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In triangle $ABC$, $AB=c$, $BC=a$, $CA=b$, $\angle A = \alpha$, $\angle B = \beta$, and $\angle C = \gamma$ where all angles are measured in degrees. Suppose that $b^2-a^2=ac$ and $\beta^2 = \alpha \gamma$. Find $\lfloor 10\beta \rfloor$.

 Dec 18, 2020
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In triangle ABC, AB=c, BC=a, CA=b,
\(\angle A = \alpha\),
\(\angle B = \beta\), and \(\angle C = \gamma\)
where all angles are measured in degrees.
Suppose that \(b^2-a^2=ac\) and \(\beta^2 = \alpha \gamma\).
Find \(\lfloor 10\beta \rfloor\).

 

1.
\(\begin{array}{|rcll|} \hline b^2-a^2 &=&ac \\ (b-a)(b+a) &=& ac \\ \mathbf{(a-b)(a+b)} &=& \mathbf{-ac} \\ \hline \end{array}\)

 

Mollweide's formula

\(\begin{array}{|rcll|} \hline (a-b)\cos\left(\dfrac{\gamma}{2}\right) &=& c\cdot \sin\left(\dfrac{\alpha-\beta}{2} \right) \qquad (1) \\ (a+b)\sin\left(\dfrac{\gamma}{2}\right) &=& c\cdot \cos\left(\dfrac{\alpha-\beta}{2} \right) \qquad (2) \\ \hline \end{array}\)

 

(1) x (2)

\(\begin{array}{|rcll|} \hline (a-b)(a+b)\sin\left(\dfrac{\gamma}{2}\right)\cos\left(\dfrac{\gamma}{2}\right) &=& c^2\sin\left(\dfrac{\alpha-\beta}{2} \right)\cos\left(\dfrac{\alpha-\beta}{2} \right) \\ -ac\dfrac{\sin(\gamma)}{2} &=& c^2\dfrac{\sin(\alpha-\beta)}{2} \\ -a\sin(\gamma) &=& c\sin(\alpha-\beta) \\ a\sin(\gamma) &=& c\left(-\sin(\alpha-\beta)\right) \\ a\sin(\gamma) &=& c\left(\sin(-(\alpha-\beta))\right) \\ a\sin(\gamma) &=& c\sin(\beta-\alpha) \\ \dfrac{a}{c} &=& \dfrac{\sin(\beta-\alpha)}{\sin(\gamma)} \\ && \boxed{\dfrac{a}{c}=\dfrac{\sin(\alpha)} {\sin(\gamma)} } \\ \dfrac{\sin(\beta-\alpha)}{\sin(\gamma)} &=& \dfrac{\sin(\alpha)} {\sin(\gamma)} \\ \sin(\beta-\alpha) &=& \sin(\alpha) \\ \beta-\alpha &=& \alpha \\ \beta &=& 2\alpha \quad \text{or} \quad \mathbf{\alpha = \dfrac{\beta}{2}} \\ \hline \end{array}\)

 

2.

\(\begin{array}{|rcll|} \hline \beta^2 &=& \alpha \gamma \quad | \quad \mathbf{\alpha = \dfrac{\beta}{2}} \\ \beta^2 &=& \dfrac{\beta}{2} \gamma\\ \beta&=& \dfrac{\gamma}{2} \\ \mathbf{\gamma} &=& \mathbf{2\beta} \\ \hline \end{array}\)

 

3.

\(\begin{array}{|rcll|} \hline \mathbf{ \alpha + \beta + \gamma } &=& \mathbf{ 180^\circ } \quad | \quad \alpha = \dfrac{\beta}{2} \\ \dfrac{\beta}{2} + \beta + \gamma &=& 180^\circ \\ \dfrac{3}{2}\cdot \beta + \gamma &=& 180^\circ \quad | \quad \gamma = 2\beta \\ \dfrac{3}{2}\cdot\beta + 2\beta &=& 180^\circ \quad | \quad :2 \\ \dfrac{3}{4}\cdot\beta + \beta &=& 90^\circ \\ \dfrac{7}{4}\cdot\beta &=& 90^\circ \\ \beta &=& \dfrac{4}{7}\cdot 90^\circ \\ \beta &=& 51.4285714286^\circ \\ 10\cdot \beta &=& 514.285714286^\circ \\ \mathbf{ \lfloor 10\beta \rfloor } &=& \mathbf{ 514 } \\ \hline \end{array}\)

 

laugh

 Dec 19, 2020

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