No more right triangle :(

Now we don't get right triangles as often anymore. We can no longer use the short cut of SOHCAHTOA. Now we have to use either the law of sines or the law of cosines. You can only use the law of sines if you have a matched pair of an angle and a side. If you do not you must use the law of cosines. These two formulas are extremely useful.

The equation for law of sines is: Sin(angle)   Sin(angle2)
                                                  opp leg  =opp leg 2
                                         This equation is used by plugging in and cross multiplying

The equation for Law of Cosines is:
      (opp leg)^2=(adj leg)^2+(adj leg2)^2-2(adhleg)*(adj leg)*cos(angle between)

Now I don't know who sits around and tries to figure out this stuff but they obviously had a good reason to do it. They must have expected this to be used.

9-3

This week in advanced math we are learning lesson 9-3.. Law of Sines. When you dont have a right triangle you are going to want to use the Law of Cosines, because you can't use SOHCAHTOA. You are only going to be able to figure the triangle out correctly is if you know the opposite leg and angle value.
FORMULA:
sin(angle)/ opposite leg=sin(angle2 / opposite leg 2

REMEMBER: always cross multiply to solve.

EXAMPLE:
A=50(degrees) B=55(degrees) a=10
now you are going to draw your triangle
now subtract 50 and 55 from 180 to give you your third angle.
C= 75(degrees)
use the formula
sin75 / 10=sin50/x
xsin75= 10sin50
10sin50/ sin75
= 10.249

This is a really confusing section.

The thing I understood most this week was section 9 – 3, Law of sines. The law of sines can only be used with non –right triangles. The only time you can use law of sines is when you have or are given one side, and one angle that are opposites of each other. This is the thing that you try before you use the law of cosines.
The formula is (sin (angle) / opposite leg) = (sin (angle 2) / opposite leg)
To solve the formula you have to cross multiply.
Example 1:
Triangle ABC, angle B = 120 degrees, b = 10, c = 8.
Sin ( 120 degrees ) / 10 = sin(C) / 8
10sin ( C ) = 8 sin( 120 degrees )
10 sin ( C ) = 6.928
Sin ( C ) = 6.928 / 10
C = Sin^ -1 ( 6.928 / 10 )
C = 43.8522 degrees

9-3 is all about the Law of Sines. The law of sine is only used with NON-RIGHT TRIANGLES. You are only about to use the law of sines when you have an angle and an opposite leg value that you know.

Example: An engineer wants to determine the distances from points A to B to a inacessible point C. From direct measurement the engineer knows the AB=25cm AngleA=110° AngleB=20°. Find AC&BC

a.You draw out your triangle from what you already know. Once draw you see the you do not know legs a or b and AngleC.
b. To find angle c subtract the two angles known by 180°.
180-110-25=50°
c.Plug into formula..
sin50/25=sin110/AC
d.Cross multiply to get BCsin50=25sin10
e.Get BC by itself by dividing 50.
f. Answer for angle BC is 25sin110/sin50 which equals 30.667
-Next solve or AC-
a.ACsin50=25sin20
b.Divide to get AC alone.
c.Answer is 25sin20/sin50 which equals 11.162

9-3

This week we learned 9-3. It was explaining and how to solve with law of sines. You use this whenever you can’t use SOHCAHTOA to solve the triangle.

Law of sines is used with non-right triangles and only use when you have an angle plus opp leg values you know.


The formula for law of sines:
Sin (angle)/opp leg = Sin (angle 2)/opp leg 2

You cross multiply to solve law of sine.



Ex:
AB=25m
Find AC and BC
1. Draw the triangle.
2. Find AC.
3. Sin 20/AC= Sin 50/25.
4. You get 25 sin (20) =sin (50) x.
5. You divide sin 50 on both sides.
6. You get 11.612m which is AC.
7. Find BC
8. Sin 50/25= Sin 110/BC
9. You get BC Sin 50= 25 Sin 110.
10. Divide Sin 50 on both sides
11. You get 25 Sin 110/ Sin 50
12. You get BC= 30.667m.



Basically if you know your formula and everything 9-3 is really easy but it can be really difficult if you do not know your formula. Know your formula and everything and you should be good and comfortable with 9-3.

This week we learned 9-3. Law of Sines. You use law of sines when you cant use SOHCAHTOA to solve the triangle.
you can only use law of sines when you have an angle and opposite leg value that you know.

Formula for Law of sines:
sin(angle) / opposite leg=sin(angle2) / opposite leg 2
you cross multiply to solve.

EXAMPLE:
A=45 degrees B=60 degrees a=14
**first you must draw your triangle!
subtract 45 and 60 from 180 to find your 3rd angle.
C=75 degrees.
set it up into your formula.
sin(75) / 14=sin(45) / x.
cross multiply.
xsin75=14sin45
set it up to divide.
14sin45 / sin 75.
plug it into your calculator and you should get an answer of:
x=10.249

Its really not that hard to work these problems. The hardest part is taking the points and making a triangle to solve from word problems.