9514 1404 393
Answer:
(2) 72°
Step-by-step explanation:
In this geometry, the angle at the tangent is half the measure of the intercepted arc.
∠CBD = (arc BD)/2 = 144°/2
∠CBD = 72°
__
<em>Additional comment</em>
Consider a point X anywhere on long arc BD. The inscribed angle at X will have half the measure of short arc BD, so will be 144°/2 = 72°. This is true regardless of the position of X on long arc BD. Now, consider that X might be arbitrarily close to point B. The angle at X is still 72°.
As X approaches B, the chord XB approaches a tangent to the circle at B. Effectively, this tangent geometry is a limiting case of inscribed angle geometry.
<em>GH</em>
- <em>Step-by-step explanation:</em>
<em>Hi !</em>
<em>GH ∈ (HGY)</em>
<em>GH ∈ (GEF) } => (HGY) ∩ (GEF) = GH</em>
<em>Good luck !</em>
Answer:
4.25
Step-by-step explanation:
Given:
The height of a right cone = 14 in
Base diameter = 17 in
To find:
The diagram of the cone and its lateral surface area.
Solution:
(a)
The diagram of a right cone with height 14 in and base diameter of 17 in is shown below.
Diagram is not to scale.
(b)
We know that lateral surface of the cone is

...(i)
Where, r is the base radius, h is vertical height and l is the slant height of the cone.
Base radius of the cone = 


Now,
Putting r=8.5, h=14 and π=3.14 in (i), we get





Therefore, the lateral surface area of the cone is 437 sq. inches.