I need your help understanding how to go about the following problem,
Triangle ABC has sides AB = BC. It is inscribed in a circle with center O, radius 10.0 cm. Find the value of the angle BAC that produces a maximum area for triangle ABC.
I need your help understanding how to go about the following problem,
Triangle ABC has sides AB = BC. It is inscribed in a circle with center O, radius 10.0 cm. Find the value of the angle BAC that produces a maximum area for triangle ABC.
$Area = \dfrac{1}{2}(2y)(x+10) = y(x+10)$
$\dfrac{dA}{dx} = y + (x+10) \cdot \dfrac{dy}{dx}$
$x^2+y^2=10^2 \implies \dfrac{dy}{dx} = -\dfrac{x}{y}$ ...
$\dfrac{dA}{dx} = y + (x+10) \cdot \left(-\dfrac{x}{y} \right) = y - \dfrac{x^2}{y} - \dfrac{10x}{y}$
$y - \dfrac{x^2}{y} - \dfrac{10x}{y} = 0$
$y^2 - x^2 - 10x = 0$
$(10^2 - x^2) - x^2 - 10x = 0$
$2x^2 + 10x - 100 = 0 \implies 2(x+10)(x-5) = 0 \implies x = 5$
$m\angle{A} = \arctan\left(\dfrac{10+x}{y}\right)$
$\displaystyle R=10$
$\displaystyle a=2R \sin A$ and $\displaystyle b=2R \sin B=2R \sin (2A)$ since $\displaystyle B=180-2A$
Therefore the area is given by
$\displaystyle \frac{1}{2}(2R \sin A)* 2R \sin (2A)*\sin A$
$\displaystyle =2R^2 \sin ^2A *\sin (2A)$
differentiate and solve for $\displaystyle A$