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2 years ago

Find the slope of the line joining two points (4, 2) and (5, 3).

Mathematics

1 answer:

Maurinko [17]2 years ago

5 0

Slope = (y2-y1)/(x2-x1) = (3-2)/5-4) = 1/1 =1

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2 years ago

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

insens350 [35]

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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3 years ago

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