A road rises 1 m for every 15 m measured along the road. a) What is the angle of inclination of the road to the nearest degree? b) How far does the car travel horizontally when it travels 15 m along the road ? Give the answer to the nearest tenth of a metre.

The answer is below

Step-by-step explanation:

The road increases by a height of 1 m for every 15 m distance. The slope of a line is the ratio of its rise to run. The slope (m) is the tangent of the inclination. Hence:

Slope = rise / run = tanθ

Where θ is the angle of inclination.

a) Given rise = 1 m, run = 15 m

Slope of road (m) = rise / run

m = 1 m/ 15 m

m = 0.0667

slope = tanθ

tanθ = 0.0667

θ = tan⁻¹(0.0667)

θ = 3.81

θ ≈ 4°

b) cosθ = horizontal distance / 15 m

horizontal distance = 15 m × cos 4

Horizontal distance = 14.96

Horizontal distance = 15 m to nearest tenth

∠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.