An observer 2.5 miles from the launch pad of a space shuttle measures the angle of elevation to the base of the shuttle to be 25
◦ soon after lift-off. d Observer θ How high is the shuttle at that instant (assuming that the shuttle is still moving vertically)? Answer in units of miles.
1 answer:
Answer:
Height of shuttle at that instant = 1.17 miles
Step-by-step explanation:
Refer the given figure.
C is the position of rocket and B is the position of observer.
Here we need to find AC, that is the height of shuttle at that instant.
Using trigonometric rules on ΔABC, we have
Height of shuttle at that instant = AC = 1.17 miles
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The diagram of the pentagon is missing, so i have attached it.
Answer:
|AE| = 130 m
|DE| = 150 m
Perimeter of pentagon = 720 m
Step-by-step explanation:
From the diagram, we can find AE from pythagoras theorem;
|AE| = √(|AA'|² + 50²)
Where AA' is the length from A to the perpendicular angle.
Now, AB = 150, and A'B is parallel to 30 m. Thus, A'B = 30
AA' = AB - A'B = 150 - 30
AA' = 120
Thus;
|AE| = √(120² + 50²)
|AE| = √(14400 + 2500)
|AE| = √16900
|AE| = 130
Similarly,
|DE| = √(|DD'|² + |ED'|²)
ED' = BC - 50
ED' = 140 - 50
ED' = 90
Also, DD' is parallel to AA' and is = 120
Thus;
|DE| = √(120² + 90²)
|DE| = √22500
|DE| = 150
Perimeter of pentagon = 150 + 130 + 150 + 150 + 140 = 720
