Can you find the Angle X? | Quick & Simple Tutorial

[ProfessorLX]

Great one, I don't usually have the answer before the last stages of the tutorial but this time I had it early, I guess all your lessons are paying off! Thanks very much!

[shiraishichan]

first time i could solve one of your problems by myself!
I like your videos so much and i hope to be better at geometry so i can solve all your problems easily
~love from a 9th grader

[davidfromstow]

As always, a great explanation - and it's so easy (afterwards!). Thank you

[arthurschwieger]

Thank you for doing these simple problems. I enjoy geometry and it has been such a long time since I learned these rules, many of them are no longer in my mind. I find it a fun exerciser to try to solve these and in some cases, just to follow what you are doing and understand the steps. My son also seems to enjoy geometry and I have given him one of your problems to solve. With very little help, he was able to solve it. I bet he will solve this one with no problem. 🙂

[sameerqureshi-khcc]

Sir you ARE a math genius! Love from Pakistan 😊👍🌹

[mcorruptofficial]

Hi dear solved by other way, through crossing AD and BD but you method much easier.
At last step, consider inner angles of a parallelogram with D and C vertex
Thank you bud.

[fetauAdu]

Sorry, I don't know how to type in the Greek letters. So I am using a, b & d for Alpha, Betha & Delta respectively.
The convex angle d (ADB) is 360-109=251 (d=251)
As you stated, angles a+b=71
The sum of the interior angles of the quadrilateral ADBC, x+a+b+d (CONVEX angle D)= 360 or
x+71+251=360
Thus, x=360-(251+71)= 38

[philipkudrna]

Before watching: 180-109=71 (sum of the other 2 angles). Double the other angles-> 142. Leaves 180-142=38 for angle x.

[aviratnakumar]

Awsome. But as D is the incenter of the given traigle we can directly use the angle subtented at incenter formula.

Here angle ADB =90°+(0.5*anglr ACB)

Solving further with the given data we get x=38

[devondevon]

Let angle A = 2p (since A has two congruent angles as shown)
Let angle B = 2n (sine B has two congruent angles as shown)
Then X + 2p + 2n =180 or X + 2(p+n)=180
but 109 + p + n = 180
Therefore p+n= 71 (180-109)
X + 2(71)= 180
X + 142 =180
X = 180-142
X = 38 Answer

[praveenchauhan.]

As being a student of class 8th I came here every day bcz maths is love... I am learning great everyday from u

[math]

sure can, thanks for the challenge bro

[kafaichan]

Let ang DAB = a, ang DBA = b, a+b+109 = 180 -> a+b = 71, 2(a+b) + x = 180 x = 180-142 = 38 deg

[shreygupta]

This was too easy solved in mind within seconds 😁

[mauriciosahady]

Também pode considerar o quadrilátero côncavo, de tal forma que 360 - 109 = 251° + 71° = 322, so X = 360 - 322

[charlesbromberick]

Sometimes it seems the prof has more theorems than grains of sand on the beach. This problem is a good example of just getting started makes things easier.

[truthandreason]

109 = a+b+x ; a+b = 180-109 = 71; x = 109-(a+b) = 109-71= 38

[vt]

My head is hurting trying to figure out the triangle CAB is being intersected in half by AD and CBA by BD. How come? Can someone please explain that?

[МихайлоЛобунець-ыв]

I am in 9th form (ukrainian educational syste), solved it in like 30 secs, my answear was 38, let me see, am i right?

Thanks for the video

[bentels]

Nice little problem. Maybe a little too easy...