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zhannawk [14.2K]
3 years ago
6

A Ladder that is 32 ft long leans against a building. The angle of elevation of the ladder is 70 degrees. To the nearest tenth o

f a foot how high off the ground is the top of the ladder?
A. 20.3 ft

B. 10.9 ft

C. 26.2 ft

D. 39.1 ft

Mathematics
2 answers:
kolezko [41]3 years ago
4 0
See the picture attached to better understand the problem

we know that
in the right triangle ABC
sin 70°=opposite side angle 70°/hypotenuse


in this problem
opposite side angle 70°=AB
hypotenuse=AC----> 32 ft
sin 70°=AB/32------> AB=32*sin 70°-----> AB=30.07 ----> AB=30.1 ft


tester [92]3 years ago
3 0

Answer:

30.1 ft.

Step-by-step explanation:

Please find the attachment.

Let the top of the ladder be 'h' feet high off the ground.

We have been given that a 32 ft long ladder leans against a building. The angle of elevation of the ladder is 70 degrees.

Upon looking at our attachment we can see that the 32 ft long side is hypotenuse and side with length h feet is opposite side of the right triangle formed by ladder and building with respect to ground.

Since sine relates the opposite side of right triangle with hypotenuse, so we can set an equation as:

\text{sin}=\frac{\text{Opposite}}{\text{Hypotenuse}}        

Upon substituting our given values we will get,

\text{sin}(70^{\circ})=\frac{h}{32}        

0.939692620786=\frac{h}{32}        

0.939692620786*32=\frac{h}{32}*32        

30.070163865152=h  

h\approx 30.1    

Therefore, the top of the ladder is 30.1 feet high off the ground.      

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Two airplanes leave an airport at the same time, the first headed due north and the second at a bearing of N42^ E . At 2:00PM, t
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Answer:

The two airplanes are about 330miles apart.

Step-by-step explanation:

The diagram interpreting the question has been attached to this response.

As shown in the diagram,

i. the airplanes leave at point C.

ii. at 2.00pm the first and second airplanes are at points A and B respectively, where they are 312miles and 487miles away from the starting point C in directions due north and N42E from the point C.

iii. the points A, B and C form a triangle with sides a, b and c.

To solve for the value of c which is the distance between the two planes at 2.00pm, the cosine rule is used.

c² = a² + b² - 2abcosC            --------------(i)

where;

b  = 312miles

a = 487miles

C = 42°

Substitute these values into equation (i) and solve as follows;

c² = (487)² + (312)² - 2(312)(487)cos(42)

c² = (237169) + (97344) - 303888cos(42)

c² = (237169) + (97344) - 303888(0.7431)

c² = 334513 - 225819.1728

c² = 108693.8272

<em>Take the square root of both sides</em>

√c² = √108693.8272

c = 329.69

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