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

Answer:

$sin(C)=\frac{9}{41}$

Step-by-step 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

$sin(C)=\frac{AB}{AC}$

substitute the values

$sin(C)=\frac{18}{82}$

Simplify

$sin(C)=\frac{9}{41}$

Answer 2

Answer:

The value of trigonometric ratio $\sin C$   is $\frac{9}{41}$

Step-by-step 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 $\sin \theta=\frac{Perpendicular}{hypotenuse}$

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, $\sin C =\frac{18}{82}$

Thus, The value of trigonometric ratio $\sin C$   is $\frac{9}{41}$