Help pls (1 Viewer)

[obliviousninja]

http://snag.gy/iDHe7.jpg

find x. (pls dont circle the letter). lols

 

[integral95]

x = 60 lol

BEcause If you look at the angles at the bottom corners , they both equal to 80.

There fore you have a cyclic quadrilateral. Since the the top chords subtend equal angles.

there fore the angles subtended from the left chord are equal making x = 60

Sorry If I'm not clear

 

[obliviousninja]

x = 60 lol

BEcause If you look at the angles at the bottom corners , they both equal to 80.

There fore you have a cyclic quadrilateral. Since the the top chords subtend equal angles.

there fore the angles subtended from the left chord are equal making x = 60

Sorry If I'm not clear

Yea, i dont get you.
Its not cyclic, as the it should subtend equal angles, ie (the 10 and 20)

 

[braintic]

x = 60 lol

BEcause If you look at the angles at the bottom corners , they both equal to 80.

There fore you have a cyclic quadrilateral. Since the the top chords subtend equal angles.

there fore the angles subtended from the left chord are equal making x = 60

Sorry If I'm not clear

To be a cyclic quad, ONE chord must subtend equal angles at two points, NOT two different chords.

Assuming you are referring to the quadrilateral formed by lopping off the top triangle, it is definitely NOT cyclic. The almost horizontal line halfway up the triangle subtends different angles at the two lower vertices.

Edit: I just realised that I interpreted this wrongly, and it is even worse than I thought. The two equal sides of the triangle (which subtend the equal angles) would not even be chords of the quadrilateral you are calling cyclic. Or were you calling the isosceles triangle a cyclic quadrilateral? Actually, I have no idea what you were trying to say.

 

[braintic]

Not sure how to do it yet, but a Geogebra construction showed it to be 20 degrees.

 

[integral95]

yeah I just realised it all failed sigh.......

 

[obliviousninja]

BUMP. Syk, come to the rescue.

 

[Recondit]

Not enough information? I give up

 

[Menomaths]

BUMP. Syk, come to the rescue.

Sykkuno?

 

[obliviousninja]

Sykkuno?

Sy*

Sykkuno is eshays though. Taco Bell.

 

[braintic]

Still don't know how to do it. But here are my thoughts:

GIVEN that the answer is 20 degrees from my Geogebra construction, then the triangles ABE and BEF (named in corresponding order) turn out to be equilangular. (The supplied diagram is way off scale - the scale drawing makes this look plausible).

So I wonder if there is a way of proving that these two triangles are similar using side lengths, perhaps using the fact that triangle AEC is isosceles.

 

[obliviousninja]

Still don't know how to do it. But here are my thoughts:

GIVEN that the answer is 20 degrees from my Geogebra construction, then the triangles ABE and BEF (named in corresponding order) turn out to be equilangular. (The supplied diagram is way off scale - the scale drawing makes this look plausible).

So I wonder if there is a way of proving that these two triangles are similar using side lengths, perhaps using the fact that triangle AEC is isosceles.

I have no idea. My teacher suggested extending some of the lines though.

 

[fionarykim]

ahh, tried doing it got numbers everywhere
but still no idea how to do it LOL

all i got is x+y(BEC) = 130
but if x is meant to be 20, then that looks absolutely wrong LOL

 

[obliviousninja]

ahh, tried doing it got numbers everywhere
but still no idea how to do it LOL

all i got is x+y(BEC) = 130
but if x is meant to be 20, then that looks absolutely wrong LOL

i got x+y=130 like several times. lol

I did mention in the diagram that it was not to scale.

 

[Drongoski]

Looks like,triangles BEF and CEB (in addition to ABE & BEF) are similar. How to show one of these?

 

[fionarykim]

LOL yeah so did i
tried to come up wth two diff equations so i could simultanesously solve
but only got fking x+y = 130 like 10 000 times HAHA

nope i give up

 

[Sy123]

Looking at the diagram, draw a circle around points A, B, E.
Our goal is to prove that BD is a tangent to the circle, so we can satisfy angles in alternate segment, finding that

To prove that BD is a tangent, we have to show that the segments, BD, DE, and DA satisfy



This is the equation of tangent to secant, if we show this, then we subsequently prove that BD is a tangent which is the goal.

To prove that this is the tangent, we need to utilise some trigonometry and algebra.

Let



We can prove the above by showing that triangles AEC and ADC are isosceles by using angle chasing

So first, lets convert BD^2 = DE BF into algebra form, we need to prove:



This is what we need to prove.

Now, lets try and find BD^2 another way, we can do this by considering the sine rule in triangle BDC



We now need to find DC in terms of a and b.

Consider triangle ACD, draw the perpendicular from A to DC, which bisects DC, we find with this that:



(all units in degrees)



Now consider, triangle ACE, we know that AC = a, and AE=CE=b, therefore we know the relationship:







What we NEED to prove is:



And, if you plug both of those degree things into wolfram alpha, they output the same decimal. So this is correct, but we need to find a way to algebraically prove this.

I'll update this when I find a way to prove that the decimals are the same

 

[Sy123]

Divide by sin(70), convert sin 80 = cos 10, then use double angle formula



Divide by 2, convert 1 of the sin(20) into cos(70)



Use double angle formula again, use the fact that sin(180-x) = sin(x)



Use the sum to product trig identity, convert sin 70 = cos 20















(angle in alternate segment)



Difficult question.

 

[obliviousninja]

Divide by sin(70), convert sin 80 = cos 10, then use double angle formula



Divide by 2, convert 1 of the sin(20) into cos(70)



Use double angle formula again, use the fact that sin(180-x) = sin(x)



Use the sum to product trig identity, convert sin 70 = cos 20















(angle in alternate segment)



Difficult question.

Holy crap.

 

[HSC2014]

LOL, there has to be a faster method...