Geometry you need to know for college

They say many students struggle with this question, and it is a common topic every year on standardized tests like the SAT. #shorts #math #maths #mathematics

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[intrepidca]

And here we also learn not to be fooled by the diagram... since, in this diagram, angle BPC is clearly greater than 90 degrees.

[MrCardeso]

Angles PAC and PAB are 20 deg, since triangles PAC and PAB are isosceles. That makes APC and APB 140 deg each, or 280 combined. 360 - 280 = 80 deg. (and that's the solution!)

[ThePickler]

let angle BAC = x
minor BPC = 2x
major BPC = 360 - 2x
major BPC = 360 - 20 - 20 - x
major BPC = 320 - x
320 - x = 360 - 2x
320 + x = 360
x = 40
2x = 80
minor BPC = 80

[fabianpascalabt]

Starting with the Same Idea He did you could also Figure Out that ABC IS also a cool Triangle, with the equal Angles being 70 degrees each. If you now Look at the remaining Triangle PBC, you know the Angles add to 180, so 70-20=50 on the smaller ones, leaving 80 for the Angle WE were looking for

[MD-vsff]

I went the long way around. The interior angles of a quadrilateral sum to 360. Combine that with the fact that with a circle the center angle is exactly double the edge angle, you get 20, 20, X, and 360-2x for the 4 angles. Solve for X with algebra and you get 40 degrees.

Then I noticed the isoceles triangles. :)

[sabrinab]

Man it's so refreshing to see numbers and circles after all of this crappp online! Thank you! I enjoy your videos alot!

[Dazed_]

I'm sure we were taught a theorem based on which angle BPC is always twice of angle BAC or vice versa

[pkmpkm]

Or you could join B and C, which gives two triangles. And if you solve the triangles, you can get the angle

[ChaineYTXF]

Let angle BAC = y.
Therefore BPC = 2y
Let angles CBP and PCB both = x (isoceles triangle)
These produce two equations:
One in triangle BPC : *2x+2y = 180*
The other in triangle ABC: 2x+20+20+y = 180 and therefore *2x + y = 140*
Equation 1 minus Equation 2:
y = 40
(also, thanks to Equation 1: x = (180-80)/2=40 )

And the angle BPC = 2y = 80 🙂

[kenhaley]

Don't know why this was labeled "HARD". It looked pretty easy compared to most geometry puzzles on this channel.

[tovekauppi]

I used triangle ABC instead but same idea. The important bit is that you remember that the angle on the inside is twice the angle on the outside.

[aryapandey]

easy way-
let angle bac be x
then angle bpc will be 2x
in the quadrilateral
20°+20°+x°+(360-2x)°=360°
because sum of all sides of quadrilateral is equal to 360°
therefore
40°+x°+360°-2x°=360°
400°-x=360°
x°=40°
.therefore 2x=80°
😁💗

[nazzjoe]

If you extend the radius CP to form a diameter to point "X" then arc AX=40°, and if you do the same on radius BP to point "Z" then arc AZ=40°, so AX+AZ = 80° and that is the same with arc BC.

[michaeldakin]

I think I went the long way around - I started with the sum of internal angles of a quadrilateral (360deg), subtracted the given angles (leaving 320deg), then (giving angle BPC value ‘x’) calculated (360-x) + x/2 = 320… x/2=40, so x=80.

[alexkirchoff]

20 + 20 + (360 - x) + x/2 = 360, solution follows.Not always you have to think outside of the box(but I admit that's a good exercise for all the times you actually have).Cheers

[tarrySubstance]

We could assume BAC = x and BPC = 2X
The sum of angles in a 4 sided object is 360°. So 20 + 20 + X + (360 - 2X) = 360
X = 40°. 2×40 = 80°. BPC = 80°. As always a great video presented a different way of finding the solution. (Edited)

[ARKGAMING]

Wow I took a way longer route.(I didn't learn math in English so some of the terms I use here may be different in English).
I proved AB=AC cause they sit on angles of the same sizes. Then I completed the line BC and let PCB=PBC=α. Then BPC=180-2α meaning BAC=90-α
Use the sum of angles in triangle ABC to get 180=90-α+20+α+20+α=130+α
So α=50 → 180-2α=80°. Took me like 3 minutes to do it in my head and your solution is clearly more elegant.

[cdmcfall]

Angle BPC is 80 degrees or 280 degrees since neither major nor minor arc is specified in the original diagram or the problem itself. BPC subtends two arcs -- major arc CAB and minor arc BC -- with supplementary angles, each with the same central vertex and endpoints.

[randerson]

Or extend BP to point D on circle and extend CP to point E on circle. Arc ED is 40 + 40 = 80 degrees. Central angle EPD is 80 degrees. Vertical angle BPC is 80 degrees.

[dinotainx]

It can also be done the following way:-
Here, angle bac= 2angle bpc

In quadrilateral BACP, 360= 20+20+angle bac + 360-2angle bac

Then angle BAC= 40.
And angle BPC=80.