In a right triangle B = 55° 30', and b = 6.05. Find c and a.
Sighs. First off, whoever invented Trigonometry, was clearly not a very kind hearted person.
I am attempting to learn Trigonometry but things are not boding very well so far. It's Sines, that's getting me all mixed up. The site I'm using has the problem,
In a right triangle B = 55° 30', and b = 6.05. Find c and a.
Then it has under the answer,
c = b/sin B = 6.05/sin 55°30' = 7.34.
a = 4.16.
Now I see how since, sin B = b/c, c = b/sin B, which would equal 6.05/sin 55°30' , but I don't see how they get 7.34. I mean how do we know what the sin of 55°30' is, and therefore what to devide 6.05 by? What is it, I'm missing? I've gone through the lesson on the site several times, but see nothing that explains it (but I could very likely just be missing something).
I do understand however how they get a once they have the 7.34.
Sighs. First off, whoever invented Trigonometry, was clearly not a very kind hearted person.
I am attempting to learn Trigonometry but things are not boding very well so far. It's Sines, that's getting me all mixed up. The site I'm using has the problem,
In a right triangle B = 55° 30', and b = 6.05. Find c and a.
Then it has under the answer,
c = b/sin B = 6.05/sin 55°30' = 7.34.
a = 4.16.
Now I see how since, sin B = b/c, c = b/sin B, which would equal 6.05/sin 55°30' , but I don't see how they get 7.34. I mean how do we know what the sin of 55°30' is, and therefore what to devide 6.05 by? What is it, I'm missing? I've gone through the lesson on the site several times, but see nothing that explains it (but I could very likely just be missing something).
I do understand however how they get a once they have the 7.34.