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Question: In a triangle ABC, if b = $(\sqrt{3}-1)$ a and $\angle$C=30°, then the value of (A - B) is equal to ...

In a triangle ABC, if b = (3−1)(\sqrt{3}-1) a and ∠\angleC=30°, then the value of (A - B) is equal to (All symbols used have usual meaning in a triangle.)

A

30°

B

45°

C

60°

D

75°

Answer

60°

Explanation

Solution

The problem asks us to find the value of (A−B)(A-B) in a triangle ABC, given the relationship between sides aa and bb, and the angle CC.

Given:

  1. b=(3−1)ab = (\sqrt{3}-1) a
  2. ∠C=30∘\angle C = 30^\circ

We need to find (A−B)(A-B).

Step 1: Use the sum of angles in a triangle. In any triangle ABC, the sum of angles is 180∘180^\circ. A+B+C=180∘A + B + C = 180^\circ Substitute the given value of CC: A+B+30∘=180∘A + B + 30^\circ = 180^\circ A+B=180∘−30∘A + B = 180^\circ - 30^\circ A+B=150∘A + B = 150^\circ

Step 2: Use Napier's Analogy (Tangent Rule). Napier's Analogy relates the sides and angles of a triangle: b−ab+a=tan⁡(B−A2)tan⁡(B+A2)\frac{b-a}{b+a} = \frac{\tan\left(\frac{B-A}{2}\right)}{\tan\left(\frac{B+A}{2}\right)}

Step 3: Calculate the terms involving sides. From the given b=(3−1)ab = (\sqrt{3}-1)a: b−a=(3−1)a−a=(3−1−1)a=(3−2)ab-a = (\sqrt{3}-1)a - a = (\sqrt{3}-1-1)a = (\sqrt{3}-2)a b+a=(3−1)a+a=(3−1+1)a=3ab+a = (\sqrt{3}-1)a + a = (\sqrt{3}-1+1)a = \sqrt{3}a

Now, find the ratio b−ab+a\frac{b-a}{b+a}: b−ab+a=(3−2)a3a=3−23\frac{b-a}{b+a} = \frac{(\sqrt{3}-2)a}{\sqrt{3}a} = \frac{\sqrt{3}-2}{\sqrt{3}}

Step 4: Calculate the term involving the sum of angles. From Step 1, we have A+B=150∘A+B = 150^\circ. So, A+B2=150∘2=75∘\frac{A+B}{2} = \frac{150^\circ}{2} = 75^\circ.

Now, calculate tan⁡(75∘)\tan\left(75^\circ\right): tan⁡(75∘)=tan⁡(45∘+30∘)\tan\left(75^\circ\right) = \tan\left(45^\circ + 30^\circ\right) Using the tangent addition formula tan⁡(x+y)=tan⁡x+tan⁡y1−tan⁡xtan⁡y\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}: tan⁡(75∘)=tan⁡45∘+tan⁡30∘1−tan⁡45∘tan⁡30∘=1+131−1⋅13=3+133−13=3+13−1\tan\left(75^\circ\right) = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3}+1}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}} = \frac{\sqrt{3}+1}{\sqrt{3}-1} To rationalize the denominator, multiply the numerator and denominator by (3+1)(\sqrt{3}+1): tan⁡(75∘)=(3+1)(3+1)(3−1)(3+1)=(3+1)2(3)2−12=3+1+233−1=4+232=2+3\tan\left(75^\circ\right) = \frac{(\sqrt{3}+1)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{(\sqrt{3}+1)^2}{(\sqrt{3})^2 - 1^2} = \frac{3+1+2\sqrt{3}}{3-1} = \frac{4+2\sqrt{3}}{2} = 2+\sqrt{3}

Step 5: Substitute the values into Napier's Analogy. 3−23=tan⁡(B−A2)2+3\frac{\sqrt{3}-2}{\sqrt{3}} = \frac{\tan\left(\frac{B-A}{2}\right)}{2+\sqrt{3}}

Now, solve for tan⁡(B−A2)\tan\left(\frac{B-A}{2}\right): tan⁡(B−A2)=(3−2)(2+3)3\tan\left(\frac{B-A}{2}\right) = \frac{(\sqrt{3}-2)(2+\sqrt{3})}{\sqrt{3}}

Let's simplify the numerator: (3−2)(2+3)=23+(3)2−2(2)−23(\sqrt{3}-2)(2+\sqrt{3}) = 2\sqrt{3} + (\sqrt{3})^2 - 2(2) - 2\sqrt{3} =23+3−4−23= 2\sqrt{3} + 3 - 4 - 2\sqrt{3} =3−4=−1= 3 - 4 = -1

So, tan⁡(B−A2)=−13\tan\left(\frac{B-A}{2}\right) = \frac{-1}{\sqrt{3}}

Step 6: Find the angle B−A2\frac{B-A}{2}. We know that tan⁡(30∘)=13\tan(30^\circ) = \frac{1}{\sqrt{3}}. Since tan⁡(B−A2)\tan\left(\frac{B-A}{2}\right) is negative, and B−A2\frac{B-A}{2} must be an angle between −90∘-90^\circ and 90∘90^\circ (as it's half the difference of angles in a triangle), we use the property tan⁡(−x)=−tan⁡x\tan(-x) = -\tan x. So, tan⁡(B−A2)=−tan⁡(30∘)=tan⁡(−30∘)\tan\left(\frac{B-A}{2}\right) = -\tan(30^\circ) = \tan(-30^\circ) Therefore, B−A2=−30∘\frac{B-A}{2} = -30^\circ

Step 7: Calculate (A−B)(A-B). Multiply by 2: B−A=−60∘B-A = -60^\circ To find (A−B)(A-B), multiply by -1: A−B=60∘A-B = 60^\circ