I have labelled all the angles that I can work out.
But I can't think of any other way to find the other angles without being 100% sure.
Thank you!
P.S. I have attached the official question - I have scribbled all over it with lines and therefore have drew out again by hand to make things clearer.
$\endgroup$7 Answers
$\begingroup$$\triangle PTS$ is isosceles because $PT =PS (= PQ) $. Hence $\angle PST = \frac{(180 - 30)}{2}=75^{\circ}$ and $\angle TSR = 90 - 75 = 15^{\circ}$. By symmetry $\triangle STR$ is isosceles so the required angle is $(180 - 30)=150^{\circ}$
$\endgroup$ 1 $\begingroup$You could also drawn a line perpedicular to $PS $ and intersecting $T$. Now, denote the point of intersection on $PS$ as $A$ and with $QR$ as $B$.
Without losing generality, assume that $PS=1$. Thus, $AS=1-\frac{\sqrt{3}}{2}$ and $AT=\frac{1}{2}$ using the properties of a $30-60-90$ degree triangle.
Now, $\angle ATS=\angle RTB $. It follows that:
$$\tan \angle ATS=\frac{1-\frac{\sqrt{3}}{{2}}}{\frac{1}{2}}=2-\sqrt{3}$$
Thus,
$$\angle ATS=\arctan({2-\sqrt{3}})=15^o$$
Now, calculate $\angle STR=180^o-2(\angle ATS)=150^o$.
Finally conclude that $\angle TSR=\angle TRS=15^o$ by the properties of isosceles triangles.
$\endgroup$ $\begingroup$$PST$ and $QTR$ are $2$ isosceles triangles, you can easily find $\angle PTS = \angle QTR = 75^{\circ}$. Thus, $\angle STR = 360^{\circ} - 60^{\circ} - 2 \times 75^{\circ} = 150^{\circ}$.
$STR$ is isosceles triangle, thus $\angle SRT = \angle TSR = 15^{\circ}$.
$\endgroup$ 11 $\begingroup$In triangle PST and QTR: PS=PT=QR, therefore angles PTS and QTR are equal $75^0$. Angle STR is equal to $150^0$
$\endgroup$ $\begingroup$Observe that $\;\Delta RTQ\cong\Delta STP\;$ by SAS, from which $\;\angle TRQ=\angle TSP\;$ and thus also $\;\angle TRS=\angle TSR\;$ .
Now, since $\;\Delta PTQ\;$ is equilateral, if $\;TD\;$ is the height to $\;PQ\;$, then $\;|TD|=\frac{\sqrt3}2x\;,\;\;x=$ the square's side, so the height $\;TF\;$ to $\;SR\;$ is $\;\Delta TSR\;$ fulfills $\;|TF|=x-\frac{\sqrt3}2x=\frac{2-\sqrt3}2x\;$ , and then in the streaight triangle $\;\Delta TRF\;$ :
$$\tan\angle TRS=\frac{|TF|}{\frac x2}=2-\sqrt3\implies\angle TRS=15^\circ\implies\angle STR=180^\circ-2\cdot15^\circ=150^\circ\;$$
$\endgroup$ $\begingroup$Assume square has side =$2$. Draw a central line and consider half the bottom broad isosceles triangle.
$$ \cot \theta = 2 - \sqrt3, \tan \theta = 2 + \sqrt3, \theta =75^0, 2\theta = 150. $$
$\endgroup$ 2 $\begingroup$Ok, This is the way I found.
- Construct a segment parallel to SR, with midpoint T. This creates 2 right triangles.
- Extend ST or TR, creating an Isosceles triangle inside one of the new right triangles.
- Bisect the isosceles triangle into 2 congruent right triangles. You will note that both of them will be 30˚/60˚/90˚ triangles
- This shows you that angle STR has two compliments of 30˚
- Angle STR must be 120˚
Hope that helps!
$\endgroup$ 1
