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IrinaK
1 month ago
15

A shopper is standing on level ground 800 feet from the base of a 250 foot tall department store the shopper looks up and sees a

flag on the stores roof to the nearest degree what is the angle of elevation to the top of the building from the point on the ground where the shopper is standing?

Mathematics
1 answer:
zzz [12.3K]1 month ago
8 0

Response:

The angle of elevation from the position of the shopper on the ground to the top of the building is 17.35°

Explanation in detail:

Considering triangle ΔABC

AB represents the height of the building, which is 250 feet

BC measures 800 ft

The angle of elevation is calculated as \theta

\tan \theta = \frac{AB}{BC}

\tan \theta = \frac{250}{800}

\tan \theta = 0.3125

\theta = 17.35°

Thus, the angle of elevation to the height of the building from where the shopper is located on the ground is 17.35°

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Solution:-

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Here,  k represents the coefficient of proportionality.

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- Given that xcm = ycm, these points lie along the line y = x.

- Compute the mass of the lamina (m):x_c_m = \frac{M_y}{m} = \frac{\int \int {x*p(x,y)} \, dA }{\int \int {p(x,y)} \, dA } \\\\y_c_m = \frac{M_x}{m} = \frac{\int \int {y*p(x,y)} \, dA }{\int \int {p(x,y)} \, dA } \\\\m = mass = \int \int {p(x,y)} \, dA

                           

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m = mass = \int \int {p(x,y)} \, dA = \int\limits^0_a \int\limits_0 {k*(x^2 + y^2)} \, dy.dx \\\\m = k\int\limits^a_0 {(x^2y + \frac{y^3}{3}) } \, dx|\limits^a^-^x_0 = k\int\limits^a_0 {(\frac{-4x^3}{3} + 2ax^2-ax^2+\frac{a^3}{3}) } \, dx\\\\m = k* [ \frac{-x^4}{3} + \frac{2ax^3}{3} - \frac{a^2x^2}{2} +\frac{a^3x}{3}] | \limits^a_0 \\\\m = k* [ \frac{-a^4}{3} + \frac{2a^4}{3} - \frac{a^4}{2} +\frac{a^4}{3}] = \frac{ka^4}{6}

- Determine ( xcm = ycm ):

               

xcm = ycm = M_y = \int \int {x*p(x,y)} \, dA = \int\limits^0_a \int\limits_0 {k*x*(x^2 + y^2)} \, dy.dx \\\\M_y = k\int\limits^a_0 x*{(x^2y + \frac{y^3}{3}) } \, dx|\limits^a^-^x_0 = k\int\limits^a_0 x*{(\frac{-4x^3}{3} + 2ax^2-ax^2+\frac{a^3}{3}) } \, dx\\\\M_y = k* [ \frac{-4x^5}{15} + \frac{ax^4}{2} - \frac{a^2x^3}{3} +\frac{a^3x^2}{6}] | \limits^a_0 \\\\M_y = k* [ \frac{-4a^5}{15} + \frac{a^5}{2} - \frac{a^5}{3} +\frac{a^5}{6}] = \frac{ka^5}{15}( ka^5 /15 ) / ( ka^4/6) =

2a/5    

   

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Ix = Iy =  a^4 / 12

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                   Ixy = a^4 / 24

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