Please help me!!!! Anyone pleaseeeee!!!

Mathematics

1 answer:

otez555 [7]3 years ago

3 0

Answer:

69 degree

Step-by-step explanation:

Using cosine formula again :))

So we have

BC^2 = AB^2 + AC^2 - 2*AB*AC*cosA

BC^2 = 32^2 + 28^2 -2*28*32*cos56

BC = 28.3884 cm

So B:

AC^2 = AB^2 + BC^2 - 2*AB*BC*cosB

CosB=0.3559

B=69.14 degree

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A large washer has an outer radius of 10mm and a hole with a diameter of 14mm. What is the area of the top surface of the washer

choli [55]

<h2>Answer:</h2>

The area of the top surface of the washer is: 160.14 square mm.

<h2>Step-by-step explanation:</h2>

The top of the surface is in the shape of a annulus with a outer radius of 10 mm and a inner radius of 7 mm ( since the diameter of the hole is: 14 mm and we know that the radius is half of the diameter)

Now, we know that the area of the annulus region is given by:

$Area=\pi (R^2-r^2)$

where R is the outer radius and r is the inner radius.

Here we have:

$R=10\ mm\\\\and\\\\r=7\ mm$

Hence, we have:

$Area\ of\ top\ surface=\pi (10^2-7^2)\\\\i.e.\\\\Area\ of\ top\ surface=\pi (100-49)\\\\i.e.\\\\Area\ of\ top\ surface=\pi\cdot 51\\\\i.e.\\\\Area\ of\ top\ surface=160.14\ mm^2$