Question

Two banks of river are parallel and the distance between two points A and B along one bank is 500 ft. For a point C on the opposite bank ABC= 41° what the width w of the river

Answer 1

Answer:

wia^2 - b^2 +2bc -c^2 are this

Answer 2

Answer:

The width of the river is about 246.8 feet.

Step-by-step explanation:

From the above question,

They have given :

Two banks of river are parallel and the distance between two points A and B along one bank is 500 ft.

For a point C on the opposite bank ABC= 41°

We want to discover the width w of the river.

In the diagram, A and B are the two factors on the identical financial institution of the river that are five hundred toes apart. C is a factor on the contrary bank, and we are given that attitude ABC is forty one degrees.

Now we have a proper triangle ACD, the place perspective ACD is ninety degrees, and perspective ADC is forty one degrees.

Using trigonometry, we can write:

tan(ADC) = w/CD

tan(41) = w/CD

CD = w/tan(41)

Similarly, we can write:

tan(CAE) = w/CE

tan(90-41) = w/CE

CE = w/tan(49)

Since AE = AB = five hundred feet, we have:

DE = AE - AD = five hundred - CD

FE = CE

Now we can use the Pythagorean theorem to clear up for w:

DE$.^2$  + FE$.^2$  = $w^2$

(500 - CD)^2 + CE^2 = $w^2$

(500 - w/tan(41))$.^2$  + (w/tan(49))$.^2$ = $w^2$

Simplifying and fixing for w, we get:

w ≈ 246.8 feet

Therefore, the width of the river is about 246.8 feet.

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