Find the angle X | A Very Nice Geometry Problem | 2 Different Methods

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Комментарии : 33

@peterr9795 9 месяцев назад

Bisect angle D to meet AB at E. Draw line CE. Then triangle ADE is congruent to triangle CDE (SAS). Therefore angle DCE equals angle A which equals 2X, and so angle ACE is also 2X. Then triangle BCE is also congruent to triangle DCE (SAS). Since angle B is 3X, and the three triangles are congruent, angles CDE and ADE are also 3X. We thus know the 4 angles of the quadrilateral ABCD, respectively 2X, 3X, 4X and 6X. Sum of the angles of a quadrilateral is 360°. Hence 15X = 360°, and so X = 24°.

@mariopopesco 9 месяцев назад

This way i got the answer too !

@arumairajputhirasigamani2907 5 месяцев назад

Good work.. Simple and clear...

@prbprb2 9 месяцев назад

There is a pretty simple method. Draw the line parallel to AB , that connects D to BC. Write two different expressions for the distance from C to the line segment AB. With a tiny amount of work one arrives at sin(2x) + sin(pi - 7x) = sin(3x); Now there is a trick. Use 2x and 5x as good variables. Terms get rearranged to sin(2x) (1 +2 cos(5x)) =0. So 5x =120 degrees.

@B.TarunRaghavaSai 9 месяцев назад

I solved it in 30sec. Given are 2X,3X & 4X. Angle 2X & 4X are opposite to each other.If we consider that then the opposite of 3X must be 1.5X (or) 6X. For easy calculation let's try 6X first. So total is 2X+3X+4X+6X=360°. =>15X=360°. =>X=360/15 [X=24°]✓✓. I don't know whether it is just coincidence or real solution.

@gregheffley6400 8 месяцев назад

there is literally nothing real in this and u wont get marks if u wrote like this in an exam

@mohammednoordesmukh3784 8 месяцев назад

Draw CE cutting AB such that CE=CB =>

@arumairajputhirasigamani2907 5 месяцев назад

Super method.. I like it..

@geralynpinto5971 8 месяцев назад

Admittedly, the given solutions are long-drawn out. But, only from the point of view of maths, they are interesting. I have learnt much. Great thanks.

@MarieAnne. 8 месяцев назад

My solution is very similar to your first one: Draw line BD Let a = BC = CD = AD. Let b = BD Since BC = CD, then △BCD is isosceles, with ∠CBD = ∠CDB = 1/2 (180−4x) = 90−2x Using Law of Sines in △BCD, we get: BD/sin(∠BCD) = CD/sin(∠CBD) b/sin(4x) = a/sin(90−2x) b/a = sin(4x)/cos(2x) = 2 sin(2x) cos(2x) / cos(2x) = 2 sin(2x) sin(2x) = b/(2a) Drop perpendicular from D to side AB at point E. So △ADE and △BDE are right triangles. In △ADE, we get: sin(∠DAE) = DE/AD sin(2x) = DE/a b/(2a) = DE/a DE = b/2 In △BDE, we have DE = b/2, and hypotenuse BD = b So, △BDE is a 30-60-90 triangles, with ∠DBE = 30° Therefore: ∠CBD = ∠CBA − ∠BDE = 3x−30. But in previous step, we calculated ∠CBD = 90−2x 3x−30 = 90−2x 5x = 120 x = 24

@victorgorelik7383 6 месяцев назад

CE- bisector, z=angle CEB, 360=2z+10x, z=60 (law of sins), x=24.

@markwu2939 8 месяцев назад

I have a simple method for elementary students. Draw a angle bisector at C and the line intersects with AB at point E. Now we have 2 groups of congruent triangles. One is △ADE and △CDE. Another is △CDE and △CBE. Thus ∠CDE = 3x and ∠CDA = 6x. Finally, 15x = 2π. That's all.

@hongningsuen1348 4 месяца назад

There is a problem with congruence test for triangles ADE & CDE. The test is SSA which is usually not considered as a congruence test. There is no problem with congruence with triangles CDE & CBE as the test is SAS.

@markwu2939 4 месяца назад

@@hongningsuen1348 You are right. SSA is not one of the congruent conditions. However, if we can confirm the ∠DAB is acute, then the simple method can work. We know 9x < 2π, so 2x< 2/9π, representing ∠DAB < 90 degrees.

@hongningsuen1348 3 месяца назад

@@markwu2939 Further review of the problem of SSA being an ambiguous congruence test in this case shows that it is not a problem. When the non-inclusive angle 2x in SSA is an acute angle. Think of DE as the swinging arm of triangle ADE as angle ADE is not fixed. Compare it with length of AD. When DE = AD x Sin(2x), there is 1 possible position for E as triangle ADE is a right-angled triangle. When AD x Sin(2x) < DE < AD, there are 2 possible positions for E to give 2 different triangles for ADE. When DE = AD, there is 1 possible position for E as triangle ADE is an isosceles triangle. When DE > AD, there is 1 possible position for E as the other point of intersection goes to extension of BA and is not valid. I omit the condition of DE < AD x Sin(2x) as intersection point E is not possible which is not the case by construction. Anyway, even the ambiguous case of 2 possible positions for E does not affect validity of the arguement of triangle ADE being congruent to triangle CDE hence it is not a problem. I call SSA a conditional congruence test. For acute non-inclusive angle [A] in SSA, there can be 0, 1 or 2 answers in forming the triangle as shown above. The condition being the comparison between side opposite to the angle [So] and side adjacent to the angle [Sa] (So < Sa SinA, So=Sa SinA, Sa SinA < So < Sa, So = Sa, So > Sa). Actually an obtuse, not actue non-inclusive angle guarantees SSA being a congruence test as it means So > Sa (longer side opposing larger angle). Sorry your arguement for the validity of SSA in this case does not work though it does not affect the deduction of congruence. It is interesting to note that the ambiguity of SSA can be deduced from law of sines as no possible angle for invalid sine function (>1), 1 or 2 possible angles for any one valid value of sine function (acute in quadrant I, 90, and obtuse in quadrant II), and from law of cosines as no real roots, 2 equal or 2 unequal real roots for the side of unknown length from solving quadratic equation of the law. Hope you enjoy this sharing of my study.

@user-gs4cq3uk2b 9 месяцев назад

A(0) D(cos2x+sin2xi) then imaginary value of B=sin2x+sin(180°-7x)+sin(-3x)=sin2x+sin7x-sin3x=sin2x(1+2cos5x)=0 ∴cos5x=-1/2 ∴x=24°

@jeanmarcbonici9525 9 месяцев назад

I implemented the same method but without complex numbers. I agree that their use is a plus!

@RajanKumar-qi4vt 9 месяцев назад

Please make video in exam oriented solution. It's time taking. In competition govt exam there is not much time to solve in basic method

@sirak_s_nt 9 месяцев назад

In government exams you don't get such tough questions... These questions are AMC/PRMO oriented questions where the innovation to solve the problem is more difficult to get than to manage the time.

@rohitmadashri7250 9 месяцев назад

Why solve maths problems? Just buy the question papers in the market and become officers and serve the country. Don't forget to put some cash in the Swiss Bank.

@RajanKumar-qi4vt 9 месяцев назад

@@rohitmadashri7250 I think you are not understand what i want saying that In govt exam you have to solve math problem as quick as fast with limited time. And 2nd thing is govt exam is completely based on CBT. There is no question paper available in market. It's not your school exam that you should take time as much you can. Ok

@user-ru7pl4tx5v 9 месяцев назад

Καλησπέρα σας. Πολύ καλή λύση. Να κάνω μία υπόδειξη, αν μου επιτρέπεται. Στο 5:00 περίπου υπολογίσατε τη γωνία DBF=30 με την βοήθεια της τριγωνομετρίας. Μία εναλλακτική απόδειξη χρησιμοποιώντας 100% Ευκλείδεια Γεωμετρία. Φέρνω το τμήμα FE. Επειδή η FE είναι διάμεσος προς την υποτείνουσα στο ορθογώνιο τρίγωνο BFD, είναι ίση με το μισό της. Άρα: FE=DE=FD=α και το τρίγωνο DEF είναι ισόπλευρο οπότε