Can you find the angle X? | Two Methods | (Step-by-step explanation) | #math #maths #geometry

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Can you find the angle X? | Two Methods | (Step-by-step explanation) | #math #maths #geometry

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[rollbruv]

Simpler way to get x:

Draw line segment AD. Angle ADE = 90(angle inscribed in a semicircle). Consider triangle DAE, angle DAE = 13. Thus, angle DAB = 53. Now draw line segment BD. Angle DBC + angle BDC = angle DAB( LHS angles subtended by arc BC and arc CD which make up arc BD, which subtends RHS). Now consider triangle BCD. X + angle DBC + angle BDC = 180. X + 53 = 180. Therefore x = 127.

[TurquoizeGoldscraper]

I used the angle ABC to get the major arc of AC = 266° and the minor arc AC = 94°. Adding the semicircle from AE gives me the major arc CE = 274*, and the angle CDE is 137°.
The interior angles of a pentagon = (5-2) × 180° = 540°
Adding the angles we get:
66° + 133° + x + 137° + 77° = 540°
x = 540° - 66° - 133° - 137° - 77°
x = 127°

[LuisdeBritoCamacho]

I've been out during the WE. So;
1) If a semicircle is divided in 4 equal slices, then the we must have 4 isosceles triangles with measures of angles: (67, 5º + 67, 5º) + 45º = 135º + 45 º = 180º.
2) Angles centered in O must add 180º.
3) Angles inscribed must add 540º.
4) The sum of all angles is equal to 540º + 180º = 720º.
5) These are constants even if the triangles are different.
Using this knowledge:
6) Angle OAB = 66º.
7) Angle ABO = 66º.
8 Angle OBC = 133º - 66º = 67º.
9) Angle BCO = 67º.
10) Angle OCD must be equal to Angle CDO, for the triangle [OCD] is isosceles = unknown.
11) Angle ODE = 77º. As the triangle [ODE] is isosceles.
12) Angle OED = 77º.
Now, adding up all the angles:
132º + 134º + 154º + OCDº + CDOº = 540º ; 420º + OCDº + CDOº = 540º ; OCDº + CDOº = 540º - 420º ; OCDº + CDOº = 120º.
As OCDº = CDOº = 60º.
Check: 132º + 134º + 154º + 120º = 540º.
So the angle BCDº = xº = BCOº + OCDº = 67º + 60º = 127º
Just a final note:
The center angles are:
AOB = 58º
BOC = 56º
COD = 80º
DOE = 26º
Sum: 48º + 46º + 60º + 26º = 180º
Check List Completed!
Answer:
The angle x is equal to 127º.

[hongningsuen]

1. Join AC and EC to form right-angled triangle ACE with angle ACE of this triangle =90 (Thales theorem). Two cyclic quadrilaterals ABCE and ACDE are formed as well
2. Opposite interior angles of cyclic quadrilateral are supplementary (property of cyclic quadrilateral). Hence for cyclic quadrilateral ABCE angle BCE = 180 - 66 = 114 and for cyclic quadrilateral ACDE angle ACD = 180 - 77 = 103.
3. Angle x = angle BCE + angle ACE - angle ACE of triangle ACE = 114 + 103 - 90 = 127.

[pwmiles]

You don't need to know the angle at B, because x is subtended on chord BD

Make isosceles triangle ABO
AOB = 180 - 66 - 66
= 48
Make isosceles triangle DEO
DOE =180 - 77 - 77
= 26

BOD = 180 - 48 - 26
= 106

By inscribed angle theorem
x = (360 - BOD)/2
= 127

[ybodoN]

A great puzzle with many different approaches (isosceles triangles, cyclic quadrilaterals, inscribed angle theorem...) 🤩

[uwelinzbauer]

Hello! I used the second method with the interior angle sums of these four isosceles triangles. Again an interesting geometric question for us, nice to remember the interior angle sum of polygons!
Greetings, best wishes!

[ChuzzleFriends]

Connect the points B, C, & D to the center O. Since the five points A – E are on circle O, segments through them and the center are the radii of circle O. They form four isosceles triangles, so the base angles of each will be congruent by the Base Angles Theorem.
Because m∠BAE = 66°, m∠ABO = 66°. So, m∠CBO = 67°.
Then m∠BCO = 67°, so m∠DCO = m∠CDO = (x - 67)°.
And m∠AED = 77°, so m∠EDO = 77°.
With all the angle measures, we can find the measures of vertex angles.
So, by the Triangle Sum Theorem, m∠AOB = 48°, m∠BOC = 46°, and m∠DOE = 26°.
By definition of straight angles, m∠COD = 60°. Use the Triangle Sum Theorem.
[2(x - 67)]° = 120°
x - 67 = 60
x = 127
So, the value of x is 127.

[solimana-soli]

AOB = 48 and DOE = 26, angle in the center = 180+48+26=254 => X = 254/2 = 127

[jimlocke]

Inscribed <BAE intercepts arc BE. The arc's measure is twice that of the inscribed angle, or 66° x 2 = 132°. Arc AE is 180 °, leaving 48° for arc AB. Inscribed <DEA intercepts arc AD. The arc's measure is twice that of the inscribed angle, or 77° x 2 = 154°. Arc AE is 180 °, leaving 26° for arc DE. <BCD intercepts the long arc BD, which is the sum of AE, AB and DE = 180° + 48° + 26° = 254°, so its measure is half of that, or x = 127°, as PreMath also found.

Using this method, it was not necessary to know that <ABC = 133°. The solution will be same as long as point C is between B and D, since <BCD will intercept the same arc, BD.

[laxmikantbondre]

This is much simpler.
Join BE.
As AE is diameterABE is 90 by Thales theorum.
So BEA is 24. i.e. 180 - 90 - 66.
So angle BED becomes 77 - 24 = 53
Quadrilateral BEDC is cyclic quadrilateral so opposite angle sum is 180. So x + 53 = 180. So x = 127.

[Abby-hisf]

@PreMath, your Page is attracting Math lovers and sharing their ways too. Thank you for creating such fun group.

[tombufford]

At a quick glance, Using a technique learnt some time ago, ABCDEA forms a Pentagon. Then using formula (n-2) *180 = sum of interior angles. where n is number of sides. Sum of angles =(5-2) * 180 = 540 = 66 + 133 + X + D + 77. Then X + D = 264.Then drawing four isosceles triangles AOB, BOC, COD and DOE. Then angle OBA = 66 and angle OBC = 133-66 = 67 then angle OCB = 67 . Angle ODE = 77. Angle OCD = X-67 and Angle ODC = X-67, Then Angle D = 77 + X-67= x+10. Then X + X+10 =264 and X = 127 degrees.

[Achill]

If we draw straight lines between O on one end and B, C, and D on the other end, we get 4 isocecles triangles, with top point in O.
One angle at B is 66, then the other is 67, because 67=133-66.
One angle at C is 67, the other is y.
One angle at D is 77, the other is y.
Sum of angles in a pentangle is 540.
We add all angles:
66+133+67+y+y+77+77 = 540 or
2y = 120
With y=60: x=127=60+67

[giuseppemalaguti]

Posto α=CDE .risulta 66+133+x+α+77=540(è un angoli opposti di un quadrilatero sono

[ОльгаСоломашенко-ьы]

∠x=0.5 degree measure of the arc BAED, as an angle inscribed in a circle. The arc ABCD=77°*2=154°, since an angle of 77° rests on it. The degree measure of the arc is DB=180°-154°=26°. We also find the degree measure of the arc AB=180°-132°=48°. ∠x rests on an arc 48°+26°+180°=254°. ∠x=0.5*254°=127°

[robertlynch]

I think they're both variations on the same theme in a way.

From the Isosceles △ theorem, we know that an I△ always has 2 legs which are equal, and 2 angles opposite where they join. From the 3 angles-of-a-△ theorem (summing to 180° exactly) … and the definition of a circle, we know that all △ drawn inside a circle, where the spokes join at the origin (center) must be isosceles.

So these in turn “power the chain” of induction about the givens. Starting with the lower-left [∠A = 66°] The 'spoke' drawn from B → O also has interior ∠ 66°. But the open ∠ is 133°, so the leftover part must be [133° - 66° = 67°].

Similarly, drawing a spoke from C → O must also have its C interior ∠ be 67°.

Working up from the right side, same treatment. Spoke from D → O has interior ∠ 77°

The only bit left is the △COD, which as you label has 2 [α] angles and a [β] angle at the origin.

  2α = 180° - β … so
  α = 90° - ½β

I ended up using the fact that the sum-of-the-angles around any point on a line equals 180° to calculate the missing β angle.

  180° = ∑[48°, 46°, 26°] + β ... move parts around
  180° - ∑[48°, 46°, 26°] = β ... do the summation
  180° - 120° = β
  β = 60°

  α = 90° - ½ 60°
  α = 60°

Like your video … following the chain of now all-known-angles, we get

  𝒙 = 67° + 60°
  𝒙 = 127°

Yay. And yes, if you've followed my comments before, you KNOW I tried to use a trigonometric approach first! And failed!!! Straight geometry sometimes works much better.

⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅

[saadkhondoker]

even if value of <B = 133 was not given, x would still be same.

[arnavkange]

There are many more methods to solve this sum...

U can again draw a triangle and by using angles in linear pair and cyclic quadrilateral theorem properties we can find x

[prossvay]

60+2(x-67)=180
60+2x-134=180
2x=180+134-60=254
So x=254/2=127°. ❤❤❤ Thanks sir.