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mr_godi [17]
3 years ago
11

In triangle ABC, AB = 8.9 m, Angle BAC = 90° and Angle ABC = 56°. H lies on BC such that AH

Mathematics
1 answer:
12345 [234]3 years ago
6 0

Answer:

(i) The length of AH is 7.3784 m

(ii) The length of HC is 10.9389 m

Step-by-step explanation:

The sum of the angles ABC, BAC and ACB is 180º, hence, ACB = 180-90-56 = 34º. Note also that AH breaks the rectangle triangle ABC into 2 rectangle triangles AHB and AHC.

i) In AHB, AB is the hypotenuse, and the angle ABH is equal to ABC, hence it is 56º. AH is the opposite side of that angle and we know that

Sin(ABH) = opposite/hypotenuse = AH/8.9

Thus,

AH = 8.9*Sin(ABH) = 8.9*Sin(56º) = 7.3784 m

ii) In AHC, AH is also the opposite of the angle ACH (which meassures 34º), and HC is the adjacent of that angle (the hypotenuse is AC). As a result

Tan(ACH) = Tan(34º) = 0.6745 = opposite/adjacent = AH/HC = 7.3784/HC

Hence,

HC = 7.3785/0.6745 = 10.9389 m

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mote1985 [20]
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A red car is driving along the road in the direction of the police car and is 130 feet up the road from the location of the poli
pishuonlain [190]

Complete question:

A police car is located 50 feet to the side of a straight road. A red car is driving along the road in the direction of the police car and is 130 feet up the road from the location of the police car. The police radar reads that the distance between the police car and the red car is decreasing at a rate of 75 feet per second. How fast is the red car actually traveling along the road.

Answer:

The red car is traveling along the road at 80.356 ft/s

Step-by-step explanation:

Given

Police car is 50 feet side off the road

Red car is 130 feet up the road

Distance between them is decreasing at the rate of 75 feet per sec

Let x be how far the police is off the road.

Let y be how far the red car is up the road.  

Let h be the distance between the police and the red car.

This forms a right triangle so we can use the Pythagorean theorem, to solve for h

h² = x² + y²

h² = 50² + 130²

h² = 19400

h = √19400

h = 139.284 ft

Again;

Let dx/dt be how fast the police is traveling toward the road.

Let dy/dt  be how fast the red car is traveling along the road.

Let dh/dt be how fast the distance between the police and the car is decreasing.

Recall that, h² = x² + y² (now differentiate with respect to time, t)

2h(dh/dt) = 2x(dx/dt) + 2y(dy/dt)

divide through by 2

h(dh/dt) = x(dx/dt) + y(dy/dt)

since the police car is not, dx/dt = 0

and dy/dt is the how fast is the red car actually traveling along the road

139.284(75) = 50(0) + 130(dy/dt)

10446.3 = 0 + 130(dy/dt)

dy/dt = 10446.3 / 130

dy/dt = 80.356 ft/s

Therefore, the red car is traveling along the road at 80.356 ft/s

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X/-3 = -1 ,Leave X by itself so multiply by -3 on both sides.

X = -1*-3

X = 3, negative times negative equals positive.
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