Acute triangle

In the acute triangle KLM, V is the intersection of its heights and X is the heel of height to the side KL. The axis of the angle XVL is parallel to the side LM and the angle MKL is 70°. What size are the KLM and KML angles?

Correct result:

KLM =  55 °
KML =  55 °

Solution:

$$\begin{gathered} \mathrm{\angle }MKL=70^{\circ } \\ \mathrm{\angle }MKL+KLM+KML=180^{\circ } \\ \delta =90^{\circ }-KLM \\ \mathrm{\angle }KMX=90-\mathrm{\angle }MKL=90-70=20^{\circ } \\ 2\delta =90^{\circ }-\mathrm{\angle }KMX \\ \delta =(90-\mathrm{\angle }KMX)\mathrm{/}2=(90-20)\mathrm{/}2=35^{\circ } \\ \mathrm{\angle }KLM=90-\delta =90-35=55^{\circ } \end{gathered}$$

$\mathrm{\angle }KML=180-\mathrm{\angle }MKL-\mathrm{\angle }KLM=180-70-55=55^{\circ }$

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