How do you work out the angle in this square?

Question

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.

score 2 · Answer 1 · answered Mar 25 '16 at 18:28

2

$\triangle PTS$ 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 $(180 - 30)=150^{\circ}$

answered Mar 25 '16 at 18:28

Deepak

26,801

1

By the way the "more interesting" version of the problem I've seen (which is somewhat more challenging) is to start with the 15 degree angle and thereby prove that the top triangle is equilateral. This must be done without trigonometry (that's the usual stipulation). – Deepak Mar 25 '16 at 18:37

score 2 · Answer 2 · answered Mar 25 '16 at 19:35

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.

SiXUlm · Accepted Answer · 2016-03-26T08:39:12.760

1

$PST$ 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 $\angle SRT = \angle TSR = 15^{\circ}$.

edited Mar 26 '16 at 08:39

answered Mar 25 '16 at 18:05

SiXUlm

2,436

What does the term a stand for? – itshanks Mar 25 '16 at 18:10
I've edited my answer. – SiXUlm Mar 25 '16 at 18:13
Thanks. How do you know that $PST$ and $QTR$ are isosceles triangles? – itshanks Mar 25 '16 at 18:16
$PQT$ is equilateral triangle, thus $PT = PQ = a$. But $PQRS$ is a square, thus $PQ = PS = a$, thus $PT = PS = a$. Same reasoning for $QTR$. – SiXUlm Mar 25 '16 at 18:19
I think you meant in your last line $;\angle SRT=\angle\color{red}{RST}=\color{green}{15^\circ};$ – DonAntonio Mar 25 '16 at 18:28
I fail to see the relevance of the first (trigonometric) part. – Deepak Mar 25 '16 at 18:45
@Joanpemo: ah yes, you are correct. I edited my answer. Thanks! – SiXUlm Mar 25 '16 at 19:04
@Deepak: the first part is actually what the question (in the photo) asked. – SiXUlm Mar 25 '16 at 19:05
@SiXUlm I might be blind, but I don't see any part of the question asking for the area of the triangle SRT? You can immediately get the angle PST without trigonometry or computing the area. – Deepak Mar 26 '16 at 00:46
@Deepak: actually you are correct, I misread the question. Also the first part has nothing to do with the second part. I edited my answer. – SiXUlm Mar 26 '16 at 08:38
@SiXUlm Thanks. – Deepak Mar 26 '16 at 10:01

score 1 · Answer 4 · answered Mar 25 '16 at 18:14

1

In triangle PST and QTR: PS=PT=QR, therefore angles PTS and QTR are equal $75^0$. Angle STR is equal to $150^0$

answered Mar 25 '16 at 18:14

alans

6,475

score 1 · Answer 5 · answered Mar 25 '16 at 18:24

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 $\;\Delta TRF\;$ :

$$\tan\angle TRS=\frac{|TF|}{\frac x2}=2-\sqrt3\implies\angle TRS=15^\circ\implies\angle STR=180^\circ-2\cdot15^\circ=150^\circ\;$$

score 1 · Answer 6 · answered Mar 25 '16 at 18:42

1

Assume square has side =$2$. Draw a central line and consider half the bottom broad isosceles triangle.

$$ \cot \theta = 2 - \sqrt3, \tan \theta = 2 + \sqrt3, \theta =75^0, 2\theta = 150. $$

answered Mar 25 '16 at 18:42

Narasimham

40,495

It's "nicer" to do this without trigonometry although without any such stipulation, it's a perfectly valid method. But if one is disallowed the use of a calculator, how would one find the angle from that "lesser known" surd ratio? – Deepak Mar 25 '16 at 18:49
In some books trig functions of 30, 60,15, 75,18,36 degrees appear early on. – Narasimham Mar 25 '16 at 19:03

Some Guy With a Hat · Answer 7 · 2016-03-25T18:29:51.223

-3

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!

edited Mar 25 '16 at 18:29

answered Mar 25 '16 at 18:04

Some Guy With a Hat

1

1

No. They are not opposite angles. The lines are purposely crooked. – itshanks Mar 25 '16 at 18:06

How do you work out the angle in this square?

7 Answers7