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1)In rectangle ABCD, we have AD = 3 and AB = 4. Let M be the midpoint of AB, and let X be the point such that MD = MX, angle MDX = 77 degrees, and A and X lie on opposite sides of DM. Find angle XCD, in degrees.

Picture: https://latex.artofproblemsolving.com/1/b/3/1b39a7de7d5064e27d8ac7d0e00bb8b0db806199.png

 

2)H is the orthocenter of acute triangle ABC and the extensions of AH, BH, and CH intersect the circumcircle of traingle ABC at A prime, B prime and C prime. We know angle AHB : angle BHC : angle CHA = 9 : 10 : 11. Find angle AprimeBprimeCprime in degrees.

Picture: https://latex.artofproblemsolving.com/6/1/1/6119f02c59ea35f11dceab068d51811cace24c1d.png

FiestyGeco  Feb 20, 2018

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 #1
avatar+19376 
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1)
In rectangle ABCD, we have AD = 3 and AB = 4.
Let M be the midpoint of AB, and let X be the point such that MD = MX, angle MDX = 77 degrees,
and A and X lie on opposite sides of DM. Find angle XCD, in degrees.

Picture: https://latex.artofproblemsolving.com/1/b/3/1b39a7de7d5064e27d8ac7d0e00bb8b0db806199.png

 

 

 

\(\text{Let $\angle XCD = x$ } \\ \text{Let $\angle MDX = \angle DXM = 77^{\circ}$ } \\ \text{Let $\angle MDC=\angle MCD = \beta $ } \\ \text{Let $\angle MXC=\angle MCX = \gamma $ } \\ \text{Let $\angle DMC = 180^{\circ} - 2 \times \beta $} \\ \text{Let $\angle XMC = 180^{\circ} - 2 \times \gamma $} \)

 

\(\begin{array}{|rcll|} \hline \angle DMX &=& 180^{\circ} - 2 \times 77^{\circ} \\ &=& 26^{\circ} \\\\ \angle XMC &=& \angle DMC - 26^{\circ} \\ &=& 180^{\circ} - 2 \times \beta - 26^{\circ} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \angle XMC = 180^{\circ} - 2 \times \beta - 26^{\circ} &=& 180^{\circ} - 2 \times \gamma \\ 180^{\circ} - 2 \times \beta - 26^{\circ} &=& 180^{\circ} - 2 \times \gamma \quad & | \quad \gamma = \beta + x \\ 180^{\circ} - 2 \times \beta - 26^{\circ} &=& 180^{\circ} - 2 \times (\beta + x) \\ 180^{\circ} - 2 \times \beta - 26^{\circ} &=& 180^{\circ} - 2 \times \beta -2\times x \\ - 26^{\circ} &=& -2\times x \\ 26^{\circ} &=& 2\times x \quad & | \quad :2 \\ \mathbf{13^{\circ}} &=& \mathbf{ x } \\ \hline \end{array}\)

 

The angle XCD is \(\mathbf{13^{\circ}}\)

 

laugh

heureka  Feb 21, 2018
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2+0 Answers

 #1
avatar+19376 
+1
Best Answer

1)
In rectangle ABCD, we have AD = 3 and AB = 4.
Let M be the midpoint of AB, and let X be the point such that MD = MX, angle MDX = 77 degrees,
and A and X lie on opposite sides of DM. Find angle XCD, in degrees.

Picture: https://latex.artofproblemsolving.com/1/b/3/1b39a7de7d5064e27d8ac7d0e00bb8b0db806199.png

 

 

 

\(\text{Let $\angle XCD = x$ } \\ \text{Let $\angle MDX = \angle DXM = 77^{\circ}$ } \\ \text{Let $\angle MDC=\angle MCD = \beta $ } \\ \text{Let $\angle MXC=\angle MCX = \gamma $ } \\ \text{Let $\angle DMC = 180^{\circ} - 2 \times \beta $} \\ \text{Let $\angle XMC = 180^{\circ} - 2 \times \gamma $} \)

 

\(\begin{array}{|rcll|} \hline \angle DMX &=& 180^{\circ} - 2 \times 77^{\circ} \\ &=& 26^{\circ} \\\\ \angle XMC &=& \angle DMC - 26^{\circ} \\ &=& 180^{\circ} - 2 \times \beta - 26^{\circ} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \angle XMC = 180^{\circ} - 2 \times \beta - 26^{\circ} &=& 180^{\circ} - 2 \times \gamma \\ 180^{\circ} - 2 \times \beta - 26^{\circ} &=& 180^{\circ} - 2 \times \gamma \quad & | \quad \gamma = \beta + x \\ 180^{\circ} - 2 \times \beta - 26^{\circ} &=& 180^{\circ} - 2 \times (\beta + x) \\ 180^{\circ} - 2 \times \beta - 26^{\circ} &=& 180^{\circ} - 2 \times \beta -2\times x \\ - 26^{\circ} &=& -2\times x \\ 26^{\circ} &=& 2\times x \quad & | \quad :2 \\ \mathbf{13^{\circ}} &=& \mathbf{ x } \\ \hline \end{array}\)

 

The angle XCD is \(\mathbf{13^{\circ}}\)

 

laugh

heureka  Feb 21, 2018
 #2
avatar+19376 
0

2)

H is the orthocenter of acute triangle ABC and the extensions of AH, BH, and CH
intersect the circumcircle of traingle ABC at A prime, B prime and C prime.
We know angle AHB : angle BHC : angle CHA = 9 : 10 : 11.
Find angle AprimeBprimeCprime in degrees.
Picture: https://latex.artofproblemsolving.com/6/1/1/6119f02c59ea35f11dceab068d51811cace24c1d.png

 

\(\text{Let $\angle A'B'C' = \angle CAB = x$ } \\ \text{Let $\angle ACB= \alpha $ } \\ \text{Let $\angle BAC= \beta $ } \\ \text{Let $\angle CBA= \gamma $ } \\ \text{Let $\angle AHB = \gamma + \beta $} \\ \text{Let $\angle BHC = \alpha + \gamma $} \\ \text{Let $\angle CHA = \beta + \alpha $} \\ \text{Let $ \mathbf{x = 180^{\circ}-2 \times\gamma }$} \)

 

In triangle ABC we set the angles \( \alpha, \beta \text{ and } \gamma\)

 

further:

\(\mathbf{ \large{x =180^{\circ} - 2 \times \gamma} } \)

 

 

\(\begin{array}{|rcrcrcrcrcr|} \hline \angle AHB &:& \angle BHC &:& \angle CHA &=& 9 &:& 10 &:& 11 \\ \gamma + \beta &:& \alpha + \gamma &:& \beta + \alpha &=& 9 &:& 10 &:& 11 \\ \hline \end{array}\)

 

\(\begin{array}{rclrclrcl} \dfrac{\alpha + \gamma }{\beta + \alpha} &=& \dfrac{10}{11} \qquad & \qquad \dfrac{\gamma + \beta}{\beta + \alpha} &=& \dfrac{9}{11} \\\\ \dfrac{\alpha + \gamma +\gamma + \beta }{\beta + \alpha} &=& \dfrac{10+9}{11} \\\\ \dfrac{\alpha + \beta + 2\times\gamma }{\alpha + \beta } &=& \dfrac{19}{11} \quad &| \quad \alpha + \beta + \gamma = 180^{\circ} \\\\ & & \quad &| \quad \alpha + \beta = 180^{\circ} - \gamma \\\\ \dfrac{180^{\circ} - \gamma + 2\times\gamma }{180^{\circ} - \gamma } &=& \dfrac{19}{11} \\\\ \dfrac{180^{\circ} + \gamma }{180^{\circ} - \gamma } &=& \dfrac{19}{11} \\\\ 11\times(180^{\circ} + \gamma ) &=&19\times(180^{\circ} - \gamma ) \\ 11\times 180^{\circ} + 11\gamma &=&19\times 180^{\circ} - 19\gamma \\ 30\gamma &=&8\times 180^{\circ}\\ \mathbf{\gamma} &\mathbf{=}& \mathbf{48^{\circ}} \\ \end{array} \)

 

\(\begin{array}{|rcll|} \hline \angle A'B'C' = x &=& 180^{\circ} - 2 \times \gamma \\ \angle A'B'C' &=& 180^{\circ} - 2 \times \gamma \\ &=& 180^{\circ} - 2 \times 48^{\circ} \\ &=& 180^{\circ} - 96^{\circ} \\ &=& \mathbf{84^{\circ}} \\ \hline \end{array} \)

 

\(\text{The angle $A'B'C'$ in degrees is $\mathbf{84^{\circ}}$ } \)

 

\(\begin{array}{rcll} \dfrac{\alpha+\gamma}{\alpha+\beta} &=& \dfrac{10}{11} \quad & | \quad \alpha+ \beta=180^{\circ}- \gamma = 180^{\circ}-84^{\circ} = 132^{\circ} \\\\ \dfrac{\alpha+48^{\circ} }{132^{\circ}} &=& \dfrac{10}{11} \\\\ \alpha &=& \dfrac{10\times 132^{\circ}}{11} -48^{\circ} \\\\ \mathbf{ \alpha } & \mathbf{=} & \mathbf{72^{\circ}} \\ \end{array}\)

 

\(\begin{array}{rcll} \alpha+\beta &=& 132^{\circ} \\ \beta &=& 132^{\circ}-\alpha \\ \beta &=& 132^{\circ}-72^{\circ} \\ \mathbf{ \beta } & \mathbf{=} & \mathbf{60^{\circ}} \\ \end{array}\)

 

\(\begin{array}{|rcll|} \hline \angle AHB = \gamma + \beta = 108^{\circ} \\ \angle BHC = \alpha + \gamma = 120^{\circ} \\ \angle CHA = \beta + \alpha = 132^{\circ} \\ \hline \end{array} \)

 

 

laugh

heureka  Feb 22, 2018
edited by heureka  Feb 22, 2018

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