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If cos A = -√ 3 / 2 and 0°≤ A ≤180°, determine sin A.
2007-06-04 04:50:42 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
sin^2 +cos^2 = 1
sin^2 + 3/4 = 1
sin^2 = 1/4
sin A = 1/2 (Sin is + for first or second quadrant)
2007-06-04 04:57:38 · answer #1 · answered by rick c 2 · 0⤊ 0⤋
subtract arc cos â 3 / 2 from 180° because
cos (180° - x) = -cos x
cos 30° = â 3 / 2
cos (150°) = cos(180° - 30°) = -â 3 / 2
Now, sin 150° = sin(180° - 30°) = sin(30°) = sqrt(1 - cos^2 30°) = sqrt(1 - 3/4) = sqrt(1/4) = 1/2
2007-06-04 11:58:10 · answer #2 · answered by Amit Y 5 · 0⤊ 0⤋
O is origin.
Let P be point P where OP = 2
P is in 2nd quadrant.
PQ is perpendicular to x axis and Q lies on this axis.
OQ² + QP² = OP²
3 + QP² = 4
QP = 1
sin A = QP / OP = 1 / 2
2007-06-04 12:02:04 · answer #3 · answered by Como 7 · 0⤊ 0⤋
you don't have a scientific calculator to solve this? The answer or preceedures should follow in the chapter lesson it's all a matter of substitution.
2007-06-04 11:58:09 · answer #4 · answered by Tomas Q 2 · 0⤊ 0⤋