alleryIf tan(theta) = sqrt(3), then theta is
- 30 degrees
- 45 degrees
- 60 degrees
- 90 degrees
Show answer and solution
Correct answer: C. 60 degrees
tan(60) = sqrt(3).
**Standard angle values:** tan 30 = 1/sqrt(3)
tan 45 = 1; tan 60 = sqrt(3).
GRE Geometry Trigonometry Mensuration — practice questions
Explore 9 free Geometry Trigonometry Mensuration questions in GRE Quantitative Aptitude. Open each question or exercise to read its content, reference answers and available explanation.
Practise Geometry Trigonometry Mensuration with Quill →Geometry Trigonometry Mensuration questions and exercises with answers
Showing questions 1–9 of 9.
If tan(theta) = sqrt(3), then theta isRead the answer and worked solution →If sec(theta) = 5/4, then tan(theta) isRead the answer and worked solution →From a point 30 m away from the foot of a tower, the angle of elevation of the top is 45 degrees. The height of the tower isRead the answer and worked solution →A pole is broken at a height. The broken part touches ground 10 m from the foot at an angle of 30 degrees with the ground. The original total height of the pole isRead the answer and worked solution →If sin(theta) + cos(theta) = sqrt(2), then sin(theta) isRead the answer and worked solution →The distance between points (3, 4) and (0, 0) isRead the answer and worked solution →If sin(theta) + cos(theta) = sqrt(2) and theta is acute, then theta isRead the answer and worked solution →The midpoint of segment joining (2, 3) and (6, 7) isRead the answer and worked solution →The line 2x + 3y = 12 has y-interceptRead the answer and worked solution →
Worked practice questions
If sec(theta) = 5/4, then tan(theta) is
- 1/4
- 3/4
- 3/5
- 4/5
Show answer and solution
Correct answer: B. 3/4
sec^2 - tan^2 = 1. So tan^2 = 25/16 - 1 = 9/16, tan = 3/4.
From a point 30 m away from the foot of a tower, the angle of elevation of the top is 45 degrees. The height of the tower is
- 15 m
- 30 m
- 45 m
- 60 m
Show answer and solution
Correct answer: B. 30 m
tan(45) = 1 = height / 30. Height = 30 m.
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