If sin(t) = -2/48 and cos(t) < 0 what is the value of cos(t) to the nearest thousandth?
If sin(t) = -14/34 and cos(t) < 0, what is the value of tan(t) to the nearest thousandth?
If tan(t) = -15/19 and sin(t) < 0, what is sin(t) to the nearest thousandth?
If tan(t) = 11/-12 and sin(t) < 0, what is cos(t) to the nearest thousandth?
The path of a projectile fired at an inclination t0 to the horizontal with initial velocity v feet per second is a parabola. The horizontal distance R, in feet, that the projectile travels is given by R = v2sin(2t)/32.2. What is R if v = 152 and t = 74? Give your answer to the nearest thousandth.
Show the steps please!
I will do the first, then you do the remaining ones in the same way
sin t = -2/48 = -1/24 and cos t < 0
therefore t must be in quadrant III ( by CAST rule)
I make a triangle, the missing side is x
x^2 + 1^2 = 24^2
x = √575
then in III, cos t = -√575/24 = appr. -.999
or
take sin^-1 (+1/24) to get 2.388°
so in III , t = 180+2.388 = 182.388°
cos 182.388° = appr. -.999
To solve these trigonometric problems, we can use the trigonometric identity relationships and the unit circle. Here are the step-by-step solutions to each question:
1. Given sin(t) = -2/48 and cos(t) < 0, we can first find the value of cos(t). We know that sin(t)^2 + cos(t)^2 = 1. Substituting sin(t) = -2/48, we have (-2/48)^2 + cos(t)^2 = 1. Simplifying, we get 4/2304 + cos(t)^2 = 1. Rearranging the equation, cos(t)^2 = 1 - 4/2304. Solving for cos(t), we find cos(t) = ±√(1 - 4/2304). Since cos(t) < 0, we take the negative root. Evaluating this using a calculator, cos(t) ≈ -0.999 to the nearest thousandth.
2. Given sin(t) = -14/34 and cos(t) < 0, we want to find the value of tan(t) to the nearest thousandth. We can use the fact that tan(t) = sin(t)/cos(t). Substituting sin(t) = -14/34 and cos(t) < 0, we have tan(t) = (-14/34)/cos(t). Since cos(t) < 0, we divide both sides by -1 to maintain the negative sign in the numerator. Simplifying further, we have tan(t) = 14/34. Evaluating this using a calculator, tan(t) ≈ 0.412 to the nearest thousandth.
3. Given tan(t) = -15/19 and sin(t) < 0, we want to find sin(t) to the nearest thousandth. We can use the fact that tan(t) = sin(t)/cos(t), and since tan(t) is negative while sin(t) is negative, it implies that cos(t) is positive. Using the identity tan(t) = sin(t)/cos(t), we substitute tan(t) = -15/19 to obtain -15/19 = sin(t)/cos(t). Simplifying further, we have sin(t) = (-15/19) * cos(t). Since sin(t) is negative, we take cos(t) to be positive. Evaluating this using a calculator, sin(t) ≈ -0.806 to the nearest thousandth.
4. Given tan(t) = 11/-12 and sin(t) < 0, we want to find cos(t) to the nearest thousandth. We can use the fact that tan(t) = sin(t)/cos(t) and substitute tan(t) = 11/-12 to obtain (11/-12) = sin(t)/cos(t). Simplifying further, we have sin(t) = (-11/12) * cos(t). Since sin(t) is negative and cos(t) is negative, we multiply both sides by -1 to maintain the signs. Thus, sin(t) = (11/12) * cos(t). Evaluating this using a calculator, cos(t) ≈ -0.917 to the nearest thousandth.
5. Given R = v^2sin(2t)/32.2, where v = 152 and t = 74, we want to find R to the nearest thousandth. Substituting the given values, R = (152^2) * sin(2*74)/32.2. Evaluating this using a calculator, R ≈ 280.305 to the nearest thousandth.
I hope this helps! Let me know if you have any further questions.