Answer:
Step-by-step explanation:
Angle sum property of triangle: Sum of all the angles of a triangle is 180
Alternate interior angles: When two parallel lines are intersected by a transversal, the pair of angles on the inner side of each of these lines but on the opposite side of the transversal are called alternate interior angles
In ΔABC,
∠1 + 90 + 38 = 180 {angle sum property of triangle}
∠1 + 128 = 180
∠1 = 180 - 128
AB // CD and AC is transversal.
{Alternate interior angles are equal}
In ΔACD,
∠2 + ∠3 + 63 = 180 {angle sum property of triangle}
∠2 + 38 +63 = 180
∠2 + 101 =180
∠2 = 180 - 101
Given:
To find:
The quadrant of the terminal side of 
and find the value of 
.
Solution:
We know that,
In Quadrant I, all trigonometric ratios are positive.
In Quadrant II: Only sin and cosec are positive.
In Quadrant III: Only tan and cot are positive.
In Quadrant IV: Only cos and sec are positive.
It is given that,
Here cos is positive and sine is negative. So, 
must be lies in Quadrant IV.
We know that,
It is only negative because lies in Quadrant IV. So,
After substituting , we get
Therefore, the correct option is B.
Simplified would be 4xy + 3x^2 + 6x
The equation of the line that has a slope of 3 and a y-intercept of -2 is:
<span>Y=3x-2, or B.</span>
Answer:C.no solution
Step-by-step explanation:
100% correct. ツ
