how do you do trig with not right angle?

Answer 1

If you do not have right angles your trig answers will be wrong!
Sound like a joke, does it not? Well, in that case it is a serious joke :).

To find a solution start with dividing your triangle into two right angled triangles and then solve the problem. It should be easy since you know the hypotenuse (of newly formed triangle) and the adjacent to it angle.

Answer 2

All triangles, right, oblique, scalene, etc. can be solved using the generalized trigonometric functions:

Law of cosines:
(i.e. angle A is opposite to side "a")
a^2 = b^2 + c^2 - 2bc*cos(A)
b^2 = a^2 + c^2 - 2ac*cos(B)
c^2 = a^2 + b^2 - 2ab*cos(C)
If you know all three side lengths, you can find one of the angles.
Or, if y ou know two sides and an angle, you can find the other side for ANY triangle.

Law of sines:

(Sin A) / a = (Sin B) / b = (Sin C) / c
If you know two or three sides or angles, you can find the remaining sides and angles of the triangle.

Triangle Area:
If you don't know the base and height, just the 3 sides:

Area = sqrt[ s*(s - a)*(s - b)*(s - c)]
where s = (a + b + c) / 2

Answer 3

The two main formula are:

The Sine Rule:
a/sin A = b/sin B = c/sin C

The Cosine Rule:
a^2 = b^2 + c^2 - 2bc cos A

Both are easy to prove. If you find difficulty, you can search the web for proofs.

Answer 4

its easy.....u just gotta breakup the triangle into pieces of right angled triangles...and then use the usual trigo relations on those rt. angld triangles.....n then put them together

n there are many othr trigo relations for which u dont need a right angled triangle
eg: for all triangles...the ratio of length of any side, to the sine of the angle to THAT side's opposite , is constant n equal to the remaining two side's ratio...

Answer 5

Law of Cosines

c^2 = a^2 + b^2 - 2ab(cosC)
a^2 = b^2 + c^2 - 2bc(cosA)
b^2 = a^2 + c^2 - 2ac(cosB)

Law of Sines
a/sinA = b/sinB = c/sinC

Law of Tangents
(a - b)/(a + b) = tan [(A - B)/2] / tan [(A + B)/2]

a, b, and c are the sides
A, B, and C are the angles

Answer 6

there are a differnet set of trig formulas for that, not Sin=o/h cos=a/h and Tan= o/a

the formulas are escaping me at the moment, but im sure there are different formulas

Answer 7

when you go to higher class you will learn to do trigs with angles like 1080 deg 18000 deg and so on.