Solve the problem in the given cyclic quadrilateral

[chwala]

Homework Statement

Solve the problem in the given cyclic quadrilateral

Relevant Equations

cyclic quadrilateral

now for part $19.1$,
My approach is as follows, using cosine rule;
$DF= r^2 + r^2- 2r^2 cos E$
We know that angle $E$ + angle $∅$= $180^0$
$∅$ is acute, therefore angle $E$ would be negative. (If $∅=60^0$ for e.g then it follows that $E=120^0$) Thus we shall have,
$DF^2= r^2 + r^2+2r^2 cos∅$
$DF^2=2r^2+2r^2 cos ∅$
$DF= \sqrt {r^2(2+2 cos ∅)}$
$DF= r\sqrt {(2+2 cos ∅)}$ is the approach correct?

Now for part $19.2$, ...this was a bit confusing to me...but i went ahead and proved lhs=rhs
To prove $2 sin^2 ∅=1 + cos ∅$
$2 sin^2 ∅=2(1- cos^2 ∅)=2(1+cos ∅)(1-cos ∅)$ =$1 + cos ∅$
$2(1-cos ∅)=1$
$1-cos ∅=\frac {1}{2}$
→$cos ∅$=$\frac {1}{2}$
on subsitituting, $cos ∅$=$\frac {1}{2}$ on both sides of the equation, we have,
$2(1-\frac {1}{4})=1+ \frac {1}{2}$
$\frac {3}{2}$=$\frac {3}{2}$ thus proved. Is this correct approach or there is a better way to prove this?

For part $19.3$
If $cos ∅$=$\frac {1}{2}$, then $∅= 60^0$

 

[Frabjous]

I am not sure how to do it, but 19.2 is not a proof. You are just assuming that it is true. It is not true in general, just for this particular construction.

 

[chwala]

I am not sure how to do it, but 19.2 is not a proof. You are just assuming that it is true. It is not true in general, just for this particular construction.

I managed to cancel out $1+cos ∅$ on both sides of the equation, no assumptions...

 

[haruspex]

The angle theta is not just any old angle. It is determined by the geometry, but I do not see where you have used that in the proof.
How did you get $2(1+cos ∅)(1-cos ∅)$ =$1 + cos ∅$?

The hint from 19.1 is that there is another expression you can get independently for DF, and that this involves the sine of theta.

 

[chwala]

The angle theta is not just any old angle. It is determined by the geometry, but you have not attempted to use that in the 'proof', so it is doomed to failure.
How did you get $2(1+cos ∅)(1-cos ∅)$ =$1 + cos ∅$?

The hint from 19.1 is that there is another expression you can get independently for DF, and that this involves the sine of theta.

Ok let me check again...I've seen it! Bingo!

 

[chwala]

$DFG=90^0$ angles in a semi-circle...
→$sin ∅=\frac {r\sqrt (2+ 2 cos ∅)} {2r}$
$2 sin ∅=\sqrt {2+ 2 cos ∅}$
$4 sin^2∅=2(1+ cos ∅)$
$2 sin^2∅=1+ cos ∅$ Bingo

 

[Frabjous]

Your 19.1 is written a little confusing. it is just using the identity
cos(pi-angle)=-cos(angle)
You also have not shown that E+theta=Pi

 

[chwala]

Your 19.1 is written a little confusing. it is just using the identity
cos(pi-angle)=-cos(angle)
You also have not shown that E+theta=Pi

cos(pi-angle)=-cos(angle) thanks for this...i did indicate that...We know that angle $E$ + angle $∅$= $180^0$ check my post $1$...or how did you want me to show this? it is basic knowledge from the properties of cyclic quadrilaterals.

 

[Frabjous]

I know that it is true if FG = r. I do not know if it is true in general.

 

[chwala]

I know that it is true if FG = r. I do not know if it is true in general.

In any cyclic quadrilateral, angles opposite to each other will always sum up to $180^0$

 

[Frabjous]

In any cyclic quadrilateral, angles opposite to each other will always sum up to $180^0$

I learned something new or something I had forgotten

 

[Frabjous]

$DFG=90^0$ angles in a semi-circle...

What does this mean?

 

[chwala]

What does this mean?

Angles subtended in a semi circle = $90^0$

 

[chwala]

just asking to prove,
$2 sin^2∅=1 + cos ∅$,
We can re-write the above equation as $2(1-cos^2∅)=1+cos ∅$ then by using the trigonometry identity,
$cos 2∅=2cos^2 ∅ -1$
→$cos ∅ = 2 cos^2 \frac {∅}{2}-1$ then we shall have,
$2(1-cos^2 ∅)=1 + 2 cos^2 \frac {∅}{2}-1$
$(1-cos^2 ∅)= cos^2 \frac {∅}{2}$...anyway, i can see it won't work because of the introduction of half-angle.

 

[chwala]

The angle theta is not just any old angle. It is determined by the geometry, but I do not see where you have used that in the proof.
How did you get $2(1+cos ∅)(1-cos ∅)$ =$1 + cos ∅$?

The hint from 19.1 is that there is another expression you can get independently for DF, and that this involves the sine of theta.

I used the difference of two squares i.e $a^2-b^2=(a+b)(a-b)$ for the lhs of the equation.

 

[haruspex]

I used the difference of two squares i.e $a^2-b^2=(a+b)(a-b)$ for the lhs of the equation.

No, you used that for this step:

$2(1- cos^2 ∅)=2(1+cos ∅)(1-cos ∅)$

How did you then lose the $2(1-cos ∅)$ factor?

 

[chwala]

No, you used that for this step:

How did you then lose the $2(1-cos ∅)$ factor?

yes that's true, i canceled out $(1+cos ∅)$ on both sides of the equation and then remained with
$2(1-cos ∅)=1$ as indicated in post $1$ but this was not the right way of doing it...i followed your advise as indicated in post $4$. Cheers.

 

[docnet]

This is trivial because all the triangles are either isosceles equilateral or 30-60-90 triangles. The three problems can be "solved" explicitly in terms of $\theta=\pi/3$

 

[chwala]

This is trivial because all the triangles are either isosceles or 30-60-90 triangles. The three problems can be "solved" explicitly in terms of $\theta=\pi/3$
View attachment 287774

Did you mean equilateral triangle by any chance? since i can see the sides of the 'isosceles' triangle having a measure of $r$ on all its 3 sides.

...Further, the question 'directs' you to use the given angle indicated as $∅$ and from the onset we do not know the value of $∅$. Can you show me how to solve the problem by considering for instance the "isosceles triangle" as you have indicated? Its pretty obvious that we do not know any angles in your "isosceles' triangle.

 

[docnet]

...but the question 'directs' you to use the given angle indicated as $∅$ and from the onset we do not know the value of $∅$. Can you show me how to solve the problem by considering for instance the isosceles triangle as you have indicated? Its pretty obvious that we do not know any angles in the given isosceles angle.
...and did you mean equilateral triangle by any chance? since i can only see the sides of the triangle as having a measures of $r$ on all the 3 sides?

yes, i meant to say equilateral. thank you for the correction! :) edited my post ^

 

[chwala]

yes, i meant to say equilateral. thank you for the correction! :) edited my post ^

ok if its an equilateral triangle then the proof can be easily shown...Bingo!

 

[Mark44]

now for part 19.1,
My approach is as follows, using cosine rule;
$DF= r^2 + r^2- 2r^2 cos E$

On the left side it should be $(DF)^2$.

We know that angle $E$ + angle $∅$= $180^0$

You should give the reason here, since several of us have forgotten that the opposite angles in a cyclic quadrilateral (AKA quadrilateral inscribed in a circle) are congruent supplements.

$∅$ is acute, therefore angle $E$ would be negative. (If $∅=60^0$ for e.g then it follows that $E=120^0$)

No, angle E is not negative, the cosines of E and $\pi - E$ are negatives of one another. You can use the identity that $\cos(\alpha) = - \cos(\pi - \alpha)$

Thus we shall have,
$DF^2= r^2 + r^2+2r^2 cos∅$
$DF^2=2r^2+2r^2 cos ∅$
$DF= \sqrt {r^2(2+2 cos ∅)}$
$DF= r\sqrt {(2+2 cos ∅)}$ is the approach correct?

Better:
$(DF)^2 = r^2 + r^2 - 2r^2\cos(E)$
$= 2r^2 - 2r\cos(\pi - E) = 2r^2 + 2r\cos(\theta)$
$\Rightarrow DF = r\sqrt{2 + 2 \cos(\theta)}$

Minor nit: E really is just a point. A better description for the angle would be ∠DEF.

 

[docnet]

You should give the reason here, since several of us have forgotten that the opposite angles in a cyclic quadrilateral (AKA quadrilateral inscribed in a circle) are congruent add up to $\pi$

I think you meant the opposite angles in a cyclic quadrilateral are supplementary instead of congruent.

 

[Mark44]

I think you meant the opposite angles in a cyclic quadrilateral are supplementary instead of congruent.

Yep, that's what I meant. I wrote what I meant in the following line; i.e., referring to $\alpha$ and $\pi - \alpha$.
I'll edit my earlier post.