If P(3,2) and Q(7,10) are the endpoints of the diameter of a circle, what is the area of the circle in square units?
1 answer:
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Answer:
Fred C. 16.30 ft
Gary B. 169.67 m
Step-by-step explanation:
Fred the owl...
Always <u>draw a diagram first</u> (see diagram below). We assume trees are perpendicular (at 90°) to the ground. The situation forms a right-angle triangle.
We need to <u>find "x"</u>, which is the direct distance between the owl and the bird. "x" is also the hypotenuse, which is the longest side in a right-angle triangle. Since we know a side and an angle, we can find the hypotenuse using primary trigonometry ratios.
∠B will be the angle of reference (the angle we are "talking" about). The side we know is opposite to ∠B (15ft). We know it's opposite because it's not touching the angle. The side we need to find is the hypotenuse.
<u>Use the trig. ratio sine</u> because it has opposite and hypotenuse in it. The general formula is:
sinθ = opp/hyp
θ means the angle, and we know it is 67°.
Replace all the information you know:
sinB = opp/hyp
sin67° = (15ft) / x Isolate "x". Rearrange the equation.
xsin67° = (15ft) Divide both sides by sin67°
x = (15ft) / sin67°
x = 16.295... ft Exact answer in decimals
x ≈ 16.30 ft Rounded to nearest hundredth
Therefore Fred the owl is 16.30 feet from the bird.
Gary spots a monkey...
Draw a diagram first (see below). G is Gary, M is the monkey, T is the bottom of the Sears Tower. We need to <u>find "x"</u>, which is the distance between the monkey and the bottom of the tower.
The angle of reference is M, which is 69°. We need to find the adjacent side and we know the opposite side. <u>The trig. ratio that has adjacent and opposite is tangent.</u>
tanθ = opp/adj
Replace all the information you know:
tanM = opp/adj
tan69° = (442m) / x Rearrange the formula to isolate "x"
xtan69° = 442m Divide both sides by tan69°
x = 442m / tan69° Solve by dividing
x = 169.6679.... m Exact decimal answer
x ≈ 169.67m Rounded to the nearest hundredth
Therefore the monkey is 169.67 metres from the tower.
Remember all the trig. ratios using the acronym SohCahToa.
o = opposite side
a = adjacent side
h = hypotenuse side
S = sine, sin for short
C = cosine, cos for short
T = tangent, tan for short
The operation is always dividing the first side in the acronym by the second side.
sinθ = opp/hyp
cosθ = adj/hyp
tanθ = opp/adj
Its asking you to draw rectangles with sides that equal 30 cm for example
Answer:
a. θ = 30°
b. μ = √3 / 15 ≈ 0.115
Step-by-step explanation:
Draw a free body diagram for each scenario (see attached figure). The body has four forces acting on it:
- Weight pulling down
- Normal force perpendicular to the incline
- Applied force parallel up the incline
- Friction force parallel to the incline
Remember that friction opposes the direction of motion. So when the body is sliding up, friction points down the incline. And when the body is sliding down, friction points up the incline.
Now apply Newton's second law to each scenario, first in the normal direction, then in the parallel direction.
For sliding up, sum of the forces normal to the incline:
∑F = ma
N − mg cos θ = 0
N = mg cos θ
Sum of the forces parallel to the incline:
∑F = ma
P₁ − f − mg sin θ = 0
P₁ − Nμ − mg sin θ = 0
Substituting the expression for normal force:
P₁ − mgμ cos θ − mg sin θ = 0
Now for sliding down, sum of the forces normal to the incline:
∑F = ma
N − mg cos θ = 0
N = mg cos θ
And sum of the forces parallel to the incline:
∑F = ma
P₂ + f − mg sin θ = 0
P₂ + Nμ − mg sin θ = 0
Substituting the expression for normal force:
P₂ + mgμ cos θ − mg sin θ = 0
We know that P₁ = 6 kg.wt, P₂ = 4 kg.wt, and mg = 10 kg.wt.
So we have two equations and two unknowns (μ and θ):
P₁ − mgμ cos θ − mg sin θ = 0
P₂ + mgμ cos θ − mg sin θ = 0
Let's start by adding the equations together:
P₁ + P₂ − 2 mg sin θ = 0
P₁ + P₂ = 2 mg sin θ
sin θ = (P₁ + P₂) / (2 mg)
Plugging in the values:
sin θ = (6 + 4) / (2 × 10)
sin θ = 1/2
θ = 30°
Now we can plug this into either equation and find μ.
P₁ − mgμ cos θ − mg sin θ = 0
6 − (10 cos 30°) μ − 10 sin 30° = 0
6 − 5√3 μ − 5 = 0
1 = 5√3 μ
μ = √3 / 15
μ ≈ 0.115
