Question

help!?!?!
Explain the difference between using the sine ratio to solve for a missing angle in a right triangle versus using the co-secant ratio. You must use complete sentences and any evidence needed (such as an example) to prove your point of view.

Answer 1

Id get the question but here some thing that might help.

\[\huge\color{red}{Sin~ V= \frac{opp}{hyp} } \]
\[\huge\color{red}{Csc~ V= \frac{hyp}{opp} } \]
\[\huge\color{green}{Cos~ V= \frac{adj}{hyp} } \]
\[\huge\color{green}{Sec~ V= \frac{hyp}{adj} } \]
\[\huge\color{blue}{Tan~ V= \frac{opp}{adj} } \]
\[\huge\color{blue}{Cot~ V= \frac{adj}{opp} } \]

Answer 2

@hartnn help plz

Answer 3

when you know hypotenuse and another side thats opposite to missing angle in a right triangle, you can use 'sin' or 'co-secant' ratios :-

sin = opposite/hypotenuse

co-secant = 1/sin

so, there is no difference. co-secant is just reciprocal of sin. you can use either of them. However you will end up using sine ratio most of the time, as most calculators allow u to get the inverse sin easily without changing settings.

Answer 4

@beccaboo022a thx

Answer 5

you wlcme ^_^

Answer 6

@wolf1728 thx

Answer 7

u r welcome luka
here's a clickable link to the calculator:
http://www.1728.org/trigcalc.htm