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I have exams for primary teaching maths coming up. And the tasks aare always at high level, so I need some help from you. Please can you help me to solve this geometric task? w is the angle bisector. Thank you so much, Sophia

enter image description here

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what have you tried? Where are you having problems? –  user31280 Feb 23 '13 at 12:41
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Do you happen to know what $w$ is? Or are you solving for $\varepsilon$ in terms of $w$? –  paraxor Feb 23 '13 at 12:45
    
the diagram is ambiguous and no assertion can be made about the exact value of $\varepsilon$. –  user31280 Feb 23 '13 at 12:46
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I’ve virtually no doubt that the dotted angles are right angles. However, $\epsilon$ is still indeterminate unless there’s some information available about $w$. –  Brian M. Scott Feb 23 '13 at 12:55
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I bet, it is the bisector of the angle. –  Berci Feb 23 '13 at 13:17
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2 Answers 2

Labeled diagram

If $w$ is a line bisecting an angle

We can find the angle in the upper-left hand corner of the diagram to be $51^\circ$ ($180-90-39=51$). Since we're assuming $w$ bisects that angle, the two angles between $w$ and the two relevant lines of the triangle are equal. This gives them each a value of $\frac{51}{2}=25.5$. Using the parallel postulate, we can then find $\varepsilon$ to be $180-(180-90-25.5)=115.5^\circ$.

If $w$ is an angle

If we assume $w$ is an angle (which it probably is not, due to its lack of angle notation), we can solve for $\varepsilon$ in terms of $w$.

We know that $A$ is $90^\circ$ ($A = 180-90$)

We can form a triangle with the known angle and angles $A$ and $B$. Since the angles of a triangle always sum to $180^\circ$, we can find $B$ as such: $$B = 180-39-A = 180-39-90 = 51.$$

We know that $C = w$ and with the knowledge that $B$ is $51^\circ$, we can find

$$D=180-51-w = 129-w$$

Then we can find

$$\varepsilon = 180-(129-w) = 51 + w$$

If you are ever informed of the value of $w$, just plug it into the last equation. For example, if you knew $w=10$, $\varepsilon=51+10=61$.

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I think $w$ denotes a line, probably the bisector of the angle. In that case $C=90-(51/2)$. –  Berci Feb 23 '13 at 13:20
    
I will soon extend my answer to solve for $\varepsilon$ with the assumption that $w$ is the bisector of the farthest upper-left hand angle of the triangle. –  paraxor Feb 23 '13 at 14:40
    
Hey yes thats right, it´s supposed to be a bisector. Sorry i forgot to write it into the discription. thank you so much for your effort already –  Sophia Feb 23 '13 at 19:32
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up vote -1 down vote accepted

After long thinking I found the solution, its 115,5 degree. If anyone is interested in the way of solving, let me know. Best and thanks, Sophia

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