Two streets intersect at a 30-degree angle. at the intersection, there are 4 crosswalks formed that are the same length. what type of quadrilateral is formed by the crosswalks? (a) square (b) rhombus (c) rectangle (d)kite tell me the answer and how you got it. : )

The question says that the two streets Intersect at an angle of 30 degrees

and then it makes four pathways.

I have shown the scenario with the help of a diagram and attached it with answer.

So we can see that a shape is formed ABCD , which we can easily see that it is a quadrilateral .

Now as the streets intersect at 30 degrees.

So it cannot be a square .

But the sides AB , BC , CD and AD are equal , as we can see in the diagram.

Now a quadrilateral with equal sides but angle other than 90 degree , is called a RHOMBUS.

So option B , is correct , It makes a RHOMBUS

side xy = 10

sin(x) = 4/5

cos(x) = 3/5

tan(x) = 4/3

step-by-step explanation:

the answer is b.

step-by-step explanation:

by the trigonometric ratio tangent, tangent equals opposite divided by adjacent.

therefore, tan(50) equals ac/3

multiply each side by 3 to get: ac=3tan(50)

the answer is b.