Question

Let in a triangle any two sides and included angle are $\sqrt{3}-1,\sqrt{3}+1,{60}^{\circ }$ respectively, then which of the following is not true?

• A. Triangle is obtuse
• B. $\angle C={150}^{\circ }$
• C. Third side is $\sqrt{6}$
• D. None of these

Answer 1

Answer: (B)

$\angle C={150}^{\circ }$

Applying
Law of cosines, we get
$cosA=\frac{b^2+c^2-a^2}{2bc}$
Hence
$\frac{1}{2}=\frac{(\sqrt{3}+1{)}^2+(\sqrt{3}-1{)}^2-a^2}{2(3-1)}$
$2=8-a^2$
$a=\sqrt{6}$
And we know that
Since
$\angle A={60}^0$
Hence
$\angle B+\angle C={180}^0-{60}^0$
$={120}^0$
Hence
$\angle C\ne {150}^0$