(1) \( \sin Q=0,866 \) (2) \( \sin Q=1 \) (3) \( \cos Q=0,866 \) (4) \( \cos a=1 \) (5) \( \tan \theta=1 \) (6) \( \tan Q=1,703 \)

Below are one acceptable set of answers. In each case the answer is given in “general‐form” (that is, including all angles that satisfy the equation). (Note: Here we use radians; you may convert to degrees if preferred.) 1) sin Q = 0.866 Recognize that 0.866 (more exactly, 0.8660254…) is √3/2. Since   sin(π/3) = √3/2  and sin(2π/3) = √3/2, the general solution is   Q = π/3 + 2πk  or  Q = 2π/3 + 2πk   for any integer k. 2) sin Q = 1 We know that sin(π/2) = 1 and that the only solution (in one period) is at π/2. Thus,   Q = π/2 + 2πk  for any integer k. 3) cos Q = 0.866 Since cos(π/6) = √3/2 = 0.866 and also cos(11π/6) = √3/2, we have   Q = π/6 + 2πk  or  Q = 11π/6 + 2πk   for any integer k. 4) cos a = 1 Cosine equals 1 when the angle is 0 (modulo 2π). Thus,   a = 0 + 2πk,  or simply a = 2πk  for any integer k. 5) tan θ = 1 Since tan(π/4) = 1, and tangent has period π, we get   θ = π/4 + πk  for any integer k. 6) tan Q = 1.703 This is not an “exact” trigonometric value so we write the solution in terms of the arctan function. That is, one solution is   Q₀ = arctan(1.703). Then the general solution is   Q = arctan(1.703) + πk  for any integer k. If a numerical answer is needed, note that arctan(1.703) is approximately 1.04 radians (≈59.7°), so the solutions are approximately   Q ≈ 1.04 + πk radians. Any answer equivalent to the above is correct.