Solve the problem.
Problem 1 :
A 29 foot water slide has a 17 foot vertical ladder. How far is it along the ground from the end of the slide back to the base of the ladder that leads to the slide?
Solution :
Given, AC = 29 foot
AB = 17 foot
Using Pythagorean Theorem,
AC² = AB² + BC²
29² = 17² + BC²
BC² = 29² - 17²
= √841 - 289
BC = √552
BC = 23.49 ft
So, the base of the ladder is 23.5 ft.
Problem 2 :
A painter leans a 30 foot ladder against one wall of a house. At what height does the ladder touch the wall if the foot of the ladder is 10 ft from the base of the wall?
Solution :
Given, AC = 30 foot
BC = 10 foot
Using Pythagorean Theorem,
AC² = AB² + BC²
30² = AB² + 10²
AB² = 30² - 10²
= √900 - 100
AB = √800
AB = 28.28 ft
So, the height is 28.3 ft.
Problem 3 :
From a distance of 45 feet from the base of a building, the angle of elevation to the top of the building is 68˚. Estimate the height of the building to the nearest foot.
Solution :
tan 68˚ = h/45
h = 45 × tan 68˚
h = 45 × 2.46
h = 111 feet
So, the height of the building is 111 feet.
Problem 4 :
A kite is currently flying at an altitude of 15 meters above the ground. If the angle of elevation from the ground to the kite is 35˚, find the length of the kite string to the nearest meter.
Solution :
sin θ = opposite / hypotenuse
sin 35˚ = BC/AB
AB = 15/sin 35˚
AB = 15/0.57
AB = 26.3
So, the length of the kite string is 26 meters.
Problem 5 :
From a distance of 1217 feet from a spotlight, the angle of elevation to a cloud base is 43˚. Find the height of the cloud base to the nearest foot.
Solution :
tan θ = opposite / adjacent
tan 43˚ = h/1217
h = tan 43˚/ 1217
h = 0.9325 × 1217
h = 1134.9 ft
So, the height of the cloud base is 1135 ft.
Solve the right triangle using the information given. Round answers to two decimal places, if necessary.
Problem 6 :
b = 8, A = 30˚; Find a, c and B.
Solution :
Find c :
cos θ = Adjacent / hypotenuse
cos 30˚ = b/c
0.866 = 8/c
c = 8/0.866
c = 9.24
Find a :
sin θ = Opposite / hypotenuse
sin 30˚ = a/9.24
0.5 = a/9.24
a = 0.5 × 9.24
a = 4.62
Find B:
B = 90˚ - A
B = 90˚ - 30˚
B = 60˚
Problem 7 :
a = 2, A = 40˚; Find b, c and B
Solution :
Find c :
sin θ = Opposite / hypotenuse
sin 40˚ = 2/c
0.642 = 2/c
c = 2/0.642
c = 3.11
Find b :
cos θ = Adjacent / hypotenuse
cos 40˚ = b/c
0.766 = b/3.11
b = 0.766 × 3.11
b = 2.38
Find B :
B = 90˚ - A
B = 90˚ - 40˚
B = 50˚
Problem 8 :
a = 7, b = 4; Find c, A and B.
Solution :
Find c:
Using Pythagorean Theorem,
c² = a² + b²
c² = 7² + 4²
c² = 49 + 16
c = √65
c = 8.06
Find A:
tan A = a/b
tan A = 7/4
A = tan^{-1 }(7/4)
A = 60.26˚
Find B:
tan B = b/a
tan B = 4/7
B = tan^{-1 }(4/7)
B = 29.74˚
Problem 9:
a = 4, c = 9; Find b, A and B.
Solution :
Find b:
Using Pythagorean Theorem,
c² = a² + b²
b² = c² - a²
b² = 9² - 4²
b² = 81 - 16
b = √65
b = 8.06
Find A:
sin A = a/c
sin A = 4/9
A = sin^{-1} (4/9)
A = 26.39˚
Find B:
cos B = a/c
cos B = 4/9
B = cos^{-1} (4/9)B = 63.61˚
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM