The angle you want - call it <em>θ</em> - is such that
tan(<em>θ</em>) = FC / AC
Find the length of the diagonal AC, i.e. a diagonal of the rectangle ABCD. ABC forms a right triangle with legs AB = 70 and BC = 50, so
AC² = AB² + BC²
→ AC = √(70² + 50²) = 10 √74
Find FC using the given angle of the sloping face:
tan(30º) = FC / BC
→ 1/√3 = FC / 50
→ FC = 50/√3
Now solve for <em>θ</em> :
tan(<em>θ</em>) = (50/√3) / (10 √74)
→ tan(<em>θ</em>) = 5/√222
→ <em>θ</em> ≈ 18.6º
Answer:
I mean I think he is because the interest is less now, I still think it's irresponsible to take from his savings but ¯\_(ツ)_/¯
The answer is D 156 weeks
The coordinates of the endpoints of the midsegment for △LMN that is parallel to LN are (3, 4.5) and (3, 3).
Explanation:
Given that △LMN
We need to determine the coordinates of the endpoints of the midsegment for △LMN that is parallel to LN
The midsegment of the triangle parallel to side LN is the midsegment connecting the midpoint of side LM and the midpoint of side MN.
The midpoint of LM is given by

Simplifying, we get,

The midpoint of MN is given by

Thus, the coordinates of the endpoints of the midsegment for △LMN that is parallel to LN are (3, 4.5) and (3, 3).