Trigonometric Equations Question 289
Question: The lengths of the sides of a triangle are $ \alpha -\beta ,\alpha +\beta $ and $ \sqrt{3{{\alpha }^{2}}+{{\beta }^{2}}}, $ $ (\alpha >\beta >0) $ . Its largest angle is
[Roorkee 1999]
Options:
A) $ \frac{3\pi }{4} $
B) $ \frac{\pi }{2} $
C) $ \frac{2\pi }{3} $
D) $ \frac{5\pi }{6} $
Show Answer
Answer:
Correct Answer: C
Solution:
- Let $ a=\alpha -\beta ,b=\alpha +\beta ,c=\sqrt{3{{\alpha }^{2}}+{{\beta }^{2}}} $ \ $ \cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab} $
Þ $ \cos C=\frac{{{\alpha }^{2}}+{{\beta }^{2}}-2\alpha \beta +{{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta -3{{\alpha }^{2}}-{{\beta }^{2}}}{2({{\alpha }^{2}}-{{\beta }^{2}})} $
Þ $ \cos C=-\frac{({{\alpha }^{2}}-{{\beta }^{2}})}{2({{\alpha }^{2}}-{{\beta }^{2}})}=\cos ( \frac{2\pi }{3} ) $
Þ $ \angle C=\frac{2\pi }{3} $ , (largest angle).
BETA
