The grade is the ratio of rise to run, i.e. the slope aka the tangent.
[tex] \tan \theta = \dfrac{1}{10} [/tex]
[tex]\theta = \arctan 0.1 \approx 5.711^\circ[/tex]
Answer: (a) 6 degrees
For part b, the road is the hypotenuse c of a right triangle whose tangent of the small angle is 1/10. The height h or rise is the side opposite the small angle.
[tex]\sin\theta = \dfrac h c[/tex]
[tex]h = c \sin \theta[/tex]
We could just take the sine of the angle we got but let's get it from the tangent exactly.
[tex]\cos^2 \theta + \sin ^2 \theta = 1[/tex]
Dividing by squared cosine
[tex]1 + \tan ^2 \theta = 1/\cos^2 \theta = 1/(1- \sin^2 \theta)[/tex]
[tex](1- \sin^2 \theta) = 1/(1 + \tan^2 \theta)[/tex]
[tex]\sin^2 \theta = 1 - 1/(1 + \tan^2 \theta)[/tex]
[tex]\sin^2 \theta = 1 - 1/(1 + (1/10)^2) = 1-1/(101/100) = 1/101[/tex]
[tex]h = c \sin \theta = 2 \sqrt{1/101} \approx 0.199[/tex]
Answer: (b) Rise of 0.199 km