The base of a cube is parallel to the horizon. If the cube is cut by a plane to form a cross section, under what circumstance wo
uld it be possible for the cross section be a non-rectangular parallelogram?
2 answers:
This question has this set of answer choices:
a) when the plane cuts three faces of the cube, separating one corner from the others
b) when the plane passes through a pair of vertices that do not share a common face
c) when the plane is perpendicular to the base and intersects two adjacent vertical faces
d) when the plane makes an acute angle to the base and intersects three vertical faces
e) not enough information to answer the question
The right answer is the first choice: a) when the plane cuts three faces of the cube, separating one corner from the others
You can see a picture of this case in the figure attached: as you can see the cross section (in pink) is a triangle.
a) when the plane cuts three faces of the cube, separating one corner from the others
b) when the plane passes through a pair of vertices that do not share a common face
c) when the plane is perpendicular to the base and intersects two adjacent vertical faces
d) when the plane makes an acute angle to the base and intersects three vertical faces
e) not enough information to answer the question
The right answer is the first choice: a) when the plane cuts three faces of the cube, separating one corner from the others
You can see a picture of this case in the figure attached: as you can see the cross section (in pink) is a triangle.
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Answer:
"76°" is the appropriate solution.
Step-by-step explanation:
Please find attachment of the diagram according to the given query.
The given values are:
In ΔDEF,
f = 610 inches
e = 590
∠E = 70°
∠F = ?
By using the law of sines, we get
⇒
On substituting the values, we get
⇒
On applying cross multiplication, we get
⇒
On substituting the values, we get
⇒
⇒
⇒
now,
⇒
⇒
