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

[predator]

Very nice solution, thank you teacher 🙏.

[petersearls]

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.

[arthurschwieger]

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. :-)

[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.

[erichiseli]

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°

[philipkudrna]

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!

[bentels]

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!

[habilaumar]

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

[sameerqureshi-khcc]

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

[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 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.

[ZUBAIR]

Interesting to see you find the unknown from the least known

[procash]

Thanks PreMath Teacher Ji !!

Nameste from India 🙏

Wonderful problem, beautifully explained which only you can do !!

[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.

[MathsOnlineVideos]

Great solution! I did it slightly differently.

Let D = y.

<ACB = 70° (Sum of angles in 🔺️ABC)
<ACB = x + y (Exterior angle of a 🔺️).
70° = x + y
Thus, x = 70° - y

Using trigonometric ratios in 🔺️ABC, we have:
AB = 14. sin 70° = 13, 1557

Now, using the Sine Rule in 🔺️ABD, we have:

[sin20°]/7 = [sin y]/13, 1557

sin y = (13, 1557 × sin20°)/7
sin y = 0, 6428
y = arcsin (0, 6428)
y = 40°

From (1) above: x = 70° - y
Thus, x = 70° - 40°
Thus, x = 30°

[JLvatron]

Great puzzle. That rectangle diagonals was brilliant!

[murdock]

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.

[soosaifernando]

Very good explanation. THANK YOU VERY MUCH.

[montynorth]

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.

[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.

[charlesbromberick]

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.