SPIN: Not The Best At Maths, NEED HELP WITH MY TEST PLEASE

[SAINT]

Math is my favorite subject and even then I still despised trig. You on your own

 

[Paper Boi]

we don’t do ‘maths’ in America

 

[Kornball The Conqueror]

6 don't as me how

 

[KyokushinKarateMan]

Fam.. like others have said a triangle is a total of 180 degrees. 180 - (120 + 15) = 45 so angle B is 45 degrees.

then you take all that info and use the tan cos cotangent bs to find the Missing length of the triangle.

lemme see what I can find.

Edit: scratch all that sine cosine sh. To find the missing length use Pythagorean’s theorem. The side opposite of the 120 degrees is c as it’s the biggest angle

so 11.4^2 + 4.2^2 = c^2

Yeah, I knew that already and stated so in the OP. I even explained HOW I reached 45deg for angle B

And no, you CANNOT use pythagorean theroem, which I also explained in the OP. Why are you even here

 

[KyokushinKarateMan]

Math is my favorite subject and even then I still despised trig. You on your own

@CopiousX

I figured it out. Well, not the answer, not yet. It requires not the Law of Cosines, but the SINE/COSINE Rule.

A^2 = B^2 + C^2 - 2bcCOSA

So: 120^2 = (45)^2 + (15)^2 - 2(45x15)COS120

Will do it tonight. I've had enough of this sh*t for now.

 

[KyokushinKarateMan]

6 don't as me how

You don't have to know an ounce of math whatsoever to know that the unknown side couldn't possibly be no damn "6" inches long when it's clearly visibly longer than the other two sides, which are 4.2 and 11.4 inches long. Please stop trolling at a time like this man

 

[SheWantTheD]

Yeah, I knew that already and stated so in the OP. I even explained HOW I reached 45deg for angle B

And no, you CANNOT use pythagorean theroem, which I also explained in the OP. Why are you even here

Why can’t you use it?

 

[Mannish]

Here's how I got 13.98:

You want to figure out the length of the unknown segment of the triangle (X)

Draw a line C bisecting the 120 degree angle that's perpendicular to the unknown side X , splitting the unknown side into two segments (Let's call them D and E) and the 120 degree angle into two angles (let's call them F and G) .

Now you have two right triangles, one comprised of sides C and D with hypotenuse of 4.2"; the other comprised of sides C and E = X - D with a 11.4" hypotenuse

Solve for C and D by noting that the right triangle with the 4.2" hypotenuse has two angles of 45 degrees (B, which is 180 - 120 - 15, and F, which is 180 - 90 - B).
- Therefore, the two sides have to be equal length, and a^2 + b^2 = c^2 can be simplified to 2a^2 = c^2
- 2a^2 = (4.2)^2 ---> a = 2.97 = C = D

Solve for E:
- a^2 + b^2 = c^2
- C^2 + E^2 = (11.4)^2
- (2.97)^2 + E^2 = (11.4)^2 ---> E = 11.01

finally, segment X = D + E = 2.97 + 11.01 = 13.98

 

[little4209]

I hate this sh1t dumbass math you never use

 

[KyokushinKarateMan]

Here's how I got 13.98:

You want to figure out the length of the unknown segment of the triangle (X)

Draw a line C bisecting the 120 degree angle that's perpendicular to the unknown side X , splitting the unknown side into two segments (Let's call them D and E) and the 120 degree angle into two angles (let's call them F and G) .

Now you have two right triangles, one comprised of sides C and D with hypotenuse of 4.2"; the other comprised of sides C and E = X - D with a 11.4" hypotenuse

Solve for C and D by noting that the right triangle with the 4.2" hypotenuse has two angles of 45 degrees (B, which is 180 - 120 - 15, and F, which is 180 - 90 - B).
- Therefore, the two sides have to be equal length, and a^2 + b^2 = c^2 can be simplified to 2a^2 = c^2
- 2a^2 = (4.2)^2 ---> a = 2.97 = C = D

Solve for E:
- a^2 + b^2 = c^2
- C^2 + E^2 = (11.4)^2
- (2.97)^2 + E^2 = (11.4)^2 ---> E = 11.01

finally, segment X = D + E = 2.97 + 11.01 = 13.98

I got 13.96. Above, I wrote it out as the Cosine Rule. Turns out it's actually the SIN rule: a/SINA = b/SINB = c/SINC
Substitute the stuff we already know and boom, 13.96"
This sh*t took my whole weekend, and this is only the beginning of the curriculum