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Akimi4 [234]
3 years ago
8

In a right triangle, angle is an acute angle and sin angle = 4/9. Evaluate the other five trigonometric functions of angle.

Mathematics
1 answer:
kkurt [141]3 years ago
5 0

Answer:

Part 1) cos(A)=\frac{\sqrt{65}}{9}

Part 2) tan(A)=4\frac{\sqrt{65}}{65}

Part 3) cot(A)=\frac{\sqrt{65}}{4}

Part 4) sec(A)=9\frac{\sqrt{65}}{65}

Part 5) csc(A)=\frac{9}{4}

Step-by-step explanation:

Let

A------> the angle

we have that

sin(A)=\frac{4}{9}

step 1

Find the cos(A)

we know that

cos^{2}(A)+sin^{2}(A)=1

substitute the value of sin(A)

cos^{2}(A)+(\frac{4}{9})^{2}=1

cos^{2}(A)=1-(\frac{4}{9})^{2}

cos^{2}(A)=1-(\frac{16}{81})

cos^{2}(A)=(\frac{65}{81})

cos(A)=\frac{\sqrt{65}}{9}

step 2

Find the tan(A)

we know that

tan(A)=\frac{sin(A)}{cos(A)}

substitute the values

tan(A)=\frac{\frac{4}{9}}{\frac{\sqrt{65}}{9}}

tan(A)=\frac{4}{\sqrt{65}}

tan(A)=4\frac{\sqrt{65}}{65}

step 3

Find the cot(A)

we know that

cot(A)=\frac{1}{tan(A)}

we have

tan(A)=\frac{4}{\sqrt{65}}

substitute

cot(A)=\frac{1}{\frac{4}{\sqrt{65}}}

cot(A)=\frac{\sqrt{65}}{4}

step 4

Find the sec(A)

we know that

sec(A)=\frac{1}{cos(A)}

we have

cos(A)=\frac{\sqrt{65}}{9}

substitute

sec(A)=\frac{1}{\frac{\sqrt{65}}{9}}

sec(A)=\frac{9}{\sqrt{65}}

sec(A)=9\frac{\sqrt{65}}{65}

step 5

Find the csc(A)

we know that

csc(A)=\frac{1}{sin(A)}

we have

sin(A)=\frac{4}{9}

substitute the value

csc(A)=\frac{1}{\frac{4}{9}}

csc(A)=\frac{9}{4}

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