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 <em>OR </em>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
