In ΔBCD, the measure of ∠D=90°, the measure of ∠C=77°, and CD = 41 feet. Find the length of BC to the nearest foot.
1 answer:
Answer:
182 ft
Step-by-step explanation:
When you are not given a diagram, always draw one for yourself (see diagram).
When you have a right triangle (triangle with 90° angle), you can use the trigonometry ratios, You can remember them using SohCahToa. It is read like this:
<u>s</u>inθ = <u>o</u>pposite/<u>h</u>ypotenuse Soh
<u>c</u>osθ = <u>a</u>djacent/<u>h</u>ypotenuse Cah
<u>t</u>anθ = <u>o</u>pposite/<u>a</u>djacent Toa
"θ" means the angle of reference (the angle you are talking about).
We are looking for the length of BC, which is the <u>hypotenuse</u>. I labelled it "d" (lowercase D) because it is opposite to ∠D.
We know ∠C = 77°. This will be our angle of reference (replace θ).
The side we know is DC, also known as "b" (lowercase B) because it's opposite to ∠B. "b" is the <u>adjacent</u> side when θ = C because "b" is touching ∠C.
Take the general trig. formula that has <u>hypotenuse</u> and <u>adjacent</u>: (cosine ratio)
cosθ = adjacent/hypotenuse
Substitute the variables specific for this problem.
cosC = b/d
Substitute the values you know.
cos77° = (41 ft) / d
Isolate "d" to the left side
dcos77° = Multiply both sides by "d"
dcos77° = 41 ft
dcos77° / cos77° = 41 ft / cos77° Divide both sides by cos77°
d = 41 ft / cos77° Input into calculator
d = 182.261874....... ft Unrounded answer
d ≈ 182 ft Rounded to nearest foot (whole number)
Remember d = BC. It's often easier to use one letter for calculations.
Therefore the length of BC is about 182 feet.
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v=
where g = 32
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v=
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= h
So maximum height of the water h =
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Given g = 32
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h =
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The pump withe the velocity of 75 ft/sec reaches the maximum height of 87.89 feet. 87.86 is greater than the maximum height 80 feet.
So the pump will meet the fire department needs.
Answer:
a) maximum; the parabola opens downward
b) positive; it must lie above the x-axis
c) x = 1.5
Step-by-step explanation:
The x-intercepts of a function are the points where the graph of the function crosses the x-axis. The y-values there are zero.
The "differences" of a function are related to the average slope between adjacent points. Second differences are related to the rate of change of the slope of the function. When <em>second differences are negative</em>, as here, the slope of the quadratic function is decreasing, becoming more negative. We say the <em>curvature</em> of the function is <em>negatve</em>, and that it <em>opens downward</em>.
__
<h3>a, b.</h3>If the graph of the parabola opens downward, and it crosses the x-axis, it must have a <em>maximum</em> that is a <em>positive value of y</em>.
__
<h3>c.</h3>The graph of a parabola is symmetrical about its vertex. That means points on the same horizontal line are the same distance from the line of symmetry, which must go through the vertex. The x-coordinate of the vertex will be the x-coordinate of the midpoint between the two x-intercepts:
x = (-2 +5)/2 = 3/2
The x-coordinate of the vertex is x = 1.5.
______
<em>Additional comment</em>
The attachment shows a table with three evenly-spaced points on the curve. The calculations show first differences (d1) and second differences (d2). You can see that the sign of the second diffference is negative, in agreement with the given conditions.
