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Kinda Pythagorean Theorem

Let {\rm AB \Gamma} be a triangle such that {\rm AB} = 10 , {\rm B \Gamma} = 22 and \hat{\rm B} = 2 \hat{\Gamma}. Find the area of the triangle.

Solution

Since \hat{\rm A} + \hat{\rm B} +\hat{\Gamma} = \pi \Rightarrow \hat{\rm A} + 3 \hat{\Gamma} = \pi it follows from the law of sines that

    \begin{align*} \frac{\alpha}{\sin \hat{\rm A}} = \frac{\beta}{\sin \hat{\rm B}} = \frac{\gamma}{\sin \Gamma} &\Rightarrow \frac{{\rm B \Gamma}}{\sin \hat{\rm A}} = \frac{\mathrm{AB}}{\sin \hat{\Gamma}}\\ &\Rightarrow \frac{22}{\sin \left ( \pi - 3 \hat{\Gamma} \right )} = \frac{10}{\sin \hat{\Gamma}} \\ &\Rightarrow \frac{22}{\sin 3\hat{\Gamma}} = \frac{10}{\sin \hat{\Gamma}} \\ &\Rightarrow \frac{22}{\sin \hat{\Gamma} \left ( 2 \cos 2 \hat{\Gamma} +1 \right ) } = \frac{10}{\sin \hat{\Gamma}} \\ &\Rightarrow 22 = 10 \left ( 2 \cos 2 \hat{\Gamma} + 1 \right ) \\ &\Rightarrow 12 = 20 \cos 2 \hat{\Gamma} \\ &\Rightarrow \frac{3}{5} = \cos \hat{\rm B} \end{align*}

Hence \displaystyle{\sin \hat{\rm B} = \sqrt{1- \cos^2 \hat{\rm B}} = \sqrt{1-\frac{9}{25}} = \frac{4}{5}}. Thus,

    \begin{align*} \mathcal{A} &= \frac{1}{2}\cdot 10 \cdot 22 \cdot \sin \hat{\rm B} \\ &=10 \cdot 11 \cdot \frac{4}{5} \\ &= 2 \cdot 4 \cdot 11\\ &= 88 \quad \text{square meters} \end{align*}

To completely justify the title of the post we give another solution based on the following proposition:

Proposition: Let ABC be a given triangle such that \hat{A} = 2 \hat{C}. Prove that

    \[a^2 = c^2 + bc\]

Proof: We are working on the following figure.

Rendered by QuickLaTeX.com

Let AD be the bisector of B\hat{A}C. Then:

    \begin{align*} \left\{ \begin{gathered} \vartriangle ADB \sim \vartriangle ABC \\ AD = DC \\ \end{gathered} \right. &\Rightarrow \left\{ \begin{gathered} \frac{{AD}} {{AC}} = \frac{{AB}} {{BC}} \\ AD = DC \\ \end{gathered} \right. \\ &\Rightarrow \frac{{DC}} {{AC}} = \frac{{AB}} {{BC}} \\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\overset{{DC = \frac{{a \cdot b}} {{b + c}},AC = B,AB = c,BC = a}}{=\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\!=\!\Rightarrow } \frac{{\frac{{a \cdot b}} {{b + c}}}} {b} = \frac{c} {a} \\ &\Rightarrow \frac{a} {{b + c}} = \frac{c} {a} \\ &\Rightarrow a^2 = c^2 + bc \end{align*}

and the result follows.

Now, the area of the initial triangle is given by Heron’s formula.

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