TRIGONOMETRIC RATIOS OF SUPPLEMENTARY ANGLES WORKSHEET

Trigonometric Ratios of Supplementary Angles Worksheet :

Worksheet given in this section will be much useful for the students who would like to practice problems on trigonometric ratios of supplementary angles.

Before look at the worksheet, if you would like to learn trigonometric ratios of supplementary angles,

Problem 1 :

Evaluate :

sin (180° - θ)

Problem 2 :

Evaluate :

cos (180° - θ)

Problem 3 :

Evaluate :

tan (180° - θ)

Problem 4 :

Evaluate :

csc (180° - θ)

Problem 5 :

Evaluate :

sec (180° - θ)

Problem 6 :

Evaluate :

cot (180° - θ)

Problem 7 :

Evaluate :

sin (180° + θ)

Problem 8 :

Evaluate :

cos (180° + θ)

Problem 9 :

Evaluate :

tan (180° + θ)

Problem 10 :

Evaluate :

csc (180° + θ)

Problem 11 :

Evaluate :

sec (180° + θ)

Problem 12 :

Evaluate :

cot (180° + θ)

Trigonometric Ratios of Supplementary Angles Worksheet - Solutions

Problem 1 :

Evaluate :

sin (180° - θ)

Solution :

To evaluate sin (180° - θ), we have to consider the following important points.

(i)  (180° - θ) will fall in the II nd quadrant.

(ii)  When we have 180°, "sin" will not be changed as "cos"

(iii)  In the II nd quadrant, the sign of "sin" is positive.

Considering the above points, we have

sin (180° - θ)  =  sin θ

Problem 2 :

Evaluate :

cos (180° - θ)

Solution :

To evaluate cos (180° - θ), we have to consider the following important points.

(i)  (180° - θ) will fall in the II nd quadrant.

(ii)  When we have 180°, "cos" will not be changed as "sin"

(iii)  In the II nd quadrant, the sign of "cos" is negative.

Considering the above points, we have

cos (180° - θ)  =  - cos θ

Problem 3 :

Evaluate :

tan (180° - θ)

Solution :

To evaluate tan (180° - θ), we have to consider the following important points.

(i)  (180° - θ) will fall in the II nd quadrant.

(ii)  When we have 180°, "tan" will not be changed as "cot"

(iii)  In the II nd quadrant, the sign of "tan" is negative.

Considering the above points, we have

tan (180° - θ)  =  - tan θ

Problem 4 :

Evaluate :

csc (180° - θ)

Solution :

To evaluate csc (180° - θ), we have to consider the following important points.

(i)  (180° - θ) will fall in the II nd quadrant.

(ii)  When we have 180°, "csc" will not be changed as "sec"

(iii)  In the II nd quadrant, the sign of "csc" is positive.

Considering the above points, we have

csc (180° - θ)  =  csc θ

Problem 5 :

Evaluate :

sec (180° - θ)

Solution :

To evaluate sec (180° - θ), we have to consider the following important points.

(i)  (180° - θ) will fall in the II nd quadrant.

(ii)  When we have 180°, "sec" will not be changed as "csc"

(iii)  In the II nd quadrant, the sign of "sec" is negative.

Considering the above points, we have

sec (180° - θ)  =  - sec θ

Problem 6 :

Evaluate :

cot (180° - θ)

Solution :

To evaluate cot (180° - θ), we have to consider the following important points.

(i)  (180° - θ) will fall in the II nd quadrant.

(ii)  When we have 180°, "cot" will not be changed as "tan"

(iii)  In the II nd quadrant, the sign of "cot" is negative.

Considering the above points, we have

cot (180° - θ)  =  - cot θ

Problem 7 :

Evaluate :

sin (180° + θ)

Solution :

To evaluate sin (180° + θ), we have to consider the following important points.

(i)  (180° + θ) will fall in the III rd quadrant.

(ii)  When we have 180°, "sin" will not be changed as "cos"

(iii)  In the III rd quadrant, the sign of "sin" is negative.

Considering the above points, we have

sin (180° + θ)  =  - sin θ

Problem 8 :

Evaluate :

cos (180° + θ)

Solution :

To evaluate cos (180° + θ), we have to consider the following important points.

(i)  (180° + θ) will fall in the III rd quadrant.

(ii)  When we have 180°, "cos" will not be changed as "sin"

(iii)  In the III rd quadrant, the sign of "cos" is negative.

Considering the above points, we have

cos (180° + θ)  =  - cos θ

Problem 9 :

Evaluate :

tan (180° + θ)

Solution :

To evaluate tan (180° + θ), we have to consider the following important points.

(i)  (180° + θ) will fall in the III rd quadrant.

(ii)  When we have 180°, "tan" will not be changed as "cot"

(iii)  In the III rd quadrant, the sign of "tan" is positive.

Considering the above points, we have

tan (180° + θ)  =  tan θ

Problem 10 :

Evaluate :

csc (180° + θ)

Solution :

To evaluate csc (180° + θ), we have to consider the following important points.

(i)  (180° + θ) will fall in the III rd quadrant.

(ii)  When we have 180°, "csc" will not be changed as "sec"

(iii)  In the III rd quadrant, the sign of "csc" is negative.

Considering the above points, we have

csc (180° + θ)  =   - csc θ

Problem 11 :

Evaluate :

sec (180° + θ)

Solution :

To evaluate sec (180° + θ), we have to consider the following important points.

(i)  (180° + θ) will fall in the III rd quadrant.

(ii)  When we have 180°, "sec" will not be changed as "csc"

(iii)  In the III rd quadrant, the sign of "sec" is negative.

Considering the above points, we have

sec (180° + θ)  =  - sec θ

Problem 12 :

Evaluate :

cot (180° + θ)

Solution :

To evaluate cot (180° + θ), we have to consider the following important points.

(i)  (180° + θ) will fall in the III rd quadrant.

(ii)  When we have 180°, "cot" will not be changed as "tan"

(iii)  In the III rd quadrant, the sign of "cot" is positive.

Considering the above points, we have

cot (180° + θ)  =  cot θ

After having gone through the stuff given above, we hope that the students would have understood trigonometric ratios of supplementary angles

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