A piece of thread of length 21 4cm is used to f... - WAEC Mathematics 2018 Question

Length of arc, L = 21.4 - 2 x 4.2cm

= 21.4 - 8.4

= 13cm

But L = \(\frac{\theta}{360^o}\) x 2\(\pi r\)

i.e 13 = \(\frac{\theta}{360^o}\)  x 2 x \(\frac{22}{7}\) x 4.2

= 13 x 360\(^o\) x 7

= \(\theta\) x 2 x 22 x 4.2

\(\theta\) = \(\frac{13 \times 360^o \times 7}{44 \times 4.2}\)

= \(\approx\) 177.27\(^o\)

\(\approx\) 177\(^o\) (to the nearest degree)