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ELEN [110]
3 years ago
5

Which of the following is the actual length of a wall that is 2 3/4 inches long on a drawing with a scale of 1 inch= 12 feet?

Mathematics
1 answer:
makvit [3.9K]3 years ago
6 0

Answer:

33 feet

Step-by-step explanation:

Given

1 inch= 12 feet

we have to find actual length of wall when length of wall on drawing is 2 3/4 inches

to do that lets multiply LHS and RHS of above equation with  2 3/4

2 3/4 inches = (2*4 +3 )/4 = 11/4 inches

1 inch* 2 3/4= 12 feet* 2 3/4

1*11/4 inches = 12*11/4 feet = 33 feet

Thus, actual length of wall is 33 feet.

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A person on a lake sees a plane flying overhead. The angle formed by his line of sight to the plane is 39°; If the plane is flyi
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⊱─━━━━━━━⊱༻●༺⊰━━━━━━━─⊰

\large \sf \dagger \: {Problem: \dagger}

A person on a lake sees a plane flying overhead. The angle formed by his line of sight to the plane is 39°; If the plane is flying about 5000 ft, find the horizontal ground distance between the person and the plane.

⊱─━━━━━━━⊱༻●༺⊰━━━━━━━─⊰

\large \sf \dagger \: {Solution: \dagger}

  • Let x = horizontal ground distance between the person and the plane.

\qquad\bigstar \: \large \sf{Equation: tan \: 39 \degree = \frac{5000}{x} }

Solving the equation,

\large\qquad \implies\tt{tan \: 39\degree =  \frac{5000}{x}}

\large\qquad \implies \tt{x(tan \: 39 \degree) = \cancel{ x} \large{(} \frac{5000}{ \cancel{x}} \large{)}}

\large\qquad\implies \tt \cancel{ \frac{x \: tan \: 39 \degree}{tan \: 39 \degree}}  =  \frac{5000}{tan \: 39 \degree}

\large\qquad \implies \tt{x =  \frac{5000}{tan \: 39\degree}}

\large \qquad \implies \tt{x  \approx  \pmb{6174.48758}}

Checking the solution,

  • \large\tt{tan \: 39 \degree  =  \frac{5000}{x}}

  • \large \tt{ tan \: 39 \degree =  \frac{5000}{6174.48758}}

  • \large\tt{0.80978 = 0.80978}

<h3>------------------------------</h3>

\large \sf \dagger \: {Answer: \dagger}

  • The horizontal distance between the person and the plane is approximately 6174 ft.

\qquad \rule{200pt}{3pt}

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