Question

Two sides of a triangle are $\sqrt{3}+1$ and $\sqrt{3}-1$ and the included angle is $60^\circ$, then the difference of the remaining angles is

• A. $30^\circ$
• B. $45^\circ$
• C. $60^\circ$
• D. $90^\circ$

Hint:

Logic Tip: Try using Napier's Analogy for a much faster shortcut! $\tan(\frac{A-B}{2}) = \frac{(\sqrt{3}+1)-(\sqrt{3}-1)}{(\sqrt{3}+1)+(\sqrt{3}-1)}\cot(30^\circ) = \frac{2}{2\sqrt{3 \cdot \sqrt{3} = 1$. Since $\tan(\frac{A-B}{2}) = 1$, then $\frac{A-B}{2} = 45^\circ$, which immediately gives $A-B = 90^\circ$ without finding the 3rd side!

Answer 1

The Correct Option is D

Concept:
This problem can be efficiently solved using Napier's Analogy (Law of Tangents), which states: $$\tan\left(\frac{A-B}{2}\right) = \left(\frac{a-b}{a+b}\right)\cot\left(\frac{C}{2}\right)$$ Alternatively, the standard Cosine Rule followed by the Sine Rule can be used to find all sides and angles explicitly. Let's use the standard method to find the angles.
Step 1: Use the Cosine Rule to find the third side c.
Let $a = \sqrt{3}+1$, $b = \sqrt{3}-1$, and $\angle C = 60^\circ$. The Cosine Rule is: $$c^2 = a^2 + b^2 - 2ab\cos C$$ First, calculate $a^2+b^2$ and $ab$: $$a^2 + b^2 = (\sqrt{3}+1)^2 + (\sqrt{3}-1)^2 = (3 + 1 + 2\sqrt{3}) + (3 + 1 - 2\sqrt{3}) = 8$$ $$ab = (\sqrt{3}+1)(\sqrt{3}-1) = (\sqrt{3})^2 - 1^2 = 3 - 1 = 2$$ Now, substitute these into the Cosine Rule: $$c^2 = 8 - 2(2)\cos(60^\circ)$$ $$c^2 = 8 - 4\left(\frac{1}{2}\right) = 8 - 2 = 6$$ $$c = \sqrt{6}$$
Step 2: Use the Sine Rule to find angle B.
The Sine Rule states: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ Using the relation involving $b$ and $c$: $$\frac{\sqrt{3}-1}{\sin B} = \frac{\sqrt{6{\sin(60^\circ)}$$ $$\frac{\sqrt{3}-1}{\sin B} = \frac{\sqrt{6{\frac{\sqrt{3{2 = \frac{2\sqrt{6{\sqrt{3 = 2\sqrt{2}$$ Solving for $\sin B$: $$\sin B = \frac{\sqrt{3}-1}{2\sqrt{2$$ This is a standard trigonometric value. We know that $\sin(15^\circ) = \frac{\sqrt{3}-1}{2\sqrt{2$. Thus, $\angle B = 15^\circ$.
Step 3: Calculate angle A and find the required difference.
The sum of angles in a triangle is $180^\circ$: $$\angle A + \angle B + \angle C = 180^\circ$$ $$\angle A + 15^\circ + 60^\circ = 180^\circ$$ $$\angle A = 180^\circ - 75^\circ = 105^\circ$$ The question asks for the difference of the remaining angles ($\angle A - \angle B$): $$\text{Difference} = 105^\circ - 15^\circ = 90^\circ$$