Answer to Question #308700 in Trigonometry for Yonnieeee

Question

1. Find the quadrant in which P(t) lies if sec t > 0 and csc t < 0.

2. Find the values of the 6 trigonometric functions of -7π/3.

3. Find the value of tan 7π/3 .

4. If sin t=√3/2 and cos t<0, find csc t.

5. Find the six trigonometric functions of an angle whose terminal side passes

through (-4,-3).

6. Find the remaining functions given that sin ϴ=513, sec t<0.

7. Find ϴ, if cot 2ϴ= √3, ϴ∈[0°,360°].

8. A ladder leans against the side of a building with its foot ten feet from the building. How far from the ground is the top of the ladder and how long is the ladder if it makes an angle of 60° with the ground?

9. At a certain moment Ship MV Alpha is 7.5 km. west of Ship MV Omega. Ship MV Omega is sailing S 30° 20’ E at the rate of 25 km./hr., while Ship MV Alpha is sailing directly north at rate of 20 km./hr. Find the distance between the two ships and the bearing with Ship MV Omega from Ship MV Alpha after 40 minutes.

Answer 1

1)Trigonometric Functions on the Unit Circle

Given a point on the terminal side of an angle

θ in

standard position. (2)Then:

θ

P

(x, y)

3)rxysin θ =Find the six trig values of (8, ­6) if it's a point on the

terminal side of an angle in standard position.

4) r = √x2 + y2

Find the six trig values of (4, 3)

5) if it's a point on the

terminal side of an angle in standard po the six trig values of (­2, ­1) if it's a point on

terminal side of an angle in standard position.

6) sin ϴ=513, sec t<0 r = √x2 + y2

Find the six trig values of (4, 3) if it's a point on the terminal side of an angle in standard po the six trig values of (­2, ­1) if it's a point on

7) The ϴ, if cot 2ϴ= √3, ϴ∈[0°,360°] will be : cotθ+cosecθ=3⇒cot2θ​=3⇒2θ​=nπ+6π⇒θ=2nπ+3πFor θϵ(0o,360o)⇒θ=60o

8) Let AC be a ladder, AB is the wall and BC is the ground as shown in the figure.We know thatcos60o=AC BC 21=AC 2.5​AC=5 mTherefore, the length of ladder is 5 m.terminal side of an angle 9) Let x and y be the initial position of the ships A and B respectively. Let x be the distance between A and X,y be the distance between B and X,z be the distance between A and B.Then, z2=x2+y22zdtdz​=2xdtdx​+2ydtdy      .........(1)zdtdz=xdtdx​+ydtdy​Given, dtdx​=35 km/hr; dtdy(1)zdtdz=xdtdx​+ydtdy​Given, dtdx​=35 km/hr; dtdy​=25 km/hrWhen t=4 hrs, x=40;y=100 and z=x2+y2=(40)2+(100)2=1600+10000​=11600​=2029

​(1)⇒2029dtdz​=40×(35)+100×(25)dz​=20293900​=29195​ km/hr