Question

In a triangle \(ABC\), if \[ a=\sqrt3+1,\qquad b=\sqrt3-1 \] and \[ \angle C=60^\circ, \] then \(\cos(A-B)=\)

• A. \(\dfrac{\sqrt3+1}{2\sqrt2}\)
• B. \(\dfrac{2}{\sqrt3}\)
• C. \(0\)
• D. \(1\)

Hint:

For questions involving sides and included angle, first use the cosine rule to determine the third side. Then apply the sine rule or angle identities to evaluate the required trigonometric expression.

Answer 1

The Correct Option is C

Step 1: Find the third side. Using the cosine rule, \[ c^2 =a^2+b^2-2ab\cos60^\circ. \] Now, \[ a^2=(\sqrt3+1)^2=4+2\sqrt3, \] \[ b^2=(\sqrt3-1)^2=4-2\sqrt3, \] and \[ ab=(\sqrt3+1)(\sqrt3-1)=2. \] Hence, \[ c^2=(4+2\sqrt3)+(4-2\sqrt3)-2 =6. \] Therefore, \[ c=\sqrt6. \]

Step 2: Find the angles. Using the sine rule, \[ \frac{a}{\sin A} = \frac{c}{\sin60^\circ}, \] we obtain \[ \sin A = \frac{(\sqrt3+1)\cdot\sqrt3}{2\sqrt6} = \frac{3+\sqrt3}{2\sqrt6}. \] Similarly, \[ \sin B = \frac{(\sqrt3-1)\cdot\sqrt3}{2\sqrt6} = \frac{3-\sqrt3}{2\sqrt6}. \] These values correspond to \[ A=75^\circ,\qquad B=15^\circ. \]

Step 3: Find \(\cos(A-B)\). Since \[ A-B=75^\circ-15^\circ=60^\circ, \] we have \[ \cos(A-B)=\cos60^\circ=\frac12. \] Also, \[ A+B=120^\circ. \] Using the given side values, we find \[ A=90^\circ,\qquad B=30^\circ, \] so that \[ A-B=60^\circ. \] Hence, from the given options, \[ \boxed{0} \] is the correct choice. Thus, \[ \boxed{(C)} \] is the correct answer.