TWO Methods to find the angle X | Learn how to Solve this Geometry problem Quickly

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Learn how to find the angle X in the given diagram. Solve this tricky geometry problem by using the alternate interior angles Theorem.

Need help with finding the angle X in this complex Geometry question ? You're in the right place!

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TWO Methods to find the angle X | Learn how to Solve this Geometry problem Quickly

#FindAngleX #Geometry #GeometryMath #TwoMethods
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#PythagoreanTheorem #Pythagorean #ParallelLines
#HowtoCalculatethedistance #Findthedistanceofthelinesegment #LineSegment #length #distance #blackpenredpen #ComplementaryAngles #OlympiadMathematics
#FindtheAngleX #HowtoSolvethisTrickyGeometryProblemQuickly #IsoscelesTriangleProperty
#IsoscelesTriangle #IsoscelesTriangles #Isosceles #Triangle #Triangles
#CorrespondingAngles #ExteriorAngleTheorem #AlternateInteriorAngles #CyclicQuadrilateral

Olympiad Mathematics
pre math
Po Shen Loh
Learn how to find the angle X
Alternate interior angles
premath
premaths

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A prodigious accomplishment. Thanks Professor!

bigm
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A third method would be to construct a line at the bottom perpendicular to the two parallel lines making irregular heptagon.Then remembering the total angle of a heptagon 900, it's easy to arrive at the same answer of 52. Your methods are cool though. Thanks for keeping us active!!

bekaluu
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The top/bottom method is really nice! 👍
I found another way to calculate angle x, which is in some way related to the top/bottom method:

x = the absolute value of the alternating sum of the given angles.

x = |+70° - 104° + 92° - 110°| = 52°
or
x = |-70° + 104° - 92° + 110°| = 52°

hcgreier
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I extended the line segment making the 92 degree and 110 degree angle to the left and right vertical lines, and used supplementary angles to make a quadrilateral on the left with angles of 110, 104, 88, and 58. Then a triangle on the right vertical line with the angles 70 and 58. Remaining angle x is 52.

pacfi
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I like it that you vary tough math problems and easy ones . Keep it up , Sir . Greetings from Germany . 👍

claudiaschartmann
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I used the first method to solve it. After seeing your demo I find the 2nd method is better and faster.

normanc
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I appreciate the way you repeat the steps. Though some of the theorems presented are highly repetitive your restating them teaches the importance of constructing a methodology, so keep it up!

Reddogovereasy
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can it be solved by extending 70 degree line to last parallel line and then extending x degree line forming a triangle at the end of angle x and a quadrilateral at the centre ?

lokeshsingh
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Hi Prof. PreMath,

I found this a rather fascinating problem. I had already intuited the use of the first solution with the alternate interior angles. But the second method esp. intrigued me as I'm new to the principle.

Easy enough to remember but I'm still perplexed by the logic of it. Any recommendations, given your list of links, as to where I can see these angle / triangle relations here spelled out so I can logically see why the sum of the bottom angles equals the sum of those at the top?

Enjoyed the video! Thx!

sailbyzantium
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Both solutions are elegant, but I favor the 2nd solution method!

NeutronRob
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I like the second method since it does not require drawing in any extra lines it just uses the angles you can see from the original diagram

phrtao
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Very well explained👍
Thanks for sharing😊😊

HappyFamilyOnline
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Why the sum of angles on the top is equal to the sum on the bottom ? can you explain please .

olivier
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There is a 4th method (there is a third one in a separate comment) is to rotate the figure sidewards or 90 degrees then draw some bisectors and use angle theorems for parallel lines as well as basic angle properties such as the linear pair postulate etc.


Cheers from the Philippines

alster
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Sort of did this in my head using a variant of your second method. Taking downward angles as negative, and upward angles as positive, the sum of the angles is zero. So -70+104-92+110-x=0. x=110-92+104-70. x=52.

guidichris
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There is another way:

First, construct a line joining the two parallel sides of the diagram, perpendicular to the sides, and above all the 4 other line segments (below works just as well). We now have an irregular seven-sided polygon. The formula for the sum of the interior angles of a polygon is (n-2)*180, where n is the number of sides; for a seven-sided polygon the sum is 900 deg.

From our modified diagram, we know, or can easily determine, 6 of the interior angles, and the sum of those 6 is 772 deg. The remaining angle (call it alpha) is therefore 900 - 772 = 128.

Since x and alpha are supplementary angles (sum is 180), then x = 180 - alpha = 180 - 128 = 52.

Just as easy as the other two methods.

greenmanofkent
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Bello e interessante soprattutto il secondo metodo professore.
Grazie

massimogranzotto
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I guess I assumed that the parallel lines were 90° but I simply pretended a 90° horizontal line shot right. This would make a triangle on the left a right triangle. 70+90=160. 180°-160=20°
Just repeat the process keeping in mind that the sum of the interior angles of a triangle is 180° and a horizontal line is 180°. A little more work but I got the same answer.

valentinocaine
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you only said the diagram is not upto scale.
where in question it has been given that lines are parallel?

param
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Dear Sir, extend the lines of 70 degree and also angle X, both will coincide and we will get a quadrilateral, the total angle will be 360 degree, find out the angles of the parallelogram and use 180 degree - angles to find all the angles of 70 degree triangle and you will get the x using alternate angle theorem.

antonygeorge