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Question 3 (1 point)
Two walls of a canyon form the walls of a steady flowing river. From a point on the shorter wall, the angle of elevation to the top of the opposing wall is 20° and the angle of depression to the bottom of the opposing wall is 45°. The distance from the top of the shorter wall to the bottom of the opposing wall is 270 feet. Using the appropriate right triangle solving strategies, solve for the following: (Do not round intermediate calculated values. Only the final answer should be rounded to one decimal place.)

canyon walls

the height of the short wall (x)
the height of the tall wall (y)
the distance between the canyon walls (z)


x = feet (Type your answer in Blank 1)



y = feet (Type your answer in Blank 2)



z = feet (Type your answer in Blank 3)

Blank 1:
Blank 2:
Blank 3:

Question 3 1 point Two walls of a canyon form the walls of a steady flowing river From a point on the shorter wall the angle of elevation to the top of the oppo class=


Answer :

mpmeghghosh

Answer:

pls see the given image solution

View image mpmeghghosh
View image mpmeghghosh
sqdancefan

Answer:

  x = 190.9 ft

  y = 260.4 ft

  z = 190.9 ft

Step-by-step explanation:

You want the values of x, y, and z in the geometry shown.

Isosceles right triangle

The lengths marked x and z are the legs of an isosceles right triangle with hypotenuse 270 ft. We know the ratio of the side lengths of such a triangle is ...

  1 : 1 : √2

so ...

  x = z = 270/√2 ≈ 190.9 . . . feet

Tall wall

The part of y that lies below the observation point has measure x.

The part of y that lies above the observation point can be found using the tangent relation:

  Tan = Opposite/Adjacent

  tan(20°) = (y-x)/z

Solving for y and using x=z, we have ...

  z·tan(20°) = y -z . . . . . . multiply by z

  z(1+tan(20°) = y . . . . . . add z

  [tex]y=\dfrac{270}{\sqrt{2}}(1+\tan(20^\circ))\approx260.4[/tex]

The tall wall (y) is about 260.4 feet high.

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