Bethany used the steps below in an attempt to solve for angle B in triangle ABC.
4.1 cm
2.7 cm
4.5 cm
[Not drawn to scale]
4.12 - 4.52+2.72 -2(4.5)(2.7)cos B
4.1=4.5+2.7-24.3cos B
-3.1--24 3cOSB
0.1276= COS B

Given that,

The area parameters are

A = 2.7 cm, B = 4.1 cm C = 4.5 cm

We want to find the angle extended by B.

Using cosine Rule

b² = a² + c² - 2ac•Cosθ

4.1² = 2.7² + 4.5² - 2 × 2.7 × 4.5• Cosθ

16.81 = 7.29 + 20.25 - 24.3•Cosθ

16.81 - 7.29 - 20.25 = -24.3•Cosθ

-10.73 = -24.3•Cosθ

Cosθ = -10.73 / -24.3

Cosθ = 0.4416

θ = Cos~1 (0.4416)

θ = 63.8°

The angle B is 63.8°

∠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.

x is greater than 5

step-by-step explanation:

2x+7 is greater than 17

    -7                           -7

2x is greater than 10

divide by 2 each side

x is greater than 5