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Nat2105 [25]
3 years ago
5

Suppose that θ is an acute angle of a right triangle and that sec(θ)=52. Find cos(θ) and csc(θ).

Mathematics
1 answer:
insens350 [35]3 years ago
4 0

Answer:

\cos{\theta} = \dfrac{1}{52}

\csc{\theta} = \dfrac{52}{\sqrt{2703}}

Step-by-step explanation:

To solve this question we're going to use trigonometric identities and good ol' Pythagoras theorem.

a) Firstly, sec(θ)=52. we're gonna convert this to cos(θ) using:

\sec{\theta} = \dfrac{1}{\cos{\theta}}

we can substitute the value of sec(θ) in this equation:

52 = \dfrac{1}{\cos{\theta}}

and solve for for cos(θ)

\cos{\theta} = \dfrac{1}{52}

side note: just to confirm we can find the value of θ and verify that is indeed an acute angle by \theta = \arccos{\left(\dfrac{1}{52}\right)} = 88.8^\circ

b) since right triangle is mentioned in the question. We can use:

\cos{\theta} = \dfrac{\text{adj}}{\text{hyp}}

we know the value of cos(θ)=1\52. and by comparing the two. we can say that:

  • length of the adjacent side = 1
  • length of the hypotenuse = 52

we can find the third side using the Pythagoras theorem.

(\text{hyp})^2=(\text{adj})^2+(\text{opp})^2

(52)^2=(1)^2+(\text{opp})^2

\text{opp}=\sqrt{(52)^2-1}

\text{opp}=\sqrt{2703}

  • length of the opposite side = √(2703) ≈ 51.9904

we can find the sin(θ) using this side:

\sin{\theta} = \dfrac{\text{opp}}{\text{hyp}}

\sin{\theta} = \dfrac{\sqrt{2703}}{52}}

and since \csc{\theta} = \dfrac{1}{\sin{\theta}}

\csc{\theta} = \dfrac{52}{\sqrt{2703}}

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