greenspacemaniac2018
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HELP PLEASE!!! I'M DESPERATE AND I'VE HAD TO POST THIS QUESTION TWICE. 70 POINTS!!!!

In a right triangle, angles R and S are acute angles. The sin R = 0.35. What is the cos R? Round your answer to the nearest hundredth.

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Answer :

LeanderRay
Given: sin R = 0.35
We know that sin R = opposite/hypotenuse in a right triangle.
Let’s assume the opposite side is x and the hypotenuse is y.
So, sin R = x/y
Given sin R = 0.35, we have:
0.35 = x/y
To find cos R, we can use the Pythagorean identity: sin^2 R + cos^2 R = 1
Substitute sin R = 0.35 into the equation:
(0.35)^2 + cos^2 R = 1 0.1225 + cos^2 R = 1 cos^2 R = 1 - 0.1225 cos^2 R = 0.8775
Taking the square root of both sides to solve for cos R:
cos R = √0.8775 cos R ≈ 0.94
Therefore, the value of cos R rounded to the nearest hundredth is approximately 0.94.
msm555

Answer:

[tex]\sf \cos R = \boxed{0.94} [/tex]

Step-by-step explanation:

To find the cosine of angle [tex]\sf R [/tex] in the right triangle when the sine of angle [tex]\sf R [/tex] is given, we can use the Pythagorean identity [tex]\sf \sin^2 R + \cos^2 R = 1 [/tex].

Here's how to do it:

Given:

[tex]\sf \sin R = 0.35 [/tex]

Use the Pythagorean identity:

[tex]\sf \sin^2 R + \cos^2 R = 1 [/tex]

Substitute the value of [tex]\sf \sin R [/tex]:

[tex]\sf (0.35)^2 + \cos^2 R = 1 [/tex]

Solve for [tex]\sf \cos^2 R [/tex]:

[tex]\sf \cos^2 R = 1 - (0.35)^2 [/tex]

[tex]\sf \cos^2 R = 1 - 0.1225 [/tex]

[tex]\sf \cos^2 R = 0.8775 [/tex]

Take the square root of both sides to find [tex]\sf \cos R [/tex]:

[tex]\sf \cos R = \sqrt{0.8775} [/tex]

[tex]\sf \cos R = 0.9367496997597[/tex]

[tex]\sf \cos R = 0.94 \textsf{(Rounding to the nearest hundredth)} [/tex]

Therefore, [tex]\sf \cos R = \boxed{0.94} [/tex].

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