If 5cosA+3=0 and 180⁰<A<306⁰ determine by means of a diagram tan²A

Answer 1

You can calculate tan² A

5 cos A + 3 = 0

Subtract 3 to both sides

5 cos A = - 3

Divide both sides by 5

cos A = - 3 / 5

For angles 180⁰ < A < 360⁰

The cosine is negative in quadrant III and positive in quadrant IV.

Angle A lies in quadrant III ( 180⁰ < A < 270⁰ )

cos² A = 9 / 25

Use basic trigonometric identity:

sin² A + cos² A =1

sin² A + 9 / 25 = 1

Subtract 9 / 25 to both sides

sin² A = 1 - 9 / 25

sin² A = 25 / 25 - 9 / 25

sin² A = 16 / 25

tan² A = sin² A / cos² A

tan² A = ( 16 / 25 ) / ( 9 / 25 )

tan² A = 16 / 9

Answer 2

5cosA+3=0 , where A is in quadrant III or IV
cosA = -3/5 , but the cosine is negative in III, so A must be in III
now construct a diagram in quadrant III
with x = -3, y = ? and r = 5
Use Pythagoras or you should recognize the 3-4-5 right-angled triangle, so
sinA = -4/5

tan A = sinA/cosA = (-4/5) / (-3/5) = 4/3

tan^2 A = 16/9