Sunday, November 8, 2009

Taylor Reflection #12

Everything we learned this week was very simple formulas to memorize and plugging in to solve.
I also took the chapter seven test finally
I NEED SOMEONE TO SHOW ME HOW TO SOLVE WHAT WE WILL NEED TO SOLVE WITH OUT THE CALCULATOR if that makes sense,
i dont know if that makes sense but in other words... i need someone to show me how to work the problems that i work now with a calculator that i wont have a calculator for later because everthing from chapter seven i use a calculator for..

I ALSO COULD USE HELP ON APPLYING THE UNIT CIRCLE.
ive learned it and i kind of know how to apply it
but i don't know all of the times when it is applied..


What i do know is how to clear up any possible confusion for plugging into the formulas of this past week ::
to solve a right triangle you use SOHCAHTOA

The hypotenuse is opposite of the right angle.
The adjacent leg is the side that is attached to the "given angle (not right angle)"
The opposite leg is the side that is across (or opposite) the "given angle (not right angle)"

SOHCAHTOA is really a anagram to remember the way to work out a right angle triangle problem using sine cosine and tangent

SOH: sin(theta)=opposite leg/hypotenuse
CAH: cos(theta)=adjacent leg/hypotenuse
TOA: tan(theta)=opposite leg/adjacent leg

solving for a right triangle is just a matter of drawing and plugging in





#2 Laws of Sines: sin A/a = sin B/b = sin C/c

*used when pairs are known in non-right angles
*set up a proportion

Must use a "pair" for sin A over a

a "pair" is given after you take the given information from a problem and you look for an angle given and a number to represent the leg directly across from that angle. Sin B/b is solved by simple algerbra and Sin C/ c is what you"ll get to have the measurements for the rest of the triangle.



#3 Area of a non-right triangle = 1/2 (leg) (leg) sin (angle b/w)

This can possibly be the most simple formula to remember and use
legs are numbers for either the hypotenuse, the adjacent, or the opposite lines in the drawn problem. any "leg" will work for the problem.

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