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malfutka [58]
3 years ago
8

PLEASE HELP! THANKS! The measures, in degrees, of the 3 angles of a triangle are given by 2x +1, 3x - 3, and 9x. What is the mea

sure of the smallest angle? A.) 13 degrees B.) 23 degrees C.) 29 degrees D.) 40 degrees please show how to solve this I GIVE THANKS!!!
Mathematics
1 answer:
drek231 [11]3 years ago
7 0
So all of the internal angles of a triangle will add up to 180 degrees.  So if we add all of the expressions given and set them equal to 180, we can solve for x, then solve for each angle...

(2x + 1) + (3x - 3) + 9x = 180
2x + 3x + 9x + 1 - 3 = 180
14x - 2 = 180
14x = 182
x = 13

Now we can substitute 13 into each expression to find the value of each angle...

2x + 1 =
2(13) + 1 =
26 + 1 = 27 degrees

3x - 3 =
3(13) - 3 =
39 - 3 = 36 degrees

9x =
9(13) = 117 degrees

So the 3 angles are 27 degrees, 36 degrees, and 117 degrees.

The smallest is 27 degrees

** Note-- I don't see 27 degrees in the answer choices you gave, so I double checked my math (I didn't find any mistakes).  Please check to see that the question is written correctly (pay close attention to plus and minus signs).
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tangare [24]

Check the picture below.

well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.

bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.

\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}

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\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{5+1}{2}~~,~~\cfrac{1+4}{2} \right)\implies \left(3~~,~~\cfrac{5}{2} \right)

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\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{\frac{5}{2}}) \stackrel{slope}{m}\implies \cfrac{4}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{\cfrac{5}{2}}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{3})\implies y-\cfrac{5}{2}=\cfrac{4}{3}x-4 \\\\\\ y=\cfrac{4}{3}x-4+\cfrac{5}{2}\implies y=\cfrac{4}{3}x-\cfrac{3}{2}

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