# Solve the problem in the given cyclic quadrilateral

• chwala
In summary, the conversation discusses using the cosine rule to find the value of DF, and then proving the identity ##2 sin^2∅=1 + cos ∅## by using trigonometric identities. However, the approach used in the proof is incorrect and does not take into account the geometry of the problem. The correct approach involves using the sine of theta and recognizing the angles subtended in a semi-circle are equal to ##90^0##.
chwala
Gold Member
Homework Statement
Solve the problem in the given cyclic quadrilateral
Relevant Equations

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##

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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.

caz said:
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...

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
haruspex said:
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!

##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

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Delta2
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

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chwala
caz said:
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.

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

caz said:
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
chwala said:
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

chwala
chwala said:
##DFG=90^0## angles in a semi-circle...
What does this mean?

caz said:
What does this mean?
Angles subtended in a semi circle = ##90^0##

Frabjous
##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.

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haruspex said:
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.

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chwala said:
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:
chwala said:
##2(1- cos^2 ∅)=2(1+cos ∅)(1-cos ∅)##
How did you then lose the ##2(1-cos ∅)## factor?

haruspex said:
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.

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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##

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Frabjous
docnet said:
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.

chwala said:
...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
docnet said:
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!

docnet
chwala said:
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##.
chwala said:
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.
chwala said:
## ∅## 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)##
chwala said:
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.

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docnet and chwala
Mark44 said:
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.

docnet said:
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.

chwala

## 1. What is a cyclic quadrilateral?

A cyclic quadrilateral is a four-sided polygon whose vertices all lie on a single circle.

## 2. How do you solve a problem involving a cyclic quadrilateral?

To solve a problem involving a cyclic quadrilateral, you can use properties and theorems specific to cyclic quadrilaterals, such as the opposite angles being supplementary and the sum of the opposite sides being equal.

## 3. What is the most common problem involving a cyclic quadrilateral?

The most common problem involving a cyclic quadrilateral is finding the length of a side or angle based on given information about the other sides and angles.

## 4. Can a cyclic quadrilateral have all equal sides and angles?

Yes, a cyclic quadrilateral can have all equal sides and angles. This type of cyclic quadrilateral is known as a kite.

## 5. Are there any real-world applications for solving problems involving cyclic quadrilaterals?

Yes, solving problems involving cyclic quadrilaterals can be useful in fields such as engineering, architecture, and navigation. For example, knowing the properties of a cyclic quadrilateral can help in designing structures with circular shapes or in determining the path of a moving object around a circular track.

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