A cone is 8cm high and its vertical angle is 62 degrees,find the diameter of its base

I need your solution

If the radius of the base is r, then

tan(62/2) = r/8

I don't know I want to know

I wonna know wat the answer is

Please what is the answer I need it for an exam Tomorrow

A cone is 8cm high and it's vertical angle is 62° find the diameter of itbase

The answer is 9.6 but i don't no how to solve it 2 0

A cone is 8cm high and it's vertical angle is 62 find the diameter of it base please show me this answer it is urgent please

62/2=r/8 =31° tan 31°=0,6008×8 =4,8064+4,8064=9,6128=9,6cm

To find the diameter of the base of a cone, we can use the trigonometric relationship in a right triangle formed by the height of the cone, the radius of the base, and the slant height (also known as the generatrix) of the cone.

Let's start by drawing a diagram of the cone:

/|
/ |
h / | r
/ |
/ |
/ |
_________/______|_____
s r s/2

In this diagram:
- h represents the height of the cone (given as 8 cm)
- r represents the radius of the base (what we want to find)
- s represents the slant height (or generatrix) of the cone

We can begin by using the definition of sine in a right triangle (opposite/hypotenuse) to relate the height of the cone, the slant height, and the angle:

sin(angle) = h / s

Rearranging the formula to solve for s, we get:

s = h / sin(angle)

Substituting the known values into this equation, we have:

s = 8 cm / sin(62 degrees)

Next, we can use the definition of the cosine in a right triangle (adjacent/hypotenuse) to relate the slant height and the radius of the base.

cos(angle) = r / s

Rearranging the formula to solve for r, we get:

r = s * cos(angle)

Substituting the value of s we found earlier into this equation, we have:

r = (8 cm / sin(62 degrees)) * cos(62 degrees)

Evaluating this expression will give us the diameter of the base since the diameter is twice the radius:

diameter of base = 2 * r

So, by substituting the value of r into this equation, we can find the diameter of the base of the cone.