# Math Help - sin and cos and triangles oh my

1. ## sin and cos and triangles oh my

hello everyone i hope your all well , anyway im new to the forum and need a little help solving some math problems i hope you can help me, first off im not very good at math and i have a hard time explaining the math so im sorry if i dont make much since but here it goes, basicly depend on the question i have to solve for either, A, a, B, b, C or c for either angle or degree if you under stand so far and i have some answered but the ones with the first angle and degrees to solve for the others confuses me here is an example $A = 49\deg , B = 44\deg, a = 18.7$ its the 18.7 with the others that confuses me, the way my teacher said to solve it, is like this, $18.7\times \sin49\deg = 14.11 \div \sin44\deg = 20.31$, which is not even an answer, so if someone has an easier way to solve a problem like this where a number is like 18.7 (with the point 7 or point whatever) please let me know and step by step on how to solve it please and im really really sorry for the long post. thanks in advance.

2. Originally Posted by luke9511
hello everyone i hope your all well , anyway im new to the forum and need a little help solving some math problems i hope you can help me, first off im not very good at math and i have a hard time explaining the math so im sorry if i dont make much since but here it goes, basicly depend on the question i have to solve for either, A, a, B, b, C or c for either angle or degree if you under stand so far and i have some answered but the ones with the first angle and degrees to solve for the others confuses me here is an example $A = 49\deg , B = 44\deg, a = 18.7$ its the 18.7 with the others that confuses me, the way my teacher said to solve it, is like this, $18.7\times \sin49\deg = 14.11 \div \sin44\deg = 20.31$, which is not even an answer, so if someone has an easier way to solve a problem like this where a number is like 18.7 (with the point 7 or point whatever) please let me know and step by step on how to solve it please and im really really sorry for the long post. thanks in advance.
what your teacher was doing was applying the "sine rule" or "the law of sines"

see if this is any help

you may also want to do a google search for "law of sines"

if you read and still need further explanation, come back and be specific as to what bothers you

and another

3. Originally Posted by Jhevon
what your teacher was doing was applying the "sine rule" or "the law of sines"

see if this is any help

you may also want to do a google search for "law of sines"

if you read and still need further explanation, come back and be specific as to what bothers you

and another

thanks for the quick answer and i went on google and found more info thanks , got another question if you dont mind whats the best way to solve SSA?

4. Originally Posted by luke9511
thanks for the quick answer and i went on google and found more info thanks , got another question if you dont mind whats the best way to solve SSA?
SSA - do you mean you know 2 sides and 1 angle? You need to be more specific:

It depends on what you want to find (another angle or the third side) and where the two known sides are in relation to the known angle.

You need to work on understanding the sine and cosine rules and their application well enough so that you can tackle each problem on its merits rather than running through some check list .....

5. Originally Posted by mr fantastic
SSA - do you mean you know 2 sides and 1 angle? You need to be more specific:

It depends on what you want to find (another angle or the third side) and where the two known sides are in relation to the known angle.

You need to work on understanding the sine and cosine rules and their application well enough so that you can tackle each problem on its merits rather than running through some check list .....
ok well i managed to figure it out but heres another question, how do i solve the area of a triangle when 1 or 2 of the sides are in inches or meters? heres an example: a = 9 in, b = 18 m, c = 11 m

6. Originally Posted by luke9511
ok well i managed to figure it out but heres another question, how do i solve the area of a triangle when 1 or 2 of the sides are in inches or meters? heres an example: a = 9 in, b = 18 m, c = 11 m
Heron's Formula is probably the fastest and easiest way.

Alternatively, you first need to use the cosine rule to find an angle, and then use the usual formula $A = \frac{1}{2} ab \sin(C)$ where C is the angle opposite side c.

As far as the inches meters problem ...... choose a unit you want to work in, say cm ...... there are 2.54 cm in 1 inch. So 1 inch = 2.54 cm therefore 9 inches = (9)(2.54) = 22.86 cm = 0.2286 m.

7. ok if i have a question thats how much would a investment be at 10% annually for 4 years, i know how to do it cause annually is 1 year, but how would i do semiannually and quarterly on a TI-83 calculator or any calculator for that matter? thanks in advance

8. Originally Posted by luke9511
ok if i have a question thats how much would a investment be at 10% annually for 4 years, i know how to do it cause annually is 1 year, but how would i do semiannually and quarterly on a TI-83 calculator or any calculator for that matter? thanks in advance