Help needed with right triangle trig!
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Draw a right triangle ABC, with m(CAB)=∅.
Let CB=7 units, AC=8 units, so by the Pythagorean theorem:
(units)
so we have drawn a triangle where sin∅=7/8
in this triangle, cot∅=(adjacent side)/(opposite side)=
Answer: <span>square root of fifteen divided by seven</span>
Let CB=7 units, AC=8 units, so by the Pythagorean theorem:
so we have drawn a triangle where sin∅=7/8
in this triangle, cot∅=(adjacent side)/(opposite side)=
Answer: <span>square root of fifteen divided by seven</span>
The measurements of the figures that should be equal in order for the ratio of volumes between the volumes of the cone, sphere, and cylinder of 1:2:3 to be true is; The radius and the height
<h3>What is a ratio?</h3>A ratio is a comparison between the magnitudes or measurements of items, expressing the number of times one item in the ratio is contained within another item in the ratio.
The formula for the volume of a cone, is,
= (1/3)·π·r₁²·h₁
The formula for the volume of a sphere, , is
= (4/3)·π·r₂³
The formula for the volume of a cylinder, , is
= π·r₃²·h₃
The ratio of the volumes in the question is 1:2:3
Therefore; The volume of the sphere is twice the volume of the cone, and the volume of the cylinder is three times the volume of the cone, which indicates;
(4/3)·π·r₂³ = 2 × (1/3)·π·r₁²·h₁...(1)
π·r₃²·h₃ = 3 × (1/3)·π·r₁²·h₁...(2)
Equation (1) indicates that we have;
(4/3)·π·r₂³ = 2 × (1/3)·π·r₁²·h₁
Where, h₁ = 2·r₂, we get;
(4/3)·π·r₂³ = 2 × (1/3)·π·r₁²·(2·r₂) = (4/3)·π·r₁²·r₂
Therefore; r₁ = r₂, and the diameter of the sphere is equal to the height of the cone
The radius of the cone is equal to the radius of the sphere
Equation (2) indicates;
π·r₃²·h₃ = 3 × (1/3)·π·r₁²·h₁ = π·r₁²·h₁
Therefore;
r₃²·h₃ = r₁²·h₁
r₃ = r₁
h₃ = h₁
The radius of the cone = The radius of the cylinder
The height of the cone = The height of the cylinder
- The ratio of the volumes, 1 : 2 : 3 is true when the measurements of the radius and the height are equal for all three solids.
Learn more about the volume of regular solids here:
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