Question on: JAMB Mathematics - 2001
A
60o
B
100o
C
120o
D
30o
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Correct Option: C
Using cosine formula (t\(\sqrt{3}\))2 = t2 + t2 - 2t2 cos\(\theta\)
3t2 = 2t2 - 2t2 cos\(\theta\) = 2t2(1 - cos\(\theta\))
1 - cos\(\theta\) = \(\frac{3t^2}{2t^2}\) = \(\frac{3}{2}\)
cos = 1 - \(\frac{3}{2} = -\frac{1}{2}\)
\(\theta\) = cos-1(-\(\frac{1}{2}\)) = 120o and 240o
N.B 0 \(\geq\) \(\theta\) 360
3t2 = 2t2 - 2t2 cos\(\theta\) = 2t2(1 - cos\(\theta\))
1 - cos\(\theta\) = \(\frac{3t^2}{2t^2}\) = \(\frac{3}{2}\)
cos = 1 - \(\frac{3}{2} = -\frac{1}{2}\)
\(\theta\) = cos-1(-\(\frac{1}{2}\)) = 120o and 240o
N.B 0 \(\geq\) \(\theta\) 360
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