The lengths of the minor and major arcs 54cm and 126cm respectively. Calculate the angle of the major sector
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Correct Answer: Option B
Explanation:
Let 0 = angle of the minor sector
angle of the major sector = 360 - \(\theta\)(angle at a point)
2 \(\pi r\) = 54 + 126(i.e circumference of minor and major arc)
2\(\pi r = 180^o\)
r = \(\frac{180}{2\pi}\) = \(\frac{90}{\pi}\)
Lenght of ninor arc
= \(\frac{\theta}{360} \times 2 \pi r\)
54 = \(\frac{\theta}{360} \times 3 \pi r\)
\(\theta = \frac{360 \times 54}{2 \pi r}\)
but r = \(\frac{90}{\pi}\) substituting \(\frac{90}{\pi}\) for r
\(\theta = \frac{360 \times 54 \times \pi}{2 \times \pi \times 90}\)
\(\theta = 2 \times 54 = 108^o\)
angle of the major sector = 360 - 108o
= 252o
Let 0 = angle of the minor sector
angle of the major sector = 360 - \(\theta\)(angle at a point)
2 \(\pi r\) = 54 + 126(i.e circumference of minor and major arc)
2\(\pi r = 180^o\)
r = \(\frac{180}{2\pi}\) = \(\frac{90}{\pi}\)
Lenght of ninor arc
= \(\frac{\theta}{360} \times 2 \pi r\)
54 = \(\frac{\theta}{360} \times 3 \pi r\)
\(\theta = \frac{360 \times 54}{2 \pi r}\)
but r = \(\frac{90}{\pi}\) substituting \(\frac{90}{\pi}\) for r
\(\theta = \frac{360 \times 54 \times \pi}{2 \times \pi \times 90}\)
\(\theta = 2 \times 54 = 108^o\)
angle of the major sector = 360 - 108o
= 252o