Trigonometric Equations Question 285
Question: Let D be the middle point of the side BC of a triangle ABC. If the triangle ADC is equilateral, then $ a^{2}:b^{2}:c^{2} $ is equal to
[Pb. CET 2004]
Options:
A) $ 1:4:3 $
B) $ 4:1:3 $
C) $ 4:3:1 $
D) $ 3:4:1 $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ \cos 120{}^\circ =\frac{x^{2}+x^{2}-AB^{2}}{2x^{2}} $
Þ $ \frac{2x^{2}-AB^{2}}{2x^{2}}=\frac{-1}{2} $
Þ $ 4x^{2}-2AB^{2}=-2x^{2} $
Þ $ 3x^{2}=AB^{2} $ Þ $ AB=x\sqrt{3} $
Þ $ a^{2}:b^{2}:c^{2}={{(2x)}^{2}}:x^{2}:{{(x\sqrt{3})}^{2}} $ = $ 4x^{2}:x^{2}:3x^{2} $ = $ 4:1:3 $ .
BETA
