Respuesta :
Answer:
[tex]\cos{\theta} = \dfrac{1}{52}[/tex]
[tex]\csc{\theta} = \dfrac{52}{\sqrt{2703}}[/tex]
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:
[tex]\sec{\theta} = \dfrac{1}{\cos{\theta}}[/tex]
we can substitute the value of sec(θ) in this equation:
[tex]52 = \dfrac{1}{\cos{\theta}}[/tex]
and solve for for cos(θ)
[tex]\cos{\theta} = \dfrac{1}{52}[/tex]
side note: just to confirm we can find the value of θ and verify that is indeed an acute angle by [tex]\theta = \arccos{\left(\dfrac{1}{52}\right)} = 88.8^\circ[/tex]
b) since right triangle is mentioned in the question. We can use:
[tex]\cos{\theta} = \dfrac{\text{adj}}{\text{hyp}}[/tex]
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.
[tex](\text{hyp})^2=(\text{adj})^2+(\text{opp})^2[/tex]
[tex](52)^2=(1)^2+(\text{opp})^2[/tex]
[tex]\text{opp}=\sqrt{(52)^2-1}[/tex]
[tex]\text{opp}=\sqrt{2703}[/tex]
- length of the opposite side = √(2703) ≈ 51.9904
we can find the sin(θ) using this side:
[tex]\sin{\theta} = \dfrac{\text{opp}}{\text{hyp}}[/tex]
[tex]\sin{\theta} = \dfrac{\sqrt{2703}}{52}}[/tex]
and since [tex]\csc{\theta} = \dfrac{1}{\sin{\theta}}[/tex]
[tex]\csc{\theta} = \dfrac{52}{\sqrt{2703}}[/tex]
So, the required values are,
[tex]cos(\theta)=\frac{1}{52} \\csc(\theta)=\frac{100}{99}[/tex]
Given,
[tex]sec(\theta)=52[/tex]
As we know that [tex]sec(\theta)[/tex] is the reciprocal of [tex]cos(\theta)[/tex] then,
[tex]cos(\theta)=\frac{1}{sec(\theta)} \\cos(\theta)=\frac{1}{52}\\[/tex]
Also, we have the relation,
[tex]sin(\theta)=\sqrt{(1-cos^2\theta)} \\sin(\theta)=\sqrt{(1-(\frac{1}{52} )^2} \\sin(\theta)=0.99\\csc(\theta)=\frac{100}{99}[/tex]
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