Question

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

Answer 1

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

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

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