What is the law of sines in trigonometry?

Answer 1

Let a, b, and c be the lengths of the legs of a triangle opposite angles A, B, and C. Then the law of sines states that
a/(sinA)=b/(sinB)=c/(sinC)=2R,

where R is the radius of the circumcircle.

In other words,

In any triangle, the ratios of the sides to the sines of the opposite angles are equal.

Answer 2

the first question references an acute attitude because the question in straight forward words lists one attitude and the are 2 solutions for an similar ideas. The connection with an acute triangle isn't coated contained in the 2d question because the sum of the three angles might want to equivalent one hundred eighty ranges. 40 4 + 71 = one hundred and fifteen one hundred eighty - one hundred and fifteen = sixty 5 examine: 40 4 + 71 + sixty 5 = one hundred eighty

Answer 3

The law of sines is (A/sinA = B/sinB)

Answer 4

side A/sin of a= Side B/ sin of B= Side C/sin of c

I've noticed that a lot of people forget that the side in question corresponds with the angle opposite of it. That's important :)

Answer 5

para todo triángulo ABC de lados a =BC, b=AC y c=AB se cumple que: a/sin A = b/sin B = c/sin C o bien:
sin A /a = sin B /b = sin C / c, o bien:
BC/ sin A = AC/ sin B = AB / sin C

Answer 6

a/sinA = b/sinB = c/sinC

Answer 7

(sin A)/a = (sinB)/b = (sinC)/c

Answer 8

(a/sinA) = (b/sinB) = (c/sinC)