First you can find angle B:
Angle B = 180 - ( 37+54 )
= 180 - 91
= 89 degrees / Angle B = 89 degrees
Now you can plug in the numbers that you have so far :
Sin 37 = Sin 89 = Sin 54
______ ______ ______
41 b x
Then you can find out what side b is equal to:
Sin 37 = Sin 89
_______ _______ } Cross - multiply
Sin 89 x 41
____________ = b So, side b is equal to 68.
So, now you should have : Sin 37 = Sin 89 = Sin 54
_______ _______ _______
41 68 x
Now you can find out what side x is equal to:
Sin 37 = Sin 54
_____ ______ } Cross - multiply
Sin 54 x 41
___________ = x So, side x is equal to 55.
The Cosine Law:
* You need at least two sides for the cosine law to apply to a problem.
a2 = b2 + c2 - 2bcCosA
OR: [ b2 = a2 + c2 - 2acCosB ]
OR: [ c2 = a2 + b2 - 2abCosC ]
Here's an example of how to use the fromula:
Solve the triangle WTV.
w2 = v2 + t2 - 2vtCosW
w2 = 7.82 + 92 - 2 ( 7.8 x 9 ) Cos112
w2 = 60.84 + 81 - ( - 52.59 )
w2 = 194.43
The square root of 194.43 is 13.94
w = 13.94
--Now you can the same thing that you did when you were using the Sine Law:
Sin 112 = Sin T
_____ _____ } Cross - mulitply
Sin 112 x 9
_________ = T
Then do Inverse Sine to get the angle ( Depending on what type of calculator you have, the way of doing this may vary... On my calculator, I press the 2nd function button, then Sin -1, then the number I got from the last equation. ) .
T = 36.7 degrees.
To get angle V, all you do is: 180 - (112 + 36.7) = 31.3 V = 31.3 degrees
Our homework was Excersize # 19 & # 25
If you are still having trouble with the Sine & Cosine Laws, here is a website that should help you to better understand it. =)
Just a few reminders:
- There will be a quiz on Monday on what we've covered on Trigonometry .
& if anyone would like to contribute to sending flowers to Chris showing us our support, bring $2.00 to Mrs.Armstrong. She will then go & buy the flowers.
Well, I did my best. Lol. I hope this blog was helpful. The next person to blog is.. Meagan. =)
( I apologise if my images don't show up on the blog. I seem to be having some trouble uploading them. =( )
2) Then let's find the obtuse angle. Do the same thing as you did to the positive ratios.
d) cos(x) = -.970 which is 166 degrees
acute angle = 166 degrees
obtuse angle = 194 degrees
e) sin(x) = -.5 which is 30 degrees
acute angle = 30 degrees
obtuse angle = 330 degrees
f) tan(x) = -1.43 which is 55 degrees
acute angle = 55 degrees
obtuse angle = 235 degrees
And that's it! Sorry for the delay guys, there was too much in my head, although it's done now (: we also got our unit tests back then we went over some of the questions. Congratulations to the all stars! And oh yeah, I pick Desitini to blog next since the people I know didn't want to since they're "busy", haha. Oh yeah, and reminder to redo question #2 on exercise 18 using the above steps.
Ms. Armstrong, a video for this one is kind of hard to find. If anyone can help me , that'll be great :)
SOH CAH TOA
The tangent function means you need to divide the opposite by the adjacent. How would that be done on a Cartesian plane?
which can also be shown like this:
the "x" in the denominator stands for the cosine which is related to the x axis
the "y" in the numerator stands for the sine which is related to the y axis
In the example below, shows the two quadrants that contain positive tangent values:
In today's class, we somehow fit two lessons in one day simply because I had to remind her that we didn't have math class tomorrow. Urgh... **groan** The mistake is mine, but the bad is not mine. Anyhow...in the first part of class, we learned about multiplying and dividing radicals. When multiplying radicals, here is one rule, or example to follow.
That's easy enough to do though isn't it? Well of course, that example was just a review of what we've been doing in the previous days. It gets a little more complicated than that. Today, we also learned how to multiply radical binomials. Now this is pretty similar to the things we did in the first unit we did this year on polynomials and factoring them. In the following I will explain, [or try to...] the concept in a step-by-step process in the following example:
Simplify: 6√2 (√6 – 2√4)
1.) In order to simplify this, we must distribute the both the radical and the....oh...the other number, I don't quite know what the exact term is, but in this case it's the 6 right before the square root of 2. You multiply that with the same type of number of both sides of the binomial and then multiplying the radicals with the radicals resulting in...
But when you are stuck in the situation where you have a radical binomial in the denominator, then it goes something like...
It is okay to have a radical in the numerator, but only then are numbers in their simplest form is when they do not have both a negative exponent and a radical in the denominator.
However, if there is a binomial in the denominator, the "difference of squares" must be multiplied. [REVIEW of DIFFERENCE OF SQUARES: Ex. (2x-3)(2x+3) = 4x2 - 9]
An example of this might be...
"The next person who will post will be "tall guy" Chris. Yes you. Haha. Mwahahahhaha.... =D"The next person who will post will be the "small person" Jamie. Yes you. Haha. Mwahahahhaha....=D