Find the numerical value of the log expression.

you would use (d.) the law of cosines or a^2 = b^2 + c^2 – 2bc(cos a)

, b^2 = a^2 + c^2 – 2ac(cos b), or c^2 = a^2 + b^2 - 2ab(cos c). you wouldn't use b. the law of sines as it is mostly effective when 1side and 2angles are given. a and c aren't the answer for they are the improper versions of the law of cosines. hope this is .

∠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.

answer: sine

step-by-step explanation: