Q.8. The approximate value pof sin (60o10”) is ( given that \sqrt{3}=1.732 ,1o=0.0175o)
A. 0.08660243
B. 0.0008660243
C. 0.8660243
D. 0.008660243
Answer :- C. 0.8660243
Explanation :- Let f(x) = \sin x
f''(x) = \cos xHere, a = 60^\circ and
h = 10^{\prime\prime} = \left(\frac{1}{360}\right)^{\circ} = \frac{1}{360} \times 0.0175^{c} = 0.000049^{c} f(a) = \sin(60^\circ) = \frac{\sqrt{3}}{2} = \frac{1.732}{2} = 0.866 f'(a) = \cos(60^\circ) = \frac{1}{2} = 0.5 f(a + h) \approx f(a) + h f'(a) \sin(60^\circ \times 10^{\prime\prime}) \approx 0.866 + 0.000049 \times 0.5 \approx 0.866024