WORK OUT THIS
sin(A+B+C)

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sin (A + B + C) = sin [( A + B) + C]

                    = sin (A + B) cos C + cos (A + B) sin C, [applying the formula of sin (α + β)]

                     = (sin A cos B + cos A sin B) cos C + (cos A cos B - sin A sin B) sin C, [applying the formula of sin (α + β) and cos (α + β)]

                      = sin A cos B cos C + sin B cos C cos A + sin C cos A cos B - sin A sin B sin C, [applying distributive property]                

                      = cos A cos B cos C (tan A + tan B + tan C - tan A tan B tan C)

therefore the expansion of sin (A + B + C) = cos A cos B cos C (tan A + tan B + tan C - tan A tan B tan C).

  n=   0.0000 - 0.3750 i  

  n=   0.0000 + 0.3750 i

step-by-step explanation:

one of these should be correct

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.