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astraxan [27]
3 years ago
14

Two sides of a triangle have measures 3 ft. and 6 ft. Also, these sides form a vertex whose angle measures 60 degrees. Calculate

the missing attributes of the triangle. Statements: x² + 3² = 6², where x is the 3rd side of the triangle. x² = 3² +6² + 2⋅3⋅6⋅cos 60°, where x is the 3rd side of the triangle. x² = 3² + 6² − 2⋅3⋅6⋅cos60°, where x is the 3rd side of the triangle. (sin 60°)/3 = (sinθ)/6, where θ is an unknown angle.
Mathematics
1 answer:
Tanya [424]3 years ago
8 0

Answer:

c² = 3² +6² - 2⋅3⋅6⋅cos 60

c = 27ft

Step-by-step explanation:

Since the angle is located in  between the sides of the sides, we will use the cosine rule to get the unknown sides

Let c be the missing sides

According to the cosine rule;

c² = a²+ b² - 2abcosC

c² = 3² +6² - 2⋅3⋅6⋅cos 60

c² = 9 + 36 - 36cos60

c² = 45 - 36cos60

c² = 45 - 36(0.5)

c² = 45 - 18

c² = 27ft

Hence the missing attribute is 27ft and the required expression is c² = 3² +6² - 2⋅3⋅6⋅cos 60

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Step-by-step explanation:

The absolute value function is a well known piecewise function (a function defined by multiple subfunctions) that is described mathematically as

                                 f(x) \ = \ |x| \ = \ \left\{\left\begin{array}{ccc}x, \ \text{if} \ x \ \geq \ 0 \\ \\ -x, \ \text{if} \ x \ < \ 0\end{array}\right\}.

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First, we simplify the expression.

                                             3\left|x \ + \ 1 \right| \ < \ 9 \\ \\ \\\-\hspace{0.2cm} \left|x \ + \ 1 \right| \ < \ 3.

We, now, can simply visualise the straight line,  y \ = \ x \ + \ 1 , as a line having its y-intercept at the point  (0, \ 1) and its <em>x</em>-intercept at the point (-1, \ 0). Then, imagine that the segment of the line where x \ < \ 0 to be reflected along the <em>x</em>-axis, and you get the graph of the absolute function y \ = \ \left|x \ + \ 1 \right|.

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Algebraically, we can solve this inequality by breaking the function into two different subfunctions (according to the definition above).

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  • Case 2 (when x \ < \ 0)

                                            -(x \ + \ 1) \ < \ 3 \\ \\ \\ \-\hspace{0.15cm} -x \ - \ 1 \ < \ 3 \\ \\ \\ \-\hspace{1cm} -x \ < \ 3 \ + \ 1 \\ \\ \\ \-\hspace{1cm} -x \ < \ 4 \\ \\ \\ \-\hspace{1.5cm} x \ > \ -4

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