Final answer:
A right triangle ABC with tan(A) = 2 can be sketched with a 1-unit adjacent side, 2-unit opposite side, and √5 hypotenuse. Points on a circle with radius √17 centered at (0,0) can be (4,1) in the first quadrant, (-4,1) in the second, (-4,-1) in the third, and (4,-1) in the fourth.
Step-by-step explanation:
To sketch a right triangle ABC with tan(A) = 2, we assume that angle A is in standard position, and its tangent is the ratio of the opposite side to the adjacent side of the triangle. Since tan(A) = 2, we can consider the lengths of the opposite side (Ay) as 2 units and the adjacent side (Ax) as 1 unit. By the Pythagorean theorem, the hypotenuse (A) would be √(2² + 1²) = √5 units. Therefore, triangle ABC could be sketched with side Ay as 2 units vertically, Ax as 1 unit horizontally, and the hypotenuse A connecting the ends of these two sides.
Given a circle with center (0,0) and a point on it (1,4), we can infer that the radius of the circle is √(1² + 4²) = √17. To find at least one point in each quadrant that lies on the circle, we can use this radius to find points equidistant from the origin in all quadrants. For instance, (4,1) would lie in the first quadrant, (-4,1) in the second, (-4,-1) in the third, and (4,-1) in the fourth, all of which would be at a radius of √17 units from the center.