[jakayaki]

Exercise 6D

8 (a) sec²θ = 3 tan

this is what i've worked out...

tan²θ + 1 = 3tanθ
tan²θ - 3tanθ + 1 = 0

= (3 ± √5)/2

so tanθ = 2.618 or 0.382
= 69.1 or 20.9

the answers at the back show = 20.9°, 69.1°, 201°, 249°

I think it's related to the "ASTC" Quadrants but I'm not quite sure how it works... Could someone please explain to me ><" Thanks in advance!

[steve2005]

Originally Posted by jakayaki

Exercise 6D

8 (a) sec²? = 3 tan

this is what i've worked out...

tan²? + 1 = 3tan?
tan²? - 3tan? + 1 = 0

= (3 ± ?5)/2

so tan? = 2.618 or 0.382
= 69.1 or 20.9

the answers at the back show = 20.9°, 69.1°, 201°, 249°

I think it's related to the "ASTC" Quadrants but I'm not quite sure how it works... Could someone please explain to me ><" Thanks in advance!

The answers in the back of the book are correct. See picture.

You have quoted ASTC usually CAST is used. The "A" goes in the first quandrant and means ALL trig are positive. The S goes in the second quantrant ( 90 - 180) and indicates the sin ratio is positive all the others are negative. ....

[Chewwy]

Originally Posted by jakayaki

Exercise 6D

8 (a) sec²θ = 3 tan

this is what i've worked out...

tan²θ + 1 = 3tanθ
tan²θ - 3tanθ + 1 = 0

= (3 ± √5)/2

so tanθ = 2.618 or 0.382
= 69.1 or 20.9

the answers at the back show = 20.9°, 69.1°, 201°, 249°

I think it's related to the "ASTC" Quadrants but I'm not quite sure how it works... Could someone please explain to me ><" Thanks in advance!

you've found answers in the first quadrant, remembering tan is also positive in the 3rd quadrant, you just need to add 180 to your first quadrant answers.

[steve2005]

Here is a CAST Diagram with a few examples. The angles go anitclockwise. So 210 comes in the third quadrant. The tan of 210 is positive and equal to tan 30

cos 330 comes in the 4th quadrant and is positve and equal to cos30
tan 100 is negative and is equal to minus tan80

[Eddie K]

Worked Solution:

sec²θ = 3tanθ
tan²θ + 1 = 3tanθ
tan²θ - 3tanθ + 1 = 0

Completing the square:

(tanθ - 3/2)² + C = 0
= (tanθ - 3/2)(tanθ - 3/2) + C = 0
= tan²θ - 3tanθ + 9/4 + C = 0
=>
9/4 + C = 1
C = -5/4
=>
(tanθ - 3/2)² - 5/4 = 0

Solving for 0° < θ < 360°:

(tanθ - 3/2)² - 5/4 = 0
(tanθ - 3/2)² = 5/4
tanθ - 3/2 = ±√(5/4)
tanθ = 3/2 ± √(5/4)
tanθ = 2.61803 and 0.38197
θ = 69.1° and 20.9°

tanθ is positive so will have solutions in the 1st and 3rd quadrants. Solutions in the 3rd quadrant are:

θ = (180° + 69.1°) and (180° + 20.9°)
θ = 249.1° and 200.9°

θ = 20.9°, 69.1°, 249.1° and 200.9°

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To find the 'other' solutions, use the root solutions. These are the ones that lie in the first quandrant. i.e. are less than 90° or π/2^c. In this case, look at the solution for tanθ. If it is positive, the other solutions will be in the 3rd quandrant, where tangents are positive. If it was negative, however, all of the solutions would lie in the 2nd and 4th quandrants, where tangents are negative. If this was the case, the root solutions would be used to find the actual solutions in these quadrants.