Can You Find Angle X? | Geometry Challenge!

[procash]

Sir, beautifully & patiently explained. You have an ocean of patience which is the prime requirement of any teacher

Even a student weak in Maths will easily understand if u take a class

Thanks

[AbulBashor-huzd]

The hardest part of solving geometry problem is to find a point like P.😂😂😂

[charlesbromberick]

very nice - designating point "P" was critical, and unfortunately I didn´t see that on my own - jajaja

[andymaczak]

I was drawing angles after angles, but no joy. Your explanation is great! Thanks!

[easternbrown]

Nice geometric solution!

I got there with tangents function. If you complete the right-angled triangle in the bottom left, with height h and base p+2d (were d is the length AD) then you see there are three right-angled triangles and by considering each one in turn that:

tan 60 = (p+2d)/h = sqrt(3)
tan 45 = (p+d)/h = 1
tan (45-x) = p/h

Multiply both sides of the second equation by two and subtract the first equation from it and you have 2-sqrt(3) = p/h
Substitute this into the third equation and you get:
x = 45 - tan-1 (2-sqrt(3))
which is 30 degrees.

[williambunter]

Sir, you are truly the Sherlock Holmes of trigonometry. You assemble the evidence point by point, and arrive at the solution step by step. It is making mathematics good fun! Thank you.

[Nmomand]

I am 72 and a retired electrical engineer. I guessed it correctly however, your step by step explanation was excellent. Thank you.

[DontRickRollMePleze]

Before these things used to be a headache for me in class. I did it cause I was forced to and did the minimum.

Now it’s my entertainment and I can’t get enough of it. I rather watch this than a series.

[siddharth]

an easier solution i think would be to just draw a line say CE parallel to AD now using alternate interior angle property we can conclude that angle ADC = angle DCE. Now since the line CE is also parallel to DB therefore we can conclude using alternate angle property that angle DBC = angle BCE. Now angle DCE= angle DCB + angle BCE = 45. Therefore, 15 + BCE = 45, therefore BCE = DBC = 30

[arumairajputhirasigamani]

In ∆CDB, angle B + 15° = 45° [exterior angle is the sum of interior opposite angles]
Therefore angle B = 30°
Let AD = DB = a
and AC = b
In ∆ABC by law of sine
2a/sin(x+15°) = b/sin30°
= b/(1/2)
= 2b
=> sin (x+15°) = a/b
In ∆ ADC
a/sin x = b/sin45°
= b/(1/√2)
= b√2
=> √2 sin x = a/b
From (1) and (2)
√2 sin x = sin (x+15°)
√2 sin x = sin x cos 15° + cos x sin 15°
(√2-cos 15°) sin x = cos x sin15°
sin x / cos x = sin 15° /(√2-cos 15°)
tan x = sin 15° / (√2-cos 15°)
tan x = 1/√3 ( pl. refer Note below)
= tan 30°
Thus the unknown angle x = 30°

Note:

sin 15° = sin (45°-30°)
= sin 45° cos 30° - cos 45° sin30°
= [1/√2] [ √3/2 - 1/2]
= (√3-1)/2√2
Similarly
cos 15 = (√3+1)/2√2
√2-cos 15 = √2 - (√3+1)/2√2
= (3-√3)/2√2
= √3(√3-1)/2√2
= √3 sin 15
Hence
sin 15° / (√2-cos 15°) = 1/√3

[aaronpatalune]

Kudos to anyone who can solve problems like this, but my hat is really off to those who can CREATE problems like this.

[badphysics]

Felt like a suspense story with a great payoff at the end! Well done good sir.

[biaohan]

3:13 to get angle PDC you don't need the 120 degree. Just use the exterior angle theorem again in the trangle PDC with angle BPD=30degree as the exterior angle, and angle PCD=15degree so angle PDC=30-15=15degree.

This is a good question that looks difficult but once the additional lines are drawn it becomes straightforward.

[jayquirk]

Great puzzle- I needed the DP portion to start before I could solve it. Thanks for providing such good content and a mix of easier and more difficult problems.

[aakashkarajgikar]

You explained way better than my last years Math teacher did! Thanks for helping me clearly understand this process!

[hasibrahman]

Please make this kind of geometry video more...and make pdf including all their rules and strategy

[Greebstreebling]

In 1970, my Maths teacher at Whitchurch high school in Cardiff, Mr Smth, was so good he helped me to learn how to do this stuff and I am forever grateful. I still haven't found a use for the integration of 1 + Tan squared (x) though. :)

[BAgodmode]

Great job demonstrating how one can use simple facts of geometry and a little deduction to reduce the problem to a simple math problem.

[jeanmarcbonici]

Well done, I didn't think it was possible to solve this problem without calculating a single length!

[harikatragadda]

*Trigonometric solution:*
Drop a perpendicular from C to meet the base line at E. Let CE= a and ∠ECA= θ. Since ∠DCE=45°, we have ED=a, and since ∠EBC= 30°, we have EB=a✓3. This gives AD= (✓3-1)a and EA=ED-AD=(2-√3)a.
Now Tanθ=EA/CE = 2-√3, which gives us θ=15°. Hence X=30°.