Two hallways, one of width 5 feet, the other of width 6 feet, meet at a right angle. See the illustration to the right. It can be shown that the length L of the ladder as a function of 0 is L(0) = 6 csc 0+5 sec 0. Complete parts (a) through (d) below. 5t- ..... (a) In calculus, the length of the longest ladder that can turn the corner is found by solving the equation 5 sec 0 tan 0 -6 csc 0 cot 0 = 0 for 0 where 0° < 0 < 90°. Solve this equation for 0 0 = (Type an integer or decimal rounded to the nearest hundredth as needed.) (b) What is the length of the longest ladder that can be carried around the corner? L = feet (Type an integer or decimal rounded to the nearest hundredth as needed.) (c) Graph L = L(0), 0° <0 < 90°, Choose the correct answer below. O A. O B. C. D. AL 20- AL AL 10- 10- 4- 8- 18- 8- 2- 6- 16- 6- 0- 4- 4- -24 14- 2- 01") 12- 0") 2- -4- 0+ 30 0- 60 90 30 60 90 30 60 90 Find the angle that minimizes the length L (Type an integor or decimal rounded to the nearest hundredth as needed.) (d) Compare the result in part (c) with the one found in part (a). Explain. OA. The answers in part (c) and part (a) are equal because the angle that minimizes the longth of tho ladder is the samo as the angle at which tho longest ladder can be carried around the corner based on the illustration. OB. The answer in part (a) is greater than the answer in part (c) because part (c) shows the angle that minimizos the length of the ladder. Part (a), on the other hand, shows the angle that maximizos the length of the ladder. OC. The answers in part (c) and part (a) are equal because L(0) = 6 csc 0+5 sec 0 5 sec 0 tan 0-6 csc 0 cot0, D. The answer in part (c) is groater than the answer in part (a) because the anglo in part (a) shows the angle at which the longth of the ladder is 0, meaning that no ladder can be carried around the corner. Part (c), on the other hand, shows the maximum length of a ladder than can be carried around the corner,

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Two hallways, one of width 5 feet, the other of width 6
feet, meet at a right angle. See the illustration to the
right. It can be shown that the length L of the ladder as a
function of 0 is L(0) = 6 csc 0 + 5 sec 0. Complete parts
(a) through (d) below.
5t-
.....
(a) In calculus, the length of the longest ladder that can turn the corner is found by solving the equation
5 sec 0 tan 0 -6 csc 0 cot 0 = 0 for 0 where 0° < 0 < 90°. Solve this equation for 0
0 =
(Type an integer or decimal rounded to the nearest hundredth as needed.)
(b) What is the length of the longest ladder that can be carried around the corner?
L=
feet
(Type an integer or decimal rounded to the nearest hundredth as needed.)
(c) Graph L = L(0), 0° <0 < 90°, Choose the correct answer below.
O A.
O B.
C.
O D.
AL
20-
AL
AL
10-
10-
4-
8-
18-
8-
2-
6-
16-
6-
0-
4-
4-
-24
130
14-
2-
0")
12-
2-
-4-
0+
30
0-
60
90
30
60
90
30
60
90
Find the angle that minimizes the length L.
(Type an integer or decimal rounded to the nearest hundredth as needed.)
(d) Compare the result in part (c) with the one found in part (a). Explain.
O A. The answers in part (c) and part (a) are equal because the angle that minimizes the longth of tho ladder is tho samo as the angle at which the longest ladder can be carried around the corner
based on the illustration.
O B. The answer in part (a) is greater than the answer in part (c) because part (c) shows the angle that minimizos the length of the ladder. Part (a), on the other hand, shows the angle that maximizos
the length of the ladder.
OC. The answers in part (c) and part (a) are equal because L (0) =6 csc 0+5 sec 0 5 sec 0 tan 0-6 csc 0 cot0.
OD. The answer in part (c) is groater than the answer in part (a) because the anglo in part (a) shows the angle at which the longth of the ladder is 0, meaning that no ladder can be carried around the
corner. Part (c), on the other hand, shows the maximum length of a ladder than can be carried around the corner.
Transcribed Image Text:Two hallways, one of width 5 feet, the other of width 6 feet, meet at a right angle. See the illustration to the right. It can be shown that the length L of the ladder as a function of 0 is L(0) = 6 csc 0 + 5 sec 0. Complete parts (a) through (d) below. 5t- ..... (a) In calculus, the length of the longest ladder that can turn the corner is found by solving the equation 5 sec 0 tan 0 -6 csc 0 cot 0 = 0 for 0 where 0° < 0 < 90°. Solve this equation for 0 0 = (Type an integer or decimal rounded to the nearest hundredth as needed.) (b) What is the length of the longest ladder that can be carried around the corner? L= feet (Type an integer or decimal rounded to the nearest hundredth as needed.) (c) Graph L = L(0), 0° <0 < 90°, Choose the correct answer below. O A. O B. C. O D. AL 20- AL AL 10- 10- 4- 8- 18- 8- 2- 6- 16- 6- 0- 4- 4- -24 130 14- 2- 0") 12- 2- -4- 0+ 30 0- 60 90 30 60 90 30 60 90 Find the angle that minimizes the length L. (Type an integer or decimal rounded to the nearest hundredth as needed.) (d) Compare the result in part (c) with the one found in part (a). Explain. O A. The answers in part (c) and part (a) are equal because the angle that minimizes the longth of tho ladder is tho samo as the angle at which the longest ladder can be carried around the corner based on the illustration. O B. The answer in part (a) is greater than the answer in part (c) because part (c) shows the angle that minimizos the length of the ladder. Part (a), on the other hand, shows the angle that maximizos the length of the ladder. OC. The answers in part (c) and part (a) are equal because L (0) =6 csc 0+5 sec 0 5 sec 0 tan 0-6 csc 0 cot0. OD. The answer in part (c) is groater than the answer in part (a) because the anglo in part (a) shows the angle at which the longth of the ladder is 0, meaning that no ladder can be carried around the corner. Part (c), on the other hand, shows the maximum length of a ladder than can be carried around the corner.
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