Question

In a $\Delta ABC$ the sides b and c are given. If there is an error $\Delta A$ in measuring the angle A, then the error $\Delta a$ in side a is given by
A. $\dfrac{S}{2a}\Delta A$
B. $\dfrac{2S}{a}\Delta A$
C. $bc\text{ }\sin A\Delta A$
D. None of these

Answer 1

Hint: For solving this problem we must use the derivative property of variables. First, we express the relation between side length and angle subtended by side of triangle by using the cosine formula. Then, by using derivatives of both sides we obtain the desired relationship between change in angle and corresponding change in side.

Complete step-by-step answer:
Consider a triangle PQR with sides p, q, r and angles P, Q, R corresponding to each side. Now, by using the cosine angle property we define the relationship between cosine P and other sides as:
$\cos P=\dfrac{{{r}^{2}}+{{q}^{2}}-{{p}^{2}}}{2qr}$.
Consider a triangle ABC with sides a, b, c and angles A, B, C corresponding to each side. Now, by using the cosine angle property we define the relationship between cosine A and other sides as:
$\begin{aligned} & \cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc} \\\ & \therefore 2bc\ \cos A={{b}^{2}}+{{c}^{2}}-{{a}^{2}}\ldots (1) \\\ \end{aligned}$
Now, using the derivative property as stated below:
$\begin{aligned} & \dfrac{d}{dx}(\cos x)=-\sin x \\\ & \dfrac{d}{dx}({{x}^{n}})=n{{x}^{n-1}} \\\ \end{aligned}$
So, taking the derivative of equation (1) with respect to side $a$ and assigning the correct notation:
$\begin{aligned} & d(2bc\cos A)=d({{b}^{2}}+{{c}^{2}}-{{a}^{2}}) \\\ & -2bc\sin AdA=-2ada \\\ & bc\sin AdA=ada \\\ & \dfrac{2}{a}\left( \dfrac{1}{2}bc\sin A \right)dA=da\ldots (2) \\\ \end{aligned}$
Now, by using sine law for a triangle we establish the relation between half perimeter and sides of the triangle.
So, by using sine property for our problem we get, $S=\dfrac{1}{2}bc\sin A$.
Putting the obtained value of equation (3) in equation (2) we simplified our expression as:
$\begin{aligned} & \dfrac{2}{a}\left( \dfrac{1}{2}bc\sin A \right)dA=da \\\ & \dfrac{2S}{a}dA=da \\\ & \Delta a=\dfrac{2S}{a}\Delta A \\\ \end{aligned}$
Therefore, option (b) is correct.

Note: The key step in this problem is expression of cosine formula and sine formula for establishing the correct relationship. Use of derivative property proves to be a big plus for this problem-solving strategy.Student should know the important relation between half perimeter and sides of triangle for solving these types of problems