Thesine rule gives C and then we have Case 7. There are either one or two solutions. Case 4: two angles and an included side given (ASA). The four-part cotangent formulae for sets (cBaC) and (BaCb) give c and b, then A follows from the sine rule. Case 5: two angles and an opposite side given (AAS). The sine rule gives b and then we have Case 7 RightTriangles. Pythagorean theorem: + =. Sines: sin = =. Cosines: cos = =. Tangents: tan = =. Now suppose we know the hypotenuse and one side, but have to find the other. For example, if = 119 and = 169, then = - - 119 = 28561 - 14161 = 14400, and the square root of 14400 is 120, so = 120. We might only know one side but we also know an Let\(\frac{sinA}{sinB}\) = \(\frac{sin(A-C)}{sin(C-B)}\), where A, B, C are angles of a triangle ABC. If the lengths of the sides opposite these angles are a, b, c respectively, then : sinA/sinB = sin(A-C)/sin(C-B) (1) b 2 - a 2 = a 2 + c 2 (2) b 2,c 2,a 2 are in A.P. (3) c 2,a 2,b 2 are in A.P. (4) a 2,b 2,c 2 are in A.P. jee; jee main; jee Step2: Calculate the given expression. Since sin A = sin B = sin C = sin D = 1. ⇒ sin A × sin B × sin C × sin D = 1 4 = 1. Hence, value of sin A × sin B × sin C × sin D is 1. Suggest Corrections. 0. Sincethe accent in the OP is put on a purely geometric solution, i can not even consider the chance to write $\cos^2 =1-\sin^2$, and rephrase the wanted equality, thus having a trigonometric function which is better suited to geometrical interpretations.. So this answer has two steps, first we reformulate the given identity in a mot-a-mot geometric manner, the geometric framework is Firstof all, recall the complex definitions of the two trigonometric functions: sin ( x) = e i x − e − i x 2 i, cos ( x) = e i x + e − i x 2. Now, the formula for sin ( 2 x) can be derived from these two. But so can it be extended by multipying each new result by 2 cos ( 2 b x): 2 sin x cos x = sin ( 2 x), 4 sin x cos x cos ( 2 x) = sin .

sin a sin b sin c formula