find the missing angle!!

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[joseluishablutzelaceijas]

An alternative (purely geometric) solution could look as follows. As angle ADB is equal to 45° and angle ACB is equal to 30°, the angle CAD is equal to 15°. If one joins point D with a point E on AC such that angle ADE is equal to 15°, then |AE| = |ED| = |DC| = |BD| and angle DEC is equal to 30°, which implies that angle BDE is equal to 60° and due to |DB| = |DE| that triangle BDE is equilateral and that the angle BEC is equal to 90°. As |AE| = |EB| and the angle BEA is equal to 90° (180° - [angle BEC]), the triangle BEC is a right and isosceles triangle and, as a consequence, the angle BAC is equal to 45°, which implies that the angle BAD is equal to 30°.

[cvkline]

I guess that wasn't a good place to stop.

[ianfowler]

I must admit I shook my head when I saw the thumbnail - not enough information! The equal sides are not given in the diagram. Anyway, I let y = AD instead, <DAC = 15 and
<ABC = 135-t. Used the sine law twice. It all worked out. I did use the exact value of sin(15) though.

[jaspermcjasper]

I've alerted the authorities to the fact that Michael Penn has been abducted and replaced by a life-form that doesn't say "And that's a good place to stop". Even as we speak they're working hard on the case.

[papafragen]

Draw a circle around D with radius |DC|. Let the intersection of the circle with AC be E.
According to Thales' theorem, triangle BCE has the angles 30, 90 and 60.
Since |BD| = |DE|, triangle BDE is equilateral.
Since angle DAE and angle EDA are 15 degrees, |AE| = |ED| = |EB| and therefore triangle BEA is isosceles.
Therefore angle theta is 45-15 = 30 degrees (base angles triangle BEA are (180-90)/2 = 45 degrees).

[ibrt]

When presented with a geometry problem, turning it into algebra is like taking a long and complicated detour to get to the destination, rather than taking the freeway. It is really much more elegant and satisfying to solve this problem using theorems from geometry.

[Generalist]

Michael if you want to improve your geometry you need to give up analytic solutions don’t fall into the trap I fell into, solve them geometrically it’s tricky but it’s worth it because you gain better insights of finding key points which leads to more elegant and less messy solutions..Another issue with analytic solutions is that when solving more complex geo problems you often stumble upon 4th and higher degree polynomials which are usually not possible to solve, for example try the generalized ladder and square problem from mind your decisions analytically-completely impossible.lastly geometric solution are just more satisfying as they are require more creativity and less brute force.

[BMusic-hrrz]

Honestly:

Without loss of generality put midpoint of equal lengths at origin, line along x axis, equal lengths = 1. Solve for intersection of lines

y = -(1/sqrt(3))(x-1)
y = -(x+1)

No pythagoras, no dropped perpendiculars, no tan angle addition identities

[Bodyknock]

Neat problem. 🙂 The only minor thing I noticed in the video is, the way he wrote and drew everything, he assumed point E fell outside the triangle, but at least in principle going into the problem at first you don’t actually know if E is outside the triangle or if it lies on the line segment BD.

I don’t think this is a big deal though, it just means that y in the first part of the proof might be negative as well as the corresponding small angle. That doesn’t seem to change any of the algebra though, and in fact as shown in the video when you solve for x/y you get a positive ratio instead of a negative one which means y is in fact positive and therefore outside the triangle.

[urnoob]

the way u wrote the bracket for the last tangent equation there was wrong

[xunningyue]

That is a very classic geometry problem. Here is one alternative solution in pure geometric construction. Suppose point P be the symmetric point of D about the line AC, we get a regular ΔCPD and BD=CD=PD. Note that ∠DBP = ∠DPB = ∠CDP/2 = 30⁰, and ∠DAP = 2∠CAD = 30⁰ = ∠DBP, it follows that A/B/D/P are co-cyclic. Therefore, the θ = ∠BAD = ∠BPD = 30⁰

[urnoob]

without drawing anything extra

just use law of sines, u get two equations and u solve for the theta variable only

a/sin(A) = b/sin(B), A = theta + 15degrees, B = 180degrees - 45degrees - theta, sin(B) further simplified to sin(theta + 45degrees) by removing the negatives

so u have 2 equations

a/sin(theta + 15degrees) = b/sin(theta +45degrees)

b/sin(135degrees) = 0.5a/sin(15degrees)

substitute a or b, and the other will also be eliminated, leaving theta to be solved for

no calculators needed, just the table for 30 45 60 degrees and the angle sum formula

u need to find the ratio of sin to cos and then solve for tan from the table, giving 30 degrees as the answer (need to simplify and rationalize all the square roots to get the table value)

[CM_France]

Hi,

10:33 : missing "and that's a good place to stop".

[tassiedevil]

Impressed by the pure geometry solutions posted. Like several others, I was drawn to the sine rule approach. Using it on triangle ACD to show that |AD|=|AC|/Sqrt[2] and then applying it to triangles ABD and ABC and eliminating sine of angle ABD leaves Rather than hammering this out with addition formulae and tan or cot, in this case (particularly recalling sin(45) =1/Sqrt[2]), theta is 30 degrees by inspection.

[graf_paper]

I solved the problem in a totally different way, using the sin and cos laws to complete each triangle. It definitely felt like a less clever solution but it was super satisfying and a really great way to review a bunch of elementary trig. Definitely keeping this problem in my back pocket for students!!

[Alan-zftt]

The worst problems like this usually include a sketch that looks to be a right angle but may be 87 to 93 degrees or even worse: 88 to 92 degrees
Just a suggestion going forward: is it reasonable to do several (linked?) videos on this problem and solution solving?
My reasoning is that viewers and student may (will?) find it helpful to see several ways to solve this problem.
I hope that is a reasonable way for learners to find their preferred working methods aware that several options do exist bringing awareness of wider solution solving in general

[WalterMcKenzie-tw]

Much easier by using law of sines. For the larger triangle, Sin (a+15)/2x = sin 30/z and using the smaller triangle, sin 45/z = sin a/x. Simplfying, we get inverse cot of 1.732 which is 30 degrees. ATAGPTS

[deebd]

My only marginally different solution was to set the x in the video to 1, without loss of generality, which then immediately gave me tan 60° = sqrt(3) = (y + 2)/(y + 1), and therefor y = (sqrt(3) - 1) / 2.
This then implied that atan(angle EAB) = y / (y + 1) = 2 - sqrt(3), and therefor angle EAB = 15°, and therefor theta = 45° - 15° = 30°.

[BMusic-hrrz]

Set the two equal lengths = 1 and solve for the line intersection point coordinates instead.

[Cutestar]

Can we solve this by drawing a parallel line to BC passing through point A and then using supplementary and complementary angle rules for parallel lines..I am getting stuck though idk