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TRIGONOMETRIC QUIZ
01 . If cos θ = 12/13, what is tan θ ?
A. ±12/13
B. ±5/12
C. 13/12
D. ±13/12
Answer B
Explanation :
Cos θ = base / hypotenuse =12/13
Using Pythagoras,
Perpendicular = √[13² – 12²] = 5
So, tan θ = perpendicular/base = ± 5/12
or, tan θ = ± 5/12 (depends on quadrant)
02 . If cosec θ = 2, cot θ = –√3, what is tan θ?
A. –½
B. –√3/2
C. –1/√3
D. –√3/2
Answer C
Explanation cot θ = -√3 , tan θ= 1/cot θ = -1/√3
03 . If sin x = 1/2 and cos y = 1/√3, then what is tan x
tan y?
A. 2/3
B. –⅕
C. √3/2
D. √2/√3
Answer D
Explanation
sin x = perpendicular/ hypotenuse = ½ , cos y = 1/√3
Using Pythagoras,
cos x = √3/ 2 , siny =√(2/3)
So, tan x = sin x / cos x = ½ / √3/2 = 1/√3
tan y = √(2/3) / 1/√3 = √2
tan x tan y = 1×√3 / √2 = √(⅔)
TRIGONOMETRIC QUIZ
01 . If cos θ = 12/13, what is tan θ ?
A. ±12/13
B. ±5/12
C. 13/12
D. ±13/12
Answer B
Explanation :
Cos θ = base / hypotenuse =12/13
Using Pythagoras,
Perpendicular = √[13² – 12²] = 5
So, tan θ = perpendicular/base = ± 5/12
or, tan θ = ± 5/12 (depends on quadrant)
02 . If cosec θ = 2, cot θ = –√3, what is tan θ?
A. –½
B. –√3/2
C. –1/√3
D. –√3/2
Answer C
Explanation cot θ = -√3 , tan θ= 1/cot θ = -1/√3
03 . If sin x = 1/2 and cos y = 1/√3, then what is tan x
tan y?
A. 2/3
B. –⅕
C. √3/2
D. √2/√3
Answer D
Explanation
sin x = perpendicular/ hypotenuse = ½ , cos y = 1/√3
Using Pythagoras,
cos x = √3/ 2 , siny =√(2/3)
So, tan x = sin x / cos x = ½ / √3/2 = 1/√3
tan y = √(2/3) / 1/√3 = √2
tan x tan y = 1×√3 / √2 = √(⅔)
