Solving Geometry In Portuguese
△ A B C is isosceles with A B = A C and ∠ B A C = 3 0 ° . It is given that ∠ B A D = ∠ B C E = 1 0 ° . Find ∠ E D A in degrees.
The answer is 65.
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4 solutions
∠ A B C is the common notation for angles, instead of A B ^ C . I've edited your question for clarity.
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Ok, no problem. When I learned the notations (my teacher was one of the best of Brazil), there was a difference between them: ∠ A B C was the "figure" and A B ^ C was the "measure". Is it wrong?
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Mathematical notation greatly depends on where you learnt it. There is no "right or wrong" answer.
Hat notation is less popular nowadays though.
I use
△
A
B
C
to denote the triangle ABC. Not certain if that's what you meant by "figure".
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@Calvin Lin – Yes, that´s it. Not sure about the correct word for it. I learned (and I teach) "figures" (or "things") are congruents to each other (e.g. ∠ A B C ≅ ∠ P Q R ), and measures are equals (measure of ∠ A B C = measure of ∠ P Q R ).
So nowadays we should use the same notation in both cases? (e.g. ∠ A B C ≅ ∠ P Q R and ∠ A B C = ∠ P Q R = 3 0 ° ) ?
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@Eloy Machado – I think most people use ∠ A B C = ∠ P Q R to denote both angles are equal in value or magnitude.
and △ A B C ≅ △ P Q R to denote two triangles are congruent to each other...
Hope this helps :)
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@Eddie The Head – well....the notation of congruent triangles never was in doubt, but thank you for your help. :-)
Still waiting for response from Calvin to my last question.
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@Eloy Machado – By the way, I didn´t like the new name of my problem. It sounds a bit pejorative and probably does not attract people to solve it.... :-(
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@Eloy Machado – I've updated the title. I didn't intend for it to be pejorative. I love the fact that mathematics can be written in different languages but understood by all.
Notation is extremely localized, and it greatly depends on whom you learnt it from.
The statement
∠
A
B
C
=
∠
X
Y
Z
is understood to refer to the measure of the angle. People often say "What is
∠
A
B
C
?", when asking for the size.
You would notice that in questions I pose, I often ask "What is the measure (in degrees) of angle
A
B
C
".
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@Calvin Lin – Thanks @Calvin Lin ! I was very curious because all "my life" I believed there were differences between the figure angle (Two rays sharing a common endpoint) and the measure of the angle, with distinct notation for them.
Now, I am really convinced there are no need for this difference. We know 2 angles are congruent when they have the same measure. But we don´t need a notation for congruence of angle because we can always use the notation of measure of the angle.
Do you agree?
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@Eloy Machado – yes
@Eloy Machado – To Solve this, I don't know about cyclic quadrilaterals, so I drew the diagram out and then mirrored it along the center and the connected E to a new F horizontally (where angleECF = 65 and FEC = 10), and BF mirrors CE, and they bisect each other on the mirror line. I then filled in all the know angles and worked it out from there.
What is the proof that the quadrilateral is cyclic?
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One necessary and sufficient condition for a convex quadrilateral to be cyclic is that an angle between a side and a diagonal is equal to the angle between the opposite side and the other diagonal.
Since ∠ E A D = ∠ B C E = 1 0 ° , A E D C is a cyclic quadrilateral.
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thanks .......
How is that, though? Im sorry but I really want to know..
Easy
65
sir what is the integral of "xtanx" i.e.-\int { xtanxdx=??` }
Umm, it can easily be solved with similarity, we can find all the unknown angles and come to find out that triangle AEY is similar to triangle CD(intersection of AD and CE). And the sides C(intersection) and (intersection)A are the sides of triangle A(intersection)C with the sides in the ratio of, say, a:b, and if we look at the smaller triangle D(intersection)E we can see that the two sides are I'm the ratio of the sides bases, this also in the ratio of a:b and both have an included angle of 95, therefore are similar, and therefore angle ACE=angle EDA=65
i have the same answer with you!@
i solved it in 3 minutes!
thanks for nice challenge!
very good one sir
It took me 1 0 minutes.
60 not 65
ABC is an isoscales triangle, with angle EAD=angle ECD, so the quadrilateral AEDC is a cyclic one. Each of the two equal angles is equal to 75 degrees. Therefore ECA=65 degrees, now angle AEC+angle DEC= 180-75=105. considering triangle AED and given that the sum of all triangles is equal to 180 degrees, therefore we get x=65.
AEDC IS A CYCLIC QUADRILATERAL SO DO IT ACCORDINGLY
BCA = CBA = 75º. The upper right triangle is 20º + 95º + 65º. The upper left triangle is 10º + 85º + 85º. So, the desired triangle is 95º + 20º + (EDA) 65º. I used angles opposite by the vertex and supplementary angles to solve this problem.
Please, clarify your argument for the 20º in "desired triangle"?
what is this??
Sure, as I found the upper left triangle and the B angle, I can affirm by supplementary angles that 180º - 85º - 75º = 20º. "The desired triangle" is the triangle with the unknown angle x. By angles opposite by the vertex, I found the other angle of this triangle is 95º. Then, I concluded by the rule about internal angles of a triangle must be 180º - 95º - 20º = 65º.
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Hi Henrique! If I understood correcty, to discover ∠ D E C = 2 0 ° you affirmed ∠ B E D = 7 5 ° . If that is what you said, why did you know ∠ B E D = 7 5 ° ?
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Because triangle BED is isosceles.
We know △ A B C is isosceles, so B C ^ A = C B ^ A = 7 5 ° . Then, E C ^ A = 6 5 ° .
Since B A ^ D = B C ^ E = 1 0 ° we know A E D C is a cyclic quadrilateral. Then, E D ^ A = E C ^ A = 6 5 °