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2 years ago

In ABC, what is the measure of angle B?

Mathematics

1 answer:

NeX [460]2 years ago

6 0

Answer:

Step-by-step explanation:

The sum of two remote interior angles (a remote interior angle is the interior angle that is not supplementary to the exterior angle given -- in this case A and B are remote angles) equals the exterior angle.

In This case A + B = 140

A = x + 2

B = 2x

x +2 + 2x = 140

3x + 2 = 140 Subtract 2 from both sides

3x = 138 Divide by 3

x = 138/3

x = 46

<B = 2x

<B = 2*46

<B = 92

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The coordinates G(7,3), H(9, 0), (5, -1) form what type of polygon?

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Answer:

Is an acute triangle

Step-by-step explanation:

we know that

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

G(7,3), H(9, 0), I(5, -1)

step 1

Find the distance GH

substitute in the formula

$d=\sqrt{(0-3)^{2}+(9-7)^{2}}$

$d=\sqrt{(-3)^{2}+(2)^{2}}$

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step 2

Find the distance IH

substitute in the formula

$d=\sqrt{(0+1)^{2}+(9-5)^{2}}$

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step 3

Find the distance GI

substitute in the formula

$d=\sqrt{(-1-3)^{2}+(5-7)^{2}}$

$d=\sqrt{(-4)^{2}+(-2)^{2}}$

$GI=\sqrt{20}\ units$

step 4

Verify what type of triangle is the polygon

we know that

If applying the Pythagoras Theorem

$c^{2}=a^{2}+b^{2}$ ----> is a right triangle

$c^{2}> a^{2}+b^{2}$ ----> is an obtuse triangle

$c^{2}< a^{2}+b^{2}$ ----> is an acute triangle

where

c is the greater side

we have

$c=\sqrt{20}\ units$

$a=\sqrt{17}\ units$

$b=\sqrt{13}\ units$

substitute

$c^{2}= (\sqrt{20})^{2}=20$

$a^{2}+b^{2}=(\sqrt{17})^{2}+(\sqrt{13})^{2}=30$

therefore

$c^{2}< a^{2}+b^{2}$

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