Can you find the angle X? | Triangles in a quadrilateral | (Step-by-step explanation) | #math #maths

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Can you find the angle X? | Triangles in a quadrilateral | (Step-by-step explanation) | #math #maths

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[Abby-hisf]

Great review of basic concepts and their application

[phungpham]

After lot of calculation we have 2 isosceles triangles DAB and CAB and angle CAB=30 degrees. J ust draw the circle of which the center is A and radius=AD=AC=AD we can see that angle x= 1/2 angle CAB= 30/2= 15 degrees.

[jamestalbott]

Thank you! Love the basic concepts at the beginning!

[uwelinzbauer]

It took me time to find and recognize the isosceles triangles, but then knowing this fact was like a revelation, and helped a lot. The remaining duty was quickly completed.
Thanks for sharing this nice puzzle 😊

[jimlocke]

At 6:50, we have found that points B, C, and D are equally distant from A. Therefore, we can construct a circle with its center at A and passing through points B, C and D. We note that <BAC is a central angle and has measure 30°, so arc BC also measures 30°. x = <CDB is an inscribed angle which intercepts the same arc, so equals (30°)/2 = 15°, as PreMath also found.

[Achill]

I would solve this trigonometrically.
All angles inside the quadrilateral can be calculated directly, except CDB, or x, and DCA, but for their sum:
CDB + DCA = 65
Furthermore, the product of the sines of all left angle at each point is equal to the product of all right angles:


The two equations with two unknowns are independent from each other and allow the calculation of tan(x).

[alster]

Okay, I got it!

I also noticed that <BEC and <AED are vertical angles

[quigonkenny]

As ∠CEB is 65°, ∠BEA and ∠DEC are both 180°-65° = 115°.

As ∠ABE is 35° and ∠BEA is 115°, ∠EAB is 180°-(35°+115°) = 30°

As ∠ABD is 35° and ∠DAB is 30°+80° = 110°, ∠BDA is 180°-(35°+110°) = 35°

As ∠DAE is 80° and ∠EDA is 35°, ∠ADE is 180°-(80°+35°) = 65°

As external sngle ∠CBF is 105°, internal angle ∠ABC is 180°-105° = 75°

As ∠ABC is 75° and ∠ABE is 35°, ∠EBC is 75°-35° = 40°

As ∠EBC is 40° and ∠CEB is 65°, ∠BCE is 180°-(40°+65°) = 75°

As ∠BCA and ∠ABC are both 75°, ∆CAB is isoceles and CA = AB

As ∠ABD and ∠BDA are both 35°, ∆DAB is isoceles and DA = AB

As DA = AC, ∆DAC is isoceles and ∠ACD = ∠CDA = 35°+x

∠ACD + ∠CDA + ∠DAC = 180°
35° + x + 35° + x + 80° = 180°
150° + 2x = 180°
2x = 30°
x = 15°

[bogdananitei]

It would be very interesting the solving of a problem where we don't have isoscels anymore.

[amitpanchal]

Im a math teacher, i really like your question.

[nandisaand]

I like how you started the video with a "concept review". Often you just dive in making computations without really explaining why or what your plan is for solving the problem, which can ultimately be more useful and impactful than showing the actual solution.

[reinamaeda]

I wonder if another method is possible to solve for x value. 7:21

[danielepatane]

to the point E we have 65 gr opposit to vertex 80+65 = 145 internal triangle is always 180gr so 180-145=35

[giuseppemalaguti]

Io ho usato varie volte il teorema dei seni, oltre a trovare gli angoli che si vedono facilmente

[WhiteGandalfs]

At first glance, i didn't recognize the equality of adjacent angles, thus tried to firstly paint an exact representation of the whole structure on paper, using tangens to draw the lines. Then landet at a solution equivalent to that of giuseppemalaguti435. Thus, the whole thing comes out as another example of "many ways to Rome" :D

[hanswust]

Why is AB = AD?

Thanks to whoever may answer my question.