The Blue Angels, the Navy's elite flying squadron, have performed for more than 427 million fans since their inception in 1946.
Suppose the two Blue Angel jets at points P and A in the diagram have elevations of 1555 feet and 1000 feet, respectively. Both are flying east toward the viewing crowd at T. From T, the measure of the angle of elevation of jet P is 5°, and the measure of the angle of elevation of jet A is 2.4°.
(a) How far from the crowd is each of the jets?
Plane P __
Plane A __
(b) Find the distance d between the nose tip of jet A and the nose tip of jet P.
To solve this problem, we can use trigonometric ratios and the concept of similar triangles. Let's start by labeling the diagram and assigning variable names to the distances we need to find.
(a) Let's call the distance from the crowd to jet P as x, and the distance from the crowd to jet A as y.
In the diagram, we have two right triangles: triangle PTM and triangle ATN.
Using trigonometric ratios, we can relate the angles of elevation to the distances:
For triangle PTM:
tan(5°) = PM / x
PM = x * tan(5°)
For triangle ATN:
tan(2.4°) = AN / y
AN = y * tan(2.4°)
Now, let's substitute the values of the angles of elevation and solve for the distances:
PM = x * tan(5°)
PM = x * 0.0874886635
PM = 0.0875x
AN = y * tan(2.4°)
AN = y * 0.0416387467
AN = 0.0416y
The total distance between jet A and jet P is given by their respective distances from the crowd:
d = PM + AN
d = 0.0875x + 0.0416y
(b) To find the distance d between the nose tip of jet A and the nose tip of jet P, we can use the Pythagorean theorem on triangle ATN:
TN^2 = AN^2 + AT^2
Since TN is the same as y, and we know AN from part (a), we can solve this equation for AT:
AT^2 = TN^2 - AN^2
AT^2 = y^2 - (0.0416y)^2
AT^2 = y^2 - 0.00173296y^2
AT^2 = 0.99826704y^2
AT = √(0.99826704y^2)
AT = 0.9991337y
Now, we can substitute this value of AT into the equation for d:
d = PM + AN
d = 0.0875x + 0.0416y
d = 0.0875x + 0.0416(0.9991337y)
d = 0.0875x + 0.0414y
Therefore, the final answers to the questions are:
(a) The distance from the crowd to jet P is given by 0.0875 times the value of x, and the distance from the crowd to jet A is given by 0.0414 times the value of y.
(b) The distance between the nose tip of jet A and the nose tip of jet P is given by 0.0875 times the value of x plus 0.0414 times the value of y.
To solve this problem, we can use trigonometry, specifically the tangent function. We'll use the given angles of elevation and the heights of the jets to calculate the distances.
(a) Let's start by finding the distance from the crowd to each jet.
For jet P:
We have the angle of elevation of 5° and the elevation of 1555 feet.
We can use the tangent function to find the distance:
tan(5°) = Opposite/Adjacent
tan(5°) = 1555 feet/Distance from crowd to jet P
To find the distance from the crowd to jet P, we rearrange the equation:
Distance from crowd to jet P = 1555 feet / tan(5°)
Similarly, for jet A:
We have the angle of elevation of 2.4° and the elevation of 1000 feet.
Using the tangent function:
tan(2.4°) = 1000 feet/Distance from crowd to jet A
Rearranging the equation:
Distance from crowd to jet A = 1000 feet / tan(2.4°)
Now, we can substitute the values and calculate the distances:
Distance from crowd to jet P = 1555 feet / tan(5°)
Distance from crowd to jet A = 1000 feet / tan(2.4°)
(b) To find the distance between the nose tips of jets A and P, we can use the Pythagorean theorem. The elevation of the two jets will represent the height of the right triangle, and the distances we calculated in part (a) will represent the legs of the triangle.
Using the Pythagorean theorem:
hypotenuse^2 = leg1^2 + leg2^2
In our case, leg1 is the distance from the crowd to jet P, and leg2 is the distance from the crowd to jet A. We want to find the hypotenuse, which represents the distance between the nose tips of the jets.
Let's calculate the distance between the nose tips:
d^2 = (Distance from crowd to jet P)^2 + (Distance from crowd to jet A)^2
Now, we can substitute the values and solve for d:
d^2 = (1555 feet / tan(5°))^2 + (1000 feet / tan(2.4°))^2
Take the square root of both sides to find the value of d:
d = √[(1555 feet / tan(5°))^2 + (1000 feet / tan(2.4°))^2]
By evaluating this expression, you will find the distance d between the nose tip of jet A and the nose tip of jet P.
a) Tan5=(1550)/distance
similar for plane A
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