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
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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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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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