The area of a triangle is computed using the formula s=12bc sin A. If the relative errors made in measuring b, c and calculating S are respectively 0.02, 0.01 and 0.13 the approximate error in A when A=π/6, is

A
0.05 radians
B
0.01 radians
C
0.05 degree
D
0.01 degree

Solution:

We have,

$\begin{array}{l} S = \frac{1}{2} b c \sin A \\ \Rightarrow d S = \frac{1}{2} d (b c \sin A) \\ \Rightarrow d S = \frac{1}{2} {c \sin A d b + b \sin A b c + b c \cos A d A} \\ \Rightarrow \frac{d S}{S} = \frac{d b}{b} + \frac{d c}{c} + \cot A d A \\ \Rightarrow 0.13 = 0.02 + 0.01 + \cot \frac{π}{6} d A [∵ \frac{d S}{S} = 0.13, \frac{d b}{b} = 0.02] \\ \Rightarrow 0.10 = \sqrt{3} d A \\ \Rightarrow d A = \frac{0.10}{\sqrt{3}} = 0.05 radians \end{array}$

The area of a triangle is computed using the formula $s = \frac{1}{2} b c \sin A .$ If the relative errors made in measuring $b, c$ and calculating $S$ are respectively 0.02, 0.01 and 0.13 the approximate error in $A$ when $A = π / 6,$ is

Solution:

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