Question

What is the trigonometric ratio for sin C ?

Enter your answer, as a simplified fraction, in the boxes.

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triangle A B C with right angle at B. B C equals 80. A C equals 82.

Answer 1

The value of trigonometric ratio
`\sin C` is
`(9)/(41)`

Explanation:

Given : Triangle ABC with right angle at B and BC = 80 and AC = 82

We have to find the value of trigonometric ratio
`\sin C`

Consider the given triangle ABC,

We first draw the given right angled triangle ABC.

Also, given BC = 80 and AC = 82

We know, Trigonometric ratio sine gives relationship between perpendicular and hypotenuse.

That is
`\sin \theta=(Perpendicular)/(hypotenuse)`

Here, For
`\theta=C`

Perpendicular = AB and hypotenuse = AC

We first find AB

Using Pythagoras theorem,

`H^2=B^2+P^2`

H = 82, B = 80

Thus,

`82^2=80^2+P^2`

Simplify, we have,

`P^2=324`

Thus, P = 18

Thus,
`\sin C =(18)/(82)`

Thus, The value of trigonometric ratio
`\sin C` is
`(9)/(41)`

Answer 2

`sin(C)=(9)/(41)`

Explanation:

see the attached figure to better understand the problem

In the right triangle ABC

Find the measure of side AB applying the Pythagoras Theorem

`AB^(2)=AC^(2) -BC^(2)`

substitute the values

`AB^(2)=82^(2) -80^(2)`

`AB^(2)=324`

`AB=18\ units`

Find the sin(C)

we know that

The function sine of angle C is equal to divide the opposite side angle C by the hypotenuse

so

`sin(C)=(AB)/(AC)`

substitute the values

`sin(C)=(18)/(82)`

Simplify

`sin(C)=(9)/(41)`