Unit-6 Trigonometry and Bearing
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Unit-6 Trigonometry and Bearing
Answer: (sqrt 3 / 2) x (sqrt 3 / 2) = 3/4.
Answer: tan 30 = sin 30 / cos 30 = (1/2) / (sqrt 3 / 2) = 1 / sqrt 3.
Answer: Bearings are measured clockwise from North; East lies at 090 degrees.
Answer: 45 = 180/4, so 45 degrees = pi/4 radians.
Answer: 30 minutes = 30/60 = 0.5 degrees, so 25 degrees 30' = 25.5 degrees.
Answer: (1 / sqrt 2) x sqrt 2 = 1.
Answer: From 1 + tan^2 theta = sec^2 theta, tan^2 theta - sec^2 theta = -1.
Answer: tan theta = 1 / cot theta = 1/1 = 1.
Answer: sin 60 = height / 5, so height = 5 sin 60 = 5 sqrt 3 / 2 m.
Answer: The resultant direction is North-East, which has a bearing of 045 degrees.
Answer: South-West is 225 degrees clockwise from North.
Answer: tan theta = side opposite theta divided by the side adjacent to theta.
Answer: cos 45 = 1 / sqrt 2 = sqrt 2 / 2.
Answer: csc 30 = 1 / sin 30 = 1 / (1/2) = 2.
Answer: sin 30 = 1/2, a standard trigonometric ratio value.
Answer: sec theta = 1 / cos theta, so cos theta x sec theta = 1.
Answer: sin theta = sqrt(1 - cos^2 theta) = sqrt(1 - 25/169) = sqrt(144/169) = 12/13.
Answer: sin 60 - cos 60 = sqrt 3/2 - 1/2 = (sqrt 3 - 1)/2.
Answer: csc theta = 1 / sin theta, so sin theta x csc theta = 1.
Answer: 30 x (pi/180) = pi/6.