Question

The area of a triangle is computed using the formula \[{\text{S = }}\dfrac{1}{2}{\text{bc }}\sin {\text{A}}\]. If the relative errors made in measuring b, c and calculating S are \[0.02\], \[0.01\] and \[0.13\] respectively the approximate error in A when \[{\text{A = }}\dfrac{\pi }{6}\] is
A) \[0.05\] Radians
B) \[0.01\] Radians
C) \[0.05\] Degree
D) \[0.01\] Degree

Answer 1

Hint:
The formula to calculate the error is
\[\dfrac{{\Delta {\text{S}}}}{{\text{S}}} = \dfrac{{\Delta {\text{b}}}}{{\text{b}}} + \dfrac{{\Delta {\text{c}}}}{{\text{c}}} + \dfrac{{\Delta \sin {\text{A}}}}{{\sin {\text{A}}}}\] , where \[\dfrac{{\Delta {\text{S}}}}{{\text{S}}}\] is the relative error for calculating S, \[\dfrac{{\Delta {\text{b}}}}{{\text{b}}}\] is the relative in measuring b,\[\dfrac{{\Delta {\text{c}}}}{{\text{c}}}\] is the relative error in measuring c and \[\dfrac{{\Delta \sin {\text{A}}}}{{\sin {\text{A}}}}\] is the relative error for \[\sin {\text{A}}\].
Apply this formula, and then use the given conditions to find the required value.

Complete step by step solution:
Given that the area of a triangle is computed using the formula \[{\text{S = }}\dfrac{1}{2}{\text{bc }}\sin {\text{A}}\].

It is given that the relative errors made in measuring b, c and calculating S are \[0.02\], \[0.01\] and \[0.13\] respectively.

We know that error formula for the given conditions is \[\dfrac{{\Delta {\text{S}}}}{{\text{S}}} = \dfrac{{\Delta {\text{b}}}}{{\text{b}}} + \dfrac{{\Delta {\text{c}}}}{{\text{c}}} + \dfrac{{\Delta \sin {\text{A}}}}{{\sin {\text{A}}}}\], where \[\dfrac{{\Delta {\text{S}}}}{{\text{S}}}\] is the relative error for calculating S, \[\dfrac{{\Delta {\text{b}}}}{{\text{b}}}\] is the relative in measuring b,\[\dfrac{{\Delta {\text{c}}}}{{\text{c}}}\] is the relative error in measuring c and \[\dfrac{{\Delta \sin {\text{A}}}}{{\sin {\text{A}}}}\] is the relative error for \[\sin {\text{A}}\].

We will now find the value of \[\dfrac{{\Delta {\text{S}}}}{{\text{S}}}\], where \[0.13\] is the relative error in calculating S.

\[\dfrac{{\Delta {\text{S}}}}{{\text{S}}} = 0.13\]

We will now find the value of \[\dfrac{{\Delta {\text{b}}}}{{\text{b}}}\], where \[0.02\] is the relative error in measuring b.

\[\dfrac{{\Delta {\text{b}}}}{{\text{b}}} = 0.02\]

We will now find the value of \[\dfrac{{\Delta {\text{c}}}}{{\text{c}}}\], where \[0.01\] is the relative error in measuring c.

\[\dfrac{{\Delta {\text{c}}}}{{\text{c}}} = 0.01\]

Replacing \[\dfrac{\pi }{6}\] for \[{\text{A}}\] in the expression \[\sin {\text{A}}\], we get
\[\sin \dfrac{\pi }{6} = \dfrac{1}{2}\]

Substituting the above values in the above error formula, we get

\[
   \Rightarrow 0.13 = 0.02 + 0.01 + \dfrac{{\Delta \sin {\text{A}}}}{{\dfrac{1}{2}}} \\
   \Rightarrow 0.13 = 0.03 + \dfrac{{\Delta \sin {\text{A}}}}{{\dfrac{1}{2}}} \\
   \Rightarrow 0.13 = 0.03 + 2\Delta \sin {\text{A}} \\
\]

Subtracting the above equation by \[0.03\] on each of the sides, we get

\[
   \Rightarrow 0.13 - 0.03 = 0.03 + \Delta \sin {\text{A}} - 0.03 \\
   \Rightarrow 0.10 = 2\Delta \sin {\text{A}} \\
\]

Dividing the above equation by 2 on each of the sides, we get

\[
   \Rightarrow \dfrac{{2\Delta \sin {\text{A}}}}{2} = \dfrac{{0.10}}{2} \\
   \Rightarrow \Delta \sin {\text{A}} = 0.05 \\
\]
Therefore, the approximate error in A when \[{\text{A = }}\dfrac{\pi }{6}\] is \[0.05\].

Hence, option A is correct.

Note:
In solving these types of questions, you should be familiar with the formula of calculating the error and relative errors. We use the given values as derivatives to solve these types of questions because derivatives provide us a way of estimating the number of function changes when a small change occurs in any input value. Here, a student may go wrong while calculating the approximate error for \[\Delta \sin {\text{A}}\] by not taking it into the sum. Also, we are supposed to write the values properly to avoid any miscalculation.