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alukav5142 [94]
2 years ago
15

100 POINTS PLUS BRAINLY

Mathematics
2 answers:
Iteru [2.4K]2 years ago
4 0

Answer:

\sf adjacent  = 4.5 \ ft

Given:

  • opposite side: 13 ft
  • adjacent unknown
  • angle of 71°

using tan rule:

\sf tan(x) = \dfrac{opposite}{adjacent}

\hookrightarrow \sf tan(71) = \dfrac{13}{adjacent}

\hookrightarrow  \sf adjacent  = \dfrac{13}{tan(71) }

\hookrightarrow  \sf adjacent  = 4.5

daser333 [38]2 years ago
4 0

Let unknown side be x

\\ \rm\hookrightarrow tan\theta=\dfrac{P}{B}

\\ \rm\hookrightarrow tan71=\dfrac{13}{x}

\\ \rm\hookrightarrow 2.9=13/x

\\ \rm\hookrightarrow x=13/2.9

\\ \rm\hookrightarrow x=4.5

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For the first circle, let's use the pythagorean theorem

\bf \textit{using the pythagorean theorem}\\\\
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now, it just so happen that the hypotenuse on that triangle, is actually 17, but we used the pythagorean theorem to find it, and the pythagorean theorem only works for right-triangles.

 so if the hypotenuse is actually 17, that means that triangle there is actually a right-triangle, meaning that the radius there, and the outside line there, are both meeting at a right-angle.

when an outside line touches the radius line, and they form a right-angle, the outside line is indeed a tangent line, since the point of tangency is always a right-angle with the radius.



now, let's check for second circle

\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
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well, low and behold, we didn't get our hypotenuse as 16 after all, meaning, that triangle is NOT a right-triangle, and that outside line is not touching the radius at a right-angle, therefore is NOT a tangent line.



let's check the third circle

\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
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this time, we did get our hypotenuse to 65, the triangle is a right-triangle, so the outside line is indeed a tangent line.
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