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Luden [163]
3 years ago
9

The drawing shows a person looking at a building on top of which an antenna is mounted. The horizontal distance between the pers

on’s eyes and the building is 83.2 m. In part a the person is looking at the base of the antenna, and his line of sight makes an angle of 33.3 with the horizontal. In part b the person is looking at the top of the antenna, and his line of sight makes an angle of 35.7o with the horizontal. How tall is the antenna?
Mathematics
1 answer:
prohojiy [21]3 years ago
5 0
Consider the triangles formed by the lines of sight in parts a and b. In part a, the distance from the horizontal to the base of the antenna is 82.0*tan31.0º; in part b, the distance from the horizontal to the top is 82.0*tan34.2º. The antenna height is the difference between these 

<span>h = 82.0*tan34.2º - 82.0*tan31.0º = 82.0*(tan34.2º - tan31.0º) = 6.46 m</span>
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Step-by-step explanation:

∠B = ∠B      ∠BAD = 90° - ∠B = ∠C

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3(2b+3)^{2}=36\\\frac{3(2b+3)^{2}}{3}=\frac{36}{3}\\(2b+3)^{2}=12

Taking square root both sides, we get

\sqrt{(2b+3)^{2}}=\pm \sqrt{12}\\2b+3=\pm \sqrt{4\times 3}\\2b+3=\pm 2\sqrt{3}\\2b+3-3=\pm 2\sqrt{3}-3\\2b=-3\pm 2\sqrt{3}\\b=\frac{-3\pm 2\sqrt{3}}{2}\\b=\frac{-3+ 2\sqrt{3}}{2}\textrm{ or }b=\frac{-3- 2\sqrt{3}}{2}

Therefore, the values of b are:

\frac{-3+ 2\sqrt{3}}{2}\textrm{ and }\frac{-3- 2\sqrt{3}}{2}

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MA=\sqrt{\left(\dfrac{\boxed{3}a}{2}\boxed{-}\boxed{0}\right)^2+\left(\dfrac{\boxed{b}}{2}\boxed{-}\boxed{0}\right)^2}=\\\\\\=&#10;\sqrt{\left(\dfrac{\boxed{3}a}{\boxed{2}}\right)^2+\left(\dfrac{b}{2}\right)^2}=\sqrt{\dfrac{\boxed{9}a^2}{\boxed{4}}+\dfrac{\boxed{b}^2}{\boxed{4}}}

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NB=\sqrt{\left(\dfrac{a}{2}\boxed{-}\boxed{2}a\right)^2+\left(\dfrac{b}{2}\boxed{-}\boxed{0}\right)^2}=\\\\\\=\sqrt{\left(\dfrac{a}{2}\boxed{-}\dfrac{\boxed{4}\boxed{a}}{2}\right)^2+\left(\dfrac{b}{2}-\boxed{0}\right)^2}=\\\\\\&#10;\sqrt{\left(\dfrac{-3a}{2}\right)^2+\left(\dfrac{b}{\boxed{2}}\right)^2}=\sqrt{\dfrac{\boxed{9}a^2}{\boxed{4}}+\dfrac{\boxed{b}^2}{\boxed{4}}}
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D = 25/27 g/cm³

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