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The lines $l_1$, $l_2$ and $l_3$ lie in an inclined plane $P$ and pass through a common point $A$. The line $l_2$ is a line of greatest slope in $P$. The line $l_1$ is perpendicular to $l_3$ and makes an acute angle $\alpha$ with $l_2$. The angles between the horizontal and $l_1$, $l_2$ and $l_3$ are $\pi$/6, $\beta$ and $\pi$/4, respectively. Show that $\hskip .2cm$ cos$\alpha$ sin$\beta$ = 1/2 and find the value of sin$\alpha$ sin$\beta$.

(STEP 2002 2.paper 6. question)

The solution is this (only the first part relevant): I don't understand why AB is parallel to AC, thus CAcosa= AB. Could anyone explain?

[1]: https://integralmaths.org/pluginfile.php/181540/mod_resource/content/0/02-2-06.jpg

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    Could you please link the question or include a picture of it? – 2017-02-06
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    http://www.damtp.cam.ac.uk/user/stcs/STEP/2002paperII.pdf – 2017-02-06

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The brown triangle is 45-45-90 and hence its sides are $1,\sqrt2, 1$

enter image description here

The blue triangle is 30-60-90 and hence its sides are $1, 2,\sqrt3 $

Let the length of the line of greatest slope = y

Added:enter image description here

$(\cos \alpha) \times [\sin \beta] = (\dfrac y2)\times [\dfrac 1y] = 1/2$

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    I'm having trouble seeing why the green triangle is right-angled. Could you please point it out? Will it still be if y is smaller/bigger than on the drawing? – 2017-02-06
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    @szantamano See added. It is independent of the length of y because the bottom edge and the top edge of the slope are parallel. – 2017-02-07