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Can someone help me shot this problem

Can someone help me shot this problem-example-1
User Donjuedo
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1 Answer

6 votes

Answer:

A) sin θ = 3/5

B) tan θ = 3/4

C) csc θ = 5/3

D) sec θ = 5/4

E) cot θ = 4/3

Explanation:

We are told that cos θ = 4/5

That θ is the acute angle of a right angle triangle.

To find the remaining trigonometric functions for angle θ, we need to find the 3rd side of the triangle.

Now, the identity cos θ means adjacent/hypotenuse.

Thus, adjacent side = 4

Hypotenuse = 5

Using pythagoras theorem, we can find the third side which is called opposite;

Opposite = √(5² - 4²)

Opposite = √(25 - 16)

Opposite = √9

Opposite = 3

A) sin θ

Trigonometric ratio for sin θ is opposite/hypotenuse. Thus;

sin θ = 3/5

B) tan θ

Trigonometric ratio for tan θ is opposite/adjacent. Thus;

tan θ = 3/4

C) csc θ

Trigonometric ratio for csc θ is 1/sin θ. Thus;

csc θ = 1/(3/5)

csc θ = 5/3

D) sec θ

Trigonometric ratio for sec θ is 1/cos θ. Thus;

sec θ = 1/(4/5)

sec θ = 5/4

E) cot θ

Trigonometric ratio for cot θ is 1/tan θ. Thus;

cot θ = 1/(3/4)

cot θ = 4/3

User Deepak Rajput
by
8.4k points

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