Trig help please

PiLuvah · Hodge Conjecture Offline Joined: 12 Aug 2008 Posts: 55

Guys I dont know when to use law of cosines and when to use law of sines. Can someone explain it to me and give me practice examples?

**Posted:** Sat Nov 07, 2009 5:38 pm

modularmarc101

Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos C$

Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

Notice that it depends on the information given and what is needed. I don't have any examples though.

**Posted:** Sat Nov 07, 2009 5:43 pm

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Forte

Hey Nate hows it going?

Alrite so, we use Law of Cosines when we have 2 sides and one included angle. Law of sines is used when we have 2 sides and 2 angles.

Try these:
In triangle RST, r=3√2, s=1, and the measure of angle T is 135. Find t.

(This one is harder)
The vertices of a triangle inscribed in a circle separate the circle into arcs whose measures are in the ratio 2:3:4. If the measure of the radius of the circle is 10. Find the longest side of the triangle.

Law of Sines:
In triangle CAR, a=24. Angle A is 27. And measure of C is 83. Find c to the nearest integer.

In triangle RST, Sin r is 0.4, and the measure of angle S is 30. Find the ratio of r:s.

**Posted:** Sat Nov 07, 2009 5:45 pm

PiLuvah · Hodge Conjecture Offline Joined: 12 Aug 2008 Posts: 55

Oh and I forgot how do you find the area of a triangle using trig? And hey Navdeep

And can you give a hint to how to start the 2nd problem?
I'll give the answers at once

**Posted:** Sat Nov 07, 2009 5:52 pm

modularmarc101

There are many, many, many, many ways. But most are derived from $\frac {ab}{2} \sin A$ .

Hint for second problem: The measures of the arcs add up to 360. Once you find that, notice that the measure of an inscribed angle if half of the arc it opens up to. One last thing, the FULL law of sines is

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$ ,

where $R$ is the radius of the circumcircle of the triangle.

Forte

Uhm. Draw a diagram. Label the arcs in the ratio 2:3:4. So Find each arc. A central angle is congruent to it's intercepted arc. Law of cosines.

**Posted:** Sat Nov 07, 2009 5:56 pm

PiLuvah · Hodge Conjecture Offline Joined: 12 Aug 2008 Posts: 55

i got the answer 20 but if its 20, it cant be a triangle. (triangle inequality rules) did I do something wrong?

And thanks a lot Marc. I get it now Smile

jhooper3581

Consider this link and this link for the laws of sines and cosines.

**Posted:** Sat Nov 07, 2009 6:01 pm

Forte

Actually no, thats the answer. I think the exact answer was like 19.8... something of that sort. But 20 is right. Its like when you first learned about triangles and the teacher labeled it a 1,2,3 triangle. It's perfectly fine. Do the Law of Sines now. Mr. Green

**Posted:** Sat Nov 07, 2009 6:02 pm