Question

Verify the sine law by taking particular triangle in four quadrants. ​

Answer 1

Suppose a, b and c represent the sides of a triangle ABC in magnitude and direction.

Then we have a+b+c=0 by triangular law of forces.

Taking cross product with vector a we have

a x a + a x b + a x c = 0. So a x b = c x a. Similarly, b x c = c x a.

Hence a x b = b x c = c x a

Therefore, |a x b| = |b x c| = |c x a|

So, ab sin C = bc sin A = ca sin B

Divide throughout by abc and take reciprocals

We get a/sin A = b/ sin B = c/ sin C which is the sine rule in a triangle.

Answer 2

Answer:

Verified.

Step-by-step explanation:

To find :- Verify the sine law of a triangle.

Solution :-

The sine law is defined as the ratio of sine of angle to the length of opposite side. It holds true for all the three sides of a triangle respective of their sides and angles.

The sine law  of triangle is explained in detail below:

In a triangle ABC,  sine of angle A is divided by the side a which is equal to the sine of angle B divided by the side b which is equal to the sine of angle C divided by the side c .

$\frac{sinA}{a} =\frac{sinB}{b} =\frac{sin C}{c}$

Hence, verified.

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