. The first trap is to treat $120^{\circ }$ as a base angle. It is the vertex angle, so each base angle is $30^{\circ }$. The second trap is to leave out the two outer base angles and write $\angle MAN$ as the angle at $A$ in the original right triangle.

Idea

Let $\theta =\angle BAC$. From the right triangle,

Since $A{C}^2+B{C}^2=3+4=7=A{B}^2$, the right angle is at $C$ and $AB=\sqrt{7}$ is the hypotenuse. The new triangles are built outward on $AB$ and on $AC$. A vertex angle of $120^{\circ }$ makes each base angle $30^{\circ }$. The side ratios of a $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ triangle give

Both isosceles triangles lie outside the original triangle, so

The angle-sum formula gives

The cosine law in $△AMN$ then gives

What the figure shows

• Drag φ. The two base angles stay equal to $(180^{\circ }-\phi )/2$.
• Press Back to the original, φ=120° and check that both base angles are $30^{\circ }$.
• Compare $\angle MAN$ on the figure with the exact value in the sidebar. Both outer base angles belong in the included angle used by the cosine law.