For finding angles given trigonometric values, we take the inverse of the trigonometric ratio and also use the rules to find the sign of trigonometric ratio in each quadrant.

Let us look at some examples of finding angles given trigonometric values:

**Example 1: **sin θ = 0.3924

Sine value is positive, hence the angle θ can be in first or second quadrant.

θ = (0.3924) = 23°6′

This angle is in the first quadrant. To find the angle in the second quadrant, we use 180 – θ.

So 180 – 23°6′ = 156°54′

Hence, the solutions or angles are 23°6′ and 156°54′

**Example 2: **sin θ = – 0.3924

Sine value is negative, hence the angle θ must be in the third or fourth quadrant.

θ = (-0.3924) = -23°6′ = 360° – θ

When you find the inverse of a negative trigonometric ratio, the calculator will always give the angle in the negative to represent 360° – θ, which is in the fourth quadrant (for sine and tangent’s negative values).

The required angles in our question is in the third quadrant, i.e. 180 + θ; and in the fourth quadrant, i.e. 360 – θ.

Therefore the two angles are:

180° + θ = 180° + 23°6′ = 203°6′

360° – θ = 360° – 23°6′ = 336°54′

**Note:**** **Another method to find the angle when trigonometric ratio value is negative is to ignore the negative sign and find the trigonometric ratio’s inverse for the value. Then we find where the trigonometric ratio is negative i.e., in which quadrant and find the angle accordingly.

Let us take the same example from above, viz. sin θ = -0.3924.

Sine is negative and in the third and fourth quadrant. We first find the value of:

(0.3924) by ignoring the negative sign and getting θ = 23°6′.

Since sine is negative and in the third and fourth quadrant, we find the value in

Third quadrant = 180° + θ = 180° + 23°6′ = 203°6′; and

Fourth quadrant = 360° – θ = 360° – 23°6′ = 336°54′

**Example 3: **tan θ = 0.25

tan is positive and is in the first and third quadrants.

θ = (0.25) = 14°2′

Therefore the two angles are:

θ and 180° + theta, i.e. 14°2′ and 180° + 14°2′

Hence the angles are: 14°2′ and 194°2′

**Example 4: ** sec θ = -2.6843

We know that sec θ =

Hence sec θ is negative where cos θ is negative.

cos θ is negative in the second and third quadrant. Hence sec θ is negative in the second and third quadrant.

Now sec θ = = -2.6843

Therefore cos θ = = -0.3725366

cos θ = -0.3725366

θ = (+0.3725366) = 68°8′

In the second quadrant, the angle is 180° – θ = 180° – 68°8′ = 111°52′.

Now since (+0.3725366) = 68°8′, θ is in the third quadrant.

So θ = 180° + 68°8′ = 248°8′

**Example 5:** If sin A = and tan A is positive, find the value of A.

For sine to be negative and tangent to be positive, the angle must be in the third quadrant.

θ = = 22°37′

Required angle is in the third quadrant, i.e. 180 + θ

So A = 180° + θ = 180° + 22°37′ = 202°37′