# how to solve angles of trigonometry in general

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### 2 Answers

The request of problem is a bit vague since it does not specify if you need to find the angle in a triangle or you just need to find the angle when the value of a trigonmetric function is given.

Considering that the problem provides the value of a trigonometric function and you just want to know the value of the angle, you need to use the inverse trigonometric functions such that:

`sin alpha = 1/2 => alpha = (-1)^n*arcsin(1/2) + npi`

`alpha = (-1)^n*(pi/6) + npi`

`cos alpha = sqrt2/2 => alpha = +-arccos(sqrt2/2) + 2npi`

`alpha = +-pi/4 + 2npi`

`tan alpha = sqrt3 => alpha = arctan sqrt3 + npi`

`alpha = pi/3 + npi`

`cot alpha = 1 => alpha = arc cot 1 + npi`

`alpha = pi/4 + npi`

**Hence, you may evaluate an angle if the value of trigonometric function is given using inverse trigonometric functions and the general relations described in the examples above.**

For right angled triangles,

sin (angle)=oppsite side/hypotenuse.

cos(angle)=adjacent side/hypotenuse.

tan(angle)=opposit side/adjacent side.

SOH CAH TOA. A way to remember.

For normal triangles,

'A' being an angle, 'a' being the opposite side to 'A'

'B' being an angle, 'b' being the opposite side to 'B'

'C'being an angle, 'c' being the opposite side to'C',

For the missing side,

a^2 = b^2 + c^2 - (2bc x cosA)

b^2 = a^2 + c^2 - (2ac x cosB)

c^2 = b^2 + a^2 - (2ba x cosC)

And for the missing angle,

cosA = b^2 + c^2 - a^2 / 2bc

cosB = a^2 + c^2 - b^2 / 2ac

cosC = b^2 + a^2 - c^2 / 2ba