Question

PLEASE HELP IMMEDIATELY!! will MARK BRAINLIEST AND GIVE 16 POINTS! (multiple choice but please try to show steps)
1.Given that 0∘≤C≤180∘, determine the value(s)
of ∠C to the nearest degree when sinC=0.9848.

A) 80°, 100°
B)100°
C)10°, 80°
D)80°

2. θ is an angle in standard position whose terminal arm is in
quadrant IV and cosθ=3/sqrt(13). Find cscθ.

A)−2/Sqrt(13)
B)−4/Sqrt(13)
C)−Sqrt(13)/2
D)Sqrt(13)/4

3.If secθ=−52, and sinθ>0, what is the value of tanθ?

A) −Sqrt(21)/2

B)Sqrt(−21)

C)−2 Sqrt(21)/21

D)−5 sqrt(21/ 21

Answer 1

Answer:

1. The values of angle C to the nearest degree are 80° , 100° ⇒ (A)

2. csc Ф = -(√13)/2 ⇒ (C)

3. tan Ф = - (√21)/2 ⇒ (A)

Step-by-step explanation:

* Lets explain how to solve the problem

1.

∵ The measure of angle C is ⇒ 0° ≤ m∠C ≤ 180°

∴ ∠C lies in the first quadrant or in the second quadrant

∴ m∠C = Ф OR  m∠C = 180 - Ф, where Ф is an acute angle

∵ sin∠C = 0.9848

∴ sin Ф =  0.9848

- Use the inverse function sin^-1 to find Ф

∴ Ф = sin^-1 0.9848

∴ Ф ≅ 80°

- Lets find ∠C

∵ m∠C = Ф

∴ m∠C = 80°

∵ m∠C = 180° - Ф

∴ m∠C = 180 - 80 = 100°

* The values of angle C to the nearest degree are 80° , 100°

2.

- The terminal arm of angle Ф is in  quadrant IV

∵ In quadrant IV sin Ф , csc Ф , tan Ф , cot Ф are negative values

∵ In quadrant IV cos Ф , sec Ф are positive values

∵ cos Ф = 3/√13

∵ csc Ф = 1/(sin Ф)

- Lets use the identity sin²Ф + cos²Ф = 1 to find sin Ф

∵ sin²Ф + (3/√13)² = 1

∴ sin²Ф + 9/13 = 1 ⇒ subtract 9/13 from both sides

∴ sin²Ф = 4/13 ⇒ take √ for both sides

∵ sin Ф = ± 2/√13

- The value of the sin the IV quadrant is negative

∴ sin Ф = - 2/√13

∵ csc Ф = 1/sin Ф

∴ csc Ф = -(√13)/2

3.

∵ sec Ф = -5/2 and sin Ф > 0

∵ sec Ф = 1/cos Ф

- The value of cos Ф is negative and the value of sin Ф is positive,

 then Ф lies in the second quadrant

∵ In the second quadrant cos Ф , sec Ф , tan Ф , cot Ф are negative

  values but sin Ф and csc Ф are positive values

- Lets use the identity tan²Ф + 1 = sec²Ф to find tan Ф

∵ sec Ф = -5/2

∵ tan²Ф + 1 = sec²Ф

∴ tan²Ф + 1 = (-5/2)²

∴ tan²Ф + 1 = 25/4 ⇒ subtract 1 from both sides

∴ tan²Ф = 21/4 ⇒ take √ for both sides

∴ tan Ф = ± √21/2

∵ Ф lies in the second quadrant

∴ tan Ф = - (√21)/2