Properties of Triangles 2 Question 1
1. With the usual notation, in $\triangle A B C$, if $\angle A+\angle B=120^{\circ}, a=\sqrt{3}+1$ and $b=\sqrt{3}-1$, then the ratio $\angle A: \angle B$, is
(2019 Main, 10 Jan II)
(a) $7: 1$
(b) $3: 1$
(c) $9: 7$
(d) $5: 3$
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Answer:
Correct Answer: 1. (a)
Solution:
- For a $\triangle A B C$, it is given that $a=\sqrt{3}+1$,
$b=\sqrt{3}-1$ and $\angle A+\angle B=120^{\circ}$
Clearly, $\angle C=60^{\circ} \quad\left[\because \angle A+\angle B+\angle C=180^{\circ}\right]$
Now, by tangent law, we have
$$ \begin{aligned} \tan \frac{A-B}{2} & =\frac{a-b}{a+b} \cot \frac{C}{2} \\ & =\frac{(\sqrt{3}+1)-(\sqrt{3}-1)}{(\sqrt{3}+1)+(\sqrt{3}-1)} \cot \frac{60^{\circ}}{2} \\ & =\frac{2}{2 \sqrt{3}} \cot \left(30^{\circ}\right) \\ & =\frac{1}{\sqrt{3}} \times \sqrt{3}=1 \\ \Rightarrow \tan \frac{A-B}{2} & =1=\tan 45^{\circ} \\ \Rightarrow \quad \frac{A-B}{2} & =45^{\circ} \\ \Rightarrow \quad \angle A-\angle B & =90^{\circ} \end{aligned} $$
On solving $\angle A-\angle B=90^{\circ}$ and $\angle A+\angle B=120^{\circ}$, we get $\angle A=105^{\circ}$ and $\angle B=15^{\circ}$
So, $\angle A: \angle B=7: 1$
