Question

PLEASE HELP I ONLY HAVE 2 MORE QUESTIONS LEFT !!!!!!!!!!!! Explain the difference between using the trigonometric ratios (sin, cos, tan) to solve for a missing angle in a right triangle versus using the reciprocal ratios (sec, csc, cot). You must use complete sentences and any evidence needed (such as an example) to prove your point of view.

Answer 1

a dark day indeed where math requires words and sentences instead of good old numbers and expressions :'(

Answer 2

yep

Answer 3

Using the inverse of sin/cos/tan will give you the angle in a right triangle
the reciprocal of sec is 1/cosx so if you you were trying to find the angle youd use cosine inverse, and changing cosine inverse to secant inverse would be 1/inversecosx which would give you 1/ (angle value here) so basically the inverse of the reciprocal trig values give you the reciprocal angle while the inverse normal trig values will give you the normal angle.

Answer 4

i don't understand

Answer 5

@Synkronize

Answer 6

i will give a metal !

Answer 7

the difference would be the orientation and location of the angle of reference

Answer 8

making use of the SOH CAH TOA
you want to first identify your angle of reference
then whatever is across that angle would be the opposite side
if you were given an angle, the opposite side, you can solve for the hypotenuse
using Sin(angle) = opposite side/h
then solve for h using algebraic manipulation

Answer 9

but what else ? bc
that is not enough

Answer 10

the adjacent side is the side next to the angle of reference. it is usually perpendicular to the opposite side

if you were given the length of the adjacent side and opposite side you can solve for the angle
you can use TOA
tangent(angle) = opposite/adjacent
then obtain the arctan of the resulting ratio

Answer 11

http://mathbits.com/MathBits/TISection/Trig/inversetrig.htm you can also use for refrence