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2 years ago

15

Find the measures of the angles of each triangle (in degrees) and classify the triangle as acute/right/obtuse and scalene/isosce

les/equilateral (one from each triple)
A=5x - 10 B = 3x C= 2x - 10

1 answer:

stiks02 [169]2 years ago

8 0

Step-by-step explanation:

given :

A=5x - 10

B = 3x

C= 2x - 10

solution:

A+B+C = 180°

5x - 10 + 3x+ 2x - 10 = 180

10x -20 = 180

10x = 200

x = 20

A = 5(20)-10 = 100-10 = 90°

B = 3x = 3(20) = 60°

C = 2x-10 =2(20)-10 =40-10 = 30°

this is the right triangle and scalene triangle

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The triangle is both an Isosceles triangle and a right triangle.

Step-by-step explanation:

Given the vertices of a triangle.

$P_{1} = (- 1, 4)$

$P_{2} = (6, 2)$ and

$P_{3} = (4, - 5)$

We find the distance between all the points to determine the length of each side of the triangle.

Distance between any two points, say, $(x_1, y_1)$ and $(x_2, y_2)$ is:

$\sqrt{\bigg ( \textbf{x}_{\textbf{2}} \hspace{1mm} \textbf{- x}_{\textbf{1}} \bigg )^{\textbf{2}} \textbf{+} \bigg( \textbf{y}_{\textbf{2}} \hspace{1mm} \textbf{- y}_{\textbf{1}} \bigg)^ {\textbf{2}}$

Length between $P_1$ and $P_{2}$ , (Side 1) :

$(x_1, y_1) = (- 1, 4)$ and

$(x_2, y_2) = (6, 2)$

Distance = $\sqrt{\bigg(6 - (-1) \bigg)^{2} \hspace{1mm} + \hspace{1mm} \bigg( 2 - 4 \bigg )^2$

$= \sqrt{7^2 + 2^2}$

$= \sqrt{49 + 4}$

$= \sqrt{\textbf{53}}$

Length of Side 1 = $\sqrt{\textbf{53}}$ units.

Distance between $P_1$ and $P_2$ , (Side 2):

$(x_1, y_1) = (-1, 4)$

$(x_2, y_2) = (4, - 5)$

Distance = $\sqrt{ \bigg( 4 + 1 \bigg)^2 \hspace{1mm} + \bigg( - 5 - 4 \bigg ) ^2$

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Length of Side 2 = $\sqrt{\textbf{106}}$ units.

Distance between $P_2$ and $P_3$ , Side 3 :

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Distance = $\sqrt{ 2^2 \hspace{1mm} + \hspace{1mm} 7^2}$

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Length of Side 3 = $\sqrt{\textbf{53}}$ units.

Note that the length of Side 1 = Length of Side 3.

That means the triangle is Isosceles.

Also, For a triangle to be right angle triangle, using Pythagoras theorem we have:

(Side 1)² + (Side 3)² = (Side 2)²

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Hence, the triangle is a right - angled triangle as well.

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