Can you calculate Angle X in 1 Minute? | Step-by-Step Tutorial

Find the unknown angle x in this triangle by using diagonal properties and the exterior angles theorem. Quick and simple explanation by PreMath.com

[@predator1702]

Very nice solution, thank you teacher 🙏.

[@PreMath]

You are welcome! Thank you for your feedback! Cheers! You are awesome Predator😀

[@petersearls4443]

I love your videos. At 70 years old I’ve forgotten most of the math I learned in school. You are bringing all back to me now in a clear and concise manner. Thank you.

[@Gargaroolala]

sin 20 = BC/14. BC = 4.78 cm. Angle BCA = 70. Angle BCD = 110. Sine rule. (sin 110)/7 = (sin angle BDC)/4.78. Angle BDC = 40. x = 180 - 110 - 40 = 30.

[@PreMath]

Nice tip Thank you for your feedback! Cheers! You are awesome Garrick😀

[@matthewleitch1]

I used a similar approach, with cos 20 and then the sine rule. However, without using a calculator I got to sin[70-x]=2cos[20]sin[20], which led to sin[70-x]=sin[40], and so to 70-x=40, and finally x = 30.

[@mohammadzeyad953]

But I am in class 9th and I don't know about trigonometry so that way sir have explained was awesome

[@timeonly1401]

@@matthewleitch1 Nice use of double-angle formula: 2cos[20]sin[20] = sin[40]

[@habilaumar3806]

I expected using sine rule to solve the problem more easily without any construction 👍 thanks for the additional method

[@erichiseli3492]

I used Thales' theorem: the triangle ABC having a right angle, AB is the diameter of the circle, so the middle point E is its center. By definition, EB is the radius and also 7. By using the property that the angles of a triangle sum up to 180 and that we have a bunch of isosceles triangles, we identify the measures of the angles as follows: ABE=BAE=20° (because the triangle is isosceles) CBE=90-20=70° (because the right angle is 90°) AEB=180-20-20=140° (property of the sum of all angles of a triangle) BED is its complementary angle, that is 180-140=40° EBD=180-40-40=100° (property of the sum of all angles of a triangle) Since we know CBE is 70 and EBD is 100, we find that CBD is 100-70=30°

[@krislegends]

Using the Thales' theorem is correct. I never new it could be attached to any right triangle.

[@JLvatron]

Great puzzle. That rectangle diagonals was brilliant!

[@illyriumus2938]

Fun and a very creative solution. Loved it!

[@ZUBAIR457]

Interesting to see you find the unknown from the least known

[@soosaifernando516]

Very good explanation. THANK YOU VERY MUCH.

[@LeoJGod]

BEST TEACHER!!!

[@philipkudrna5643]

I like that you can do it without any trigonometry! I knew there must be some trick based on the fact that 7 is half of 14. But I couldn’t figure out the „semi-diagonal“ trick. Very celever and elegant!

[@PreMath]

Glad to hear that! Thank you for your feedback! Cheers! You are awesome Philip😀

[@servenserov]

*Another way:* BD=14•Sin(20°); 7/Sin(110°)=14•Sin(20°)/Sin(∡C); Sin(110°)=Cos(20°); 2•Sin(20°)•Cos(20°)=Sin(∡C); Sin(40°)=Sin(∡C); ∡C=40°; ∡x=180-110-40=30°;

[@arthurschwieger82]

It worked! Having the reminder to give the video a thumbs up near the beginning had me clicking the like/thumbs up right away on this video. :-)

[@PreMath]

Fantastic! Thanks for your thoughtfulness and wisdom. You are the best Arthur. Stay blessed😀

[@arthurschwieger82]

@@PreMath - Your channel is doing way better than mine (due to the boring nature of my content) and I want your channel to keep getting better. Keep up the great work on producing this content. :-D

[@ravikrpranavam]

Thanks for explaining the solution.

[@usman_mmalik]

your are doing excellent work. ❣

[@kamalendunchandra4561]

Thanks sir for this video 😊😊

[@montynorth3009]

A smart and simple solution. Thanks.

[@PreMath]

Glad to hear that! Thank you for your feedback! Cheers! You are awesome Monty😀

[@bentels5340]

Very nice. I am always fascinated to see how you solve these problems (mostly because the amount of trigonometry they taught me in high school you could stick in your eye and still see very well). Thanks!

[@PreMath]

Glad to hear that! Thank you for your feedback! Cheers! You are awesome Ben😀

[@williambunter3311]

Great, as always!

[@PreMath]

Excellent Thank you for your feedback! Cheers! You are awesome William😀

[@5181809]

That’s great!

[@PreMath]

Thank you! Cheers! 😀

[@maarufabdul128]

Thank u sir

[@procash1968]

Thanks PreMath Teacher Ji !! Nameste from India 🙏 Wonderful problem, beautifully explained which only you can do !!

[@PreMath]

So nice of you Pracash ji Thank you for your feedback! Cheers! You are awesome😀 Love and prayers from the USA!

[@pbondin]

I started with the realisation that the midpoint of AC (point E on your diagram) is the centre of a circle ABC (a diameter subtends a right angle when its ends are produced to any point on the circumference). Thus BDE and ABE are isosceles triangles. Thus angle ABE = 20 deg, which means, Angle BEC = 40 deg and therefore angle BDC = 40 deg. Now since triangle ABC is a right triangle, Angle ACB = 70 deg, Therefore angle CBD = (70 - 40) = 30 deg.

[@simpleman283]

I like seeing alternate ways to solve problems, 👍

[@maruthasalamoorthiviswanat153]

Nice solution Sir.

[@PreMath]

Thank you for your feedback! Cheers! You are awesome Viswanathan😀

[@mahalakshmiganapathy6455]

Superb

[@PreMath]

Thank you for sharing! Cheers! Keep rocking😀

[@sameerqureshi-kh7cc]

Sir you are not only an awesome mathematician but a soft spoken peson as well, lots of love and respect for you from Pakistan 😊👍🇵🇰

[@PreMath]

Sameer dear, you are too generous! I'm just an ordinary human being. Thanks for your kind words.You are the best! Love and prayers from the USA!

[@a_j6650]

GooD Explanation SIR**

[@PreMath]

Thank you for your feedback! Cheers! You are awesome AJ😀 Love and prayers from the USA!

[@ALEX_FROM_E]

Классная задача. Спасибо!👍

[@PreMath]

Dear Alexei, Рад это слышать! Спасибо за ваш отзыв! Ваше здоровье! Ты классный. Любовь и молитвы из США!😀

[@pranavamali05]

What a great question thaanku

[@charlesbromberick4247]

I couldn´t come up with that nice construction so I solve it with first calculating Ab using sine 20, then angle D with the law of sines, then angle ACB from 180 degree theorem, and then angle BCD using another 180 theorem and finallt "x" with 180 theoren one more time.

[@PreMath]

Great approach Charles. Many ways to solve this problem. That's the beauty of math! Thank you for sharing! Cheers! Keep rocking😀

[@ssa1591]

Медиана равна половине гипотенузы, ззачем строить прямоугольник. Сумма острых углов в прямоугольном треугольнике равнв 90°. Задача свелась к 7 класса. Через тригонометрию - это 8-9 класс.

[@murdock5537]

Excellent task that is a lot of fun! I first tried the trig functions, went well too. Much faster is the purely geometric solution: Since CDB = BEC = 2a = 40 degree; x = 100 - 70 = 30 degree.

[@242math]

got it, very well explained, thanks for sharing, happy holidays

[@PreMath]

Same to you my friend Thank you for your feedback! Cheers! You are awesome 😀

[@abdirazksomo88]

very nice

[@PreMath]

Thanks dear Cheers! Keep rocking😀

[@DB-lg5sq]

شكرا لكم على المجهودات يمكن استعمال علاقة sinفي المثلث sin 20 ..ABD و sinD نجد x=30

[@user-sk9oi9jl2g]

BC=14*sin(20). Angl( BCD)=110. BC/sin(angl(D))=BD/sin(110). 14*sin(20)*sin(110)=7*sin(D), 2*sin(20)*cos(20)=sin(D). sin(D)=sin(40), angl(D)=40, x=180-110-40=30.

[@PreMath]

Excellent Olga Thank you for your feedback! Cheers! You are awesome 😀 Love and prayers from the USA!

[@Mathematician6124]

Excellent. It was the same method I also followed.

[@s1ng23m4n]

Nice!

[@PreMath]

Thank you! Cheers! Keep rocking😀

[@Skank_and_Gutterboy]

I like this problem, there are a lot of ways to approach it.

[@govindashit6524]

Thanks sir, prey from India.

[@PreMath]

So nice of you Govinda Thank you for your feedback! Cheers! You are awesome😀 Love and prayers from the USA!

[@waheisel]

I knew I didn't have the most elegant solution when I was 10 minutes into the 1 minute solution and seemingly just getting started! I used the law of sines to get an equation for the big and for the small triangle, substituted length CD from one equation into the other, did some algebra and got cos20*cosx-sin20*sinx=2sin20*cos20. I recognized the left side as cos(x+20) using the cosine sum formula. I recognized the right side as sin(2*20) or sin(40) using the sin double angle formula. And since sin40=cos50, I got x+20=50! Not elegant. Not 1 minute. But, as PreMath says; "exciting"! Thanks PreMath for the excellent puzzle.

[@PreMath]

Excellent Thank you for your feedback! Cheers! You are awesome William😀

[@KAvi_YA666]

Nice video. Good luck.

[@PreMath]

Great ADK dear Thank you Cheers! Keep rocking😀

[@KAvi_YA666]

@@PreMath 📚📚📚📚😊😊😊📚📚📚📚

[@azumamurakami7842]

May I ask what software did you use to make your video ? thank you.

[@DDX01]

Nice I have drawn perpendicular from B to the AC line at suposed point E thus the triangles ABE and BCE are congruent and trangle BED also made a right angle triangle at the end of the day my ans was also 30 degs.I have given Thumbs up👍

[@PreMath]

Super DDX Thank you for your feedback! Cheers! You are awesome 😀

[@montynorth3009]

From triangle ABC, AB = 14 cos 20. From triangle ABD using Sin Rule. 7/sin 20 = AB/sin BDA Re-arranging, sin BDA = (AB x sin 20)/7 Substituting for AB, sin BDA = (14 cos 20 x sin 20)/7 So sin BDA = 2 x cos 20 x sin 20 Using formula sin a x cos a = 1/2 sin 2a Sin BDA = sin 40 So BDA =40 degrees X = 70 -40 = 30.

[@PreMath]

Great approach Monty. Many ways to solve this problem. That's the beauty of math! Thank you for sharing! Cheers! Keep rocking😀

[@Silver-oo8nm]

First in triangle CAB, using sin 20 = BC/14, I found out BC to be approximately 4.778 units. Then in triangle BCD, using sine rule, I found out angle CDB to be approximately 40 degrees. Last, in triangle BCD, we now know Angle BCD = 110 degrees [given] and angle CDB approximately equal to 40 degrees, that is the value that we have calculated using sine rule, so using angle sum of triangle, x = 180 - 110 - 40 = 30 degrees (Answer) It is an alternate method to find x, where we do not need to visualize any imaginary figure inside the triangle, by dropping an imaginary line.

[@tzisorey]

I definitely don't remember learning that a line from the centre-point of the hypotenuse to the right-angle of a right-triangle will be exactly half the length of the hypotenuse - but your rectangle proof is pretty clear. TIL.

[@PreMath]

Thank you for your feedback! Cheers! You are awesome 😀

[@furetetka]

Even though both math and school in general are pretty much behind me, I enjoy watching your videos. I wish those 10 years ago there was such an opportunity and maybe I would have taken a higher level math exam back then than just the regular one. Maybe now I'll up my game and start tutoring? Who knows, for now I get a lot of knowledge and pleasure from your videos in my free time. I will also add that English is not my native language so I am learning mathematical vocabulary in this language by the way.

[@PreMath]

My dear friend, we are all lifelong learners. Thank you for your nice feedback! Cheers! You are awesome Sajko.😀 Love and prayers from the USA!

[@AajanmaHindu]

Sir, Thanks for the video! Can x really be determined in 1 minute? lot of constructions are involved!

[@user-xb3sb6eg5n]

Today I missed the trick...I spend 3 mins and then I tryied the video solution...and !!!! Thanks again

[@PreMath]

No worries! Thank you for your feedback! Cheers! You are awesome 😀 Love and prayers from the USA!

[@AnonimityAssured]

I've been reflecting on this problem over the last couple of days, while nodding off to sleep. Clearly, the actual values used in it are arbitrary, even though the result is neat and familiar. The lengths 7 and 14 are wholly irrelevant to the solution, except for the fact that they are in a 1-to-2 ratio. The angle of 20 degrees could also be varied. If we call that angle θ, we can determine, by reasoning similar to that in the video, that _x_ = 90 − 3θ degrees. So, if θ is 20 degrees, θ = 90 − 3∙20 degrees = 30 degrees. That solves the problem, but it leads us to wonder what _x_ would be for other values of θ. For example, what value of θ would give _x_ = θ? (The answer is particularly neat in radians.) What value of θ would make the two line segments on the right correspond to each other? With what values of θ would _x_ equal 2θ, 3θ, and 6θ respectively? What would happen if θ were 45 degrees? Would it be possible to make all lengths in the triangles exact integers? If so, what would be the shortest lengths possible? I'm sure that there are all sorts of other spin-offs from the original problem. Any other suggestions?

[@michaelmounts1269]

👍👍

[@abhishekpatil1063]

Take point O as mipoint of AC and centre of circle.Now join OB.Now Angle BOD= 40 degrees..Also triangleBOD is isosceles with base angles=40 degrees each .So remaining angleisx+70=100i.e x= 30 degrees.

[@pratyushdutta5781]

Awesome video Sir! Thank you..

[@PreMath]

Most welcome! Thank you for your feedback! Cheers! You are awesome Dutta😀 Love and prayers from the USA!

[@susennath6035]

Excellent. Good. Today may be taken more than one minute

[@PreMath]

Thank you for sharing! Cheers! Keep rocking😀

[@chaosredefined3834]

BC/14 = sin(20) Hence, BC = 14*sin(20) BC/sin(BDC) = BD/sin(BCD). As you point out, BCD = 110, so sin(BCD) = sin(110). As sin(x) = sin(180-x), we get sin(BCD) = sin(70). Furthermore, since we already have a 20 around the place, let's use sin(x)=cos(90-x), so sin(BCD) = sin(20) 14*sin(20)/sin(BDC) = 7/cos(20) Multiply by the denominators, and we get 14*sin(20)*cos(20) = 7*sin(BDC) 2*sin(x)*cos(x) = sin(2x), so 14*sin(20)*cos(20) = 7*sin(40) 7*sin(BDC) = 7*sin(40) sin(BDC) = sin(40) So, BDC is either 40 or 140. As it is in triangle ABD where the angle ABD is larger than 90, we know that BDC must be smaller than 90. Hence BDC is 40, and therefore x is 180 - 110 - 40 = 30.

[@PreMath]

Super job Thank you for your feedback! Cheers! You are awesome 😀

[@mrdetective18]

Please bring questions on surface area and volume...

[@PreMath]

Sure, soon BTW, I've already uploaded vids on this topic in premath channel. You may check them out. You are awesome Saurabh😀

[@forgtmnot27]

Thought you'd go Thales' Theorem, so many ways to skin this cat!

[@PreMath]

Many ways to solve this problem. That's the beauty of math! Thank you for sharing! Cheers! Keep rocking😀

[@nizamuddinahmed2193]

We can apply angle in a semicircle is 90. E is the circum center.

[@syselana3946]

If we draw a circle passing from points ABC, line AC is its diameter. If we draw a circle center on D and radius 7 it will pass from point B and intersects line AD at a point E The two circles are equal (radius=7) So the angle BDA is twice the angle BAC as they "see" on the same arc. Actually at two symmetric, equal, half, arcs BD and BE So angle BDA=40 and x=30

[@syselana3946]

I found similar and simpler answers like the one of Erich Iseli. Still though he could go straight to the fact that BE=BD=7 so angle BED=angle BDE and since BED=ABE+EBA and EBA=20 => BDE=40 and x=30

[@holyshit922]

Side length of BC from sine of angle A From sum of angles and from that anles ACB and BCD are supplementary Angle BCD is 110 degrees From cosine law in triangle BCD we calculate side length of CD then from cosine law in triangle BCD we calculate cosine of x Finally we will get that x is 30 degrees

[@waynevenables8176]

Could x equal any other number but 30 ? I created an equilateral triangle using the “ 7 “ segment and my answer is 10. Please show me my error.

[@Couch-Tomato]

Because of ∠ABC=90°, you can find A,B,C are on a same circle that has center point E.

[@PreMath]

Many ways to solve this problem. That's the beauty of math! Thank you for sharing! Cheers! Keep rocking😀

[@josemath6828]

Me parece un ejercicio interesante.

[@PreMath]

¡Me alegra escuchar eso! ¡Gracias por tus comentarios! ¡Salud! Eres increíble. ¡Amor y oraciones desde EE. UU.!

[@bilalmallick8428]

Ok, so basically in solving problems like this, one needs to create an isosceles triangle, or a special triangle, to figure out other angles. Then, once a set of angles are solved , another isosceles or special triangle can be drawn to find angle in question.

[@mathfullyexplained]

Also Let E be midpoint of hypotenuse AC. Midpoint of hypotenuse is equal distance from all vertices. AE = CE = BE. Now follow the video.

[@PreMath]

Super! Thank you for your nice feedback! Cheers! You are awesome my friend.😀

[@brianstelter7067]

Solved by simultaneous equation. Simple subtraction.

[@EddieDraaisma]

Using sin(2a)=2sin(a)*cos(a), height of triangle (B to AD) equals AB*sin(20deg) = 14*cos(20deg)*sin(20deg) = 7*sin(40deg) = 7*sin(angle(BDC)) = 7*sin(70deg-x); so 40 deg = 70deg - x, so x = 30 deg!

[@PreMath]

Well done! Thank you for your feedback! Cheers! You are awesome Eddie😀

[@piman9280]

Well, that was definitely *more* than one minute! From right triangle, AB = 14cos20; from full triangle, sin(D)/14cos(20) = sin(20)/7 => sin(D)=2sin(20)cos(20) => sin(D) = sin(40) => D = 40. But angle BCA = 70 => 70 = 40 + x => x = 30. This is still more than one minute, but no construction is necessary.

[@PreMath]

Thank you for your feedback! Cheers! You are awesome Piman😀

[@jerrym9380]

还可以用三角函数求解，不过会麻烦一些

[@PreMath]

你是绝对正确的！ 感谢您的反馈意见！干杯! 你真棒杰瑞。😀

[@gloubiboulgazeblob]

Nice solution indeed, I was needlessly trying to use sin and cos laws, waisting a lot of time...

[@MK06086]

Isosceles triangle is why I am glad I am Greek

[@MathsOnlineVideos]

Great solution! I did it slightly differently. Let D = y.

[@PreMath]

Great approach Siyanda. Many ways to solve this problem. That's the beauty of math! Thank you for sharing! Cheers! Keep rocking😀

[@noneek680]

Good. I don't find yet

[@theophonchana5025]

#triangle #RightTriangle

[@luigipirandello5919]

Beautiful geometric question. Thank you, Sir.

[@PreMath]

Most welcome Luis Thank you for sharing! Cheers! Keep rocking😀

[@deepaksisodiya2234]

ARE triangle ABE AND BEC congurate or not?

[@murdock5537]

Yes, you can easliy proof it: draw the two lines of the side halves of the rectangle - you get eight equal areas that make up the total area of the rectangle. Obviously the area of ABE = BEC. 🙂

[@prithvinayak917]

Easy by sine rule

[@washingtoncostasilva625]

What if AC was not 14? Because the solution is based to this value, otherwise how could we find an answer?

[@sandanadurair5862]

BC=14*sin20 in Triangle ABC D =70-x Angle BCD=110 IN triangle BCD BC/Sin(70-x) = 7/Sin110 Sin(70-x) = BC. Sin(110)/7 Sin(70-x) = 14. Sin20. Cos20/7 (Since sin110 = cos20) Sin(70-x) = 2 Sin 20.Cos20 = Sin(40) Sin(70-x) = Sin(40) 70- x = 40 Hence X = 30

[@PreMath]

Great approach. Many ways to solve this problem. That's the beauty of math! Thank you for sharing! Cheers! Keep rocking😀

[@sharonmarshall3671]

I applied the sine formula to both triangles and got the same result

[@guyluck9253]

Maybe you could teach us some calculus. I am sure you would make it easier for us not so good mathematicians

[@nehronghamil4352]

Alternate Solution: Draw line from B perpendicular to AD call it h h=14 sin(20)*cos(20)= 7 * 2 * sin(20) * cos(20) = 7 * sin (40) (1) angle BDC = 180 - (x+110) = 70-x h = 7 * sin(70-x) (2) (1) & (2): 7 * sin (40) = 7 * sin(70-x) 70-x=40 or x=30

[@TinkeringJohn]

You thought outside the box by thinking outside the triangle.

[@PreMath]

Yes, indeed! Thank you for your feedback! Cheers! You are awesome John😀

[@theophonchana5025]

Isosceles triangle EBD

[@FranC49]

Trazando una altura BE a AD en ⊿BEC : Cos(x+20°) = 14Sin(20°)Cos(20°)/7 Cos(x+20°) = 2Sin(20°)Cos(20°) Cos(x+20°) = Sin(40°) x + 20° + 40° = 90° x = 30°

[@PreMath]

Super Fran!

[@mustafizrahman2822]

Answer is 30 degrees. I used law of sines. But I have solved it in 1 minute 11 seconds.

[@PreMath]

Great approach Mustafiz. Many ways to solve this problem. That's the beauty of math! Thank you for sharing! Cheers! Keep rocking😀

[@starpawsy]

This is very easy to calculate numerically. IF there is an analytical solution, it escapes me. But I dont care.

[@theophonchana5025]

Right triangle ABC

[@VolkGreg]

BD/sin(ےBCD) = BC/sin(ےBDC) 7/sin(20°+90°) = 14sin(20°)/sin(90°-20°-x) sin(70°-x) = 2sin(20°)cos(20°) = sin(40°) x = 30°

[@PreMath]

Great approach Greg. Many ways to solve this problem. That's the beauty of math! Thank you for sharing! Cheers! Keep rocking😀

[@agp1745]

That's why I hate high school geometry. the solution could be one of 10000 possible tricks. It is demoralizing. I won't be able to tackle any of this. Ever. Can anyone provide me with tips on how to break my cycle of despair? thanks

[@PreMath]

No worries! We are all lifelong learners. Just keep persevering and everything will be alright😀 Thank you for your feedback! Cheers! You are awesome AGP😀

[@user-ri3fq4vm8u]

Hi,CD=AD_AC

[@tzamiko1]

Applying the law of sines, I found an error of over 10%!!! The use of Trigonometry may lead to large errors.

[@PreMath]

Thank you for your feedback! Cheers! You are awesome 😀

[@pokey5428]

So, the answer to the question "can you calculate angle X in 1 minute" is NO, and apparently neither can you.

[@theophonchana5025]

x + 40° = 70° x = 30°

[@PreMath]

Very good Theo Thank you for your feedback! Cheers! You are awesome 😀

[@vidyadharjoshi5714]

The problem is over in step 4 ? Angle BED = 40, Since BE = BD so Angle BDE = 40, Angle BCD = 180 - 70 = 110. So Angle x = 180 - 110 -40 = 30.

[@srilatapn6367]

X 20degree

[@pdg3870]

I don’t doubt the math, but I do doubt the drawing itself. It is not drawn to scale. Clearly angle X is not greater than 20 degrees. Also clearly evident is that line segment BD is greater in length than 1/2 of segment AC.

[@giuseppemalaguti435]

30...teorema dei seni, 7/sin110=14sin20/sin(70-x)....semplifico x=30

[@PreMath]

Eccellente Giuseppe Grazie! Saluti! Continua a dondolare😀

[@paulklee5790]

You can’t fool me.... I saw the angle x up in that tight corner, only took a second. Try hiding it better next time.....

[@user-km7hi4so8x]

أستاذنا الفاضل ياليت تتكلم باللغة العربية فأنا متابع وأحب مادة الرياضيات .