Questions: If (a, b, c) are the sides of a triangle and (A, B, C) are the opposite angles, find (partial A / partial a, partial A / partial b, partial A / partial c) by implicit differentiation of the Law of Cosines. (partial A / partial a=) (partial A / partial b=) (partial A / partial c=)

If (a, b, c) are the sides of a triangle and (A, B, C) are the opposite angles, find (partial A / partial a, partial A / partial b, partial A / partial c) by implicit differentiation of the Law of Cosines.

(partial A / partial a=)
(partial A / partial b=)
(partial A / partial c=)

Transcript text: If $a, b, c$ are the sides of a triangle and $A, B, C$ are the opposite angles, find $\partial A / \partial a, \partial A / \partial b, \partial A / \partial c$ by implicit differentiation of the Law of Cosines. \[ \begin{array}{l} \partial A / \partial a= \\ \partial A / \partial b= \\ \partial A / \partial c= \end{array} \]

Solution

Solution Steps

To find the partial derivatives of angle $ A $ with respect to the sides $ a, b, $ and $ c $ using the Law of Cosines, follow these steps:

Law of Cosines: Start with the equation $ c^2 = a^2 + b^2 - 2ab\cos(A) $.
Implicit Differentiation: Differentiate both sides of the equation with respect to $ a $, $ b $, and $ c $ to find the partial derivatives.
Solve for $\partial A / \partial a$, $\partial A / \partial b$, and $\partial A / \partial c$: Rearrange the differentiated equations to express the partial derivatives in terms of $ a, b, c, $ and $ A $.

Step 1: Apply the Law of Cosines

The Law of Cosines for a triangle with sides $a$, $b$, and $c$ and opposite angles $A$, $B$, and $C$ is given by:

\[ c^2 = a^2 + b^2 - 2ab \cos A \]

Step 2: Differentiate with Respect to $a$

To find $\frac{\partial A}{\partial a}$, differentiate both sides of the equation with respect to $a$:

\[ 0 = 2a - 2b \cos A + 2ab \sin A \cdot \frac{\partial A}{\partial a} \]

Rearrange to solve for $\frac{\partial A}{\partial a}$:

\[ 2ab \sin A \cdot \frac{\partial A}{\partial a} = 2b \cos A - 2a \]

\[ \frac{\partial A}{\partial a} = \frac{b \cos A - a}{ab \sin A} \]

Step 3: Differentiate with Respect to $b$

To find $\frac{\partial A}{\partial b}$, differentiate both sides of the equation with respect to $b$:

\[ 0 = 2b - 2a \cos A + 2ab \sin A \cdot \frac{\partial A}{\partial b} \]

Rearrange to solve for $\frac{\partial A}{\partial b}$:

\[ 2ab \sin A \cdot \frac{\partial A}{\partial b} = 2a \cos A - 2b \]

\[ \frac{\partial A}{\partial b} = \frac{a \cos A - b}{ab \sin A} \]

Step 4: Differentiate with Respect to $c$

To find $\frac{\partial A}{\partial c}$, differentiate both sides of the equation with respect to $c$:

\[ 2c = 2ab \sin A \cdot \frac{\partial A}{\partial c} \]

Rearrange to solve for $\frac{\partial A}{\partial c}$:

\[ \frac{\partial A}{\partial c} = \frac{c}{ab \sin A} \]

Final Answer

\[ \begin{array}{l} \boxed{\frac{\partial A}{\partial a} = \frac{b \cos A - a}{ab \sin A}} \\ \boxed{\frac{\partial A}{\partial b} = \frac{a \cos A - b}{ab \sin A}} \\ \boxed{\frac{\partial A}{\partial c} = \frac{c}{ab \sin A}} \end{array} \]

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