In a triangle ABC, the sides b and c are fixe | vedclass

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(C) Using the Law of Cosines: $a^2 = b^2 + c^2 - 2bc \cos A$. Differentiating both sides with respect to $A$, we get: $2a \delta a = 2bc \sin A \delta

Vedclass · Accepted Answer

$\frac{\Delta \delta A}{2 R^2 \sin^2 A} 	imes 100$

Vedclass · Answer

(C) Using the Law of Cosines: $a^2 = b^2 + c^2 - 2bc \cos A$. Differentiating both sides with respect to $A$, we get: $2a \delta a = 2bc \sin A \delta A$. Thus, $\delta a = \frac{bc \sin A \delta A}{a}$. We know that the area of the triangle is $\Delta = \frac{1}{2} bc \sin A$, so $bc \sin A = 2\Delta$. Also, by the Sine Rule, $\frac{a}{\sin A} = 2R$, which implies $a = 2R \sin A$. Substituting these into the expression for $\delta a$: $\delta a = \frac{2\Delta \delta A}{2R \sin A} = \frac{\Delta \delta A}{R \sin A}$. The percentage error is $\frac{\delta a}{a} 	imes 100 = \frac{\Delta \delta A}{R \sin A \cdot 2R \sin A} 	imes 100 = \frac{\Delta \delta A}{2R^2 \sin^2 A} 	imes 100$.

In a $\triangle ABC$, the sides $b$ and $c$ are fixed. If there is an error of $\delta A$ in measuring angle $A$, then the percentage error in measuring the length of the side $a$ is:

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