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Ann [662]
4 years ago
14

Angle a and angle b are vertical angles with measue of angle a=x and measure of angle b=5x-80​

Mathematics
1 answer:
Gelneren [198K]4 years ago
8 0

Answer:

we know that

Vertical Angles are the angles opposite each other when two lines cross. They are always congruent to one another

in this problem

∠A and ​ ∠B ​ are vertical angles

so

∠A=∠B ​

we have

∠A=

∠B=

equate angle A and angle B

solve for x

therefore

the answer is

the measure of angle A is

Step-by-step explanation:

Hope it helps:)

Brainliest pls? Have a good day/night

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TEA [102]

Answer:

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Step-by-step explanation:

I will be writing the equation of the perpendicular bisector in the slope-intercept form which is y=mx +c, where m is the gradient and c is the y-intercept.

A perpendicular bisector is a line that cuts through the other line perpendicularly (at 90°) and into 2 equal parts (and thus passes through the midpoint of the line).

Let's find the gradient of the given line.

\boxed{gradient =  \frac{y1 -y 2}{x1 - x2} }

Gradient of given line

=  \frac{1 - ( - 5)}{3 - ( - 9)}

=  \frac{1 + 5}{3 + 9}

=  \frac{6}{12}

=   \frac{1}{2}

The product of the gradients of 2 perpendicular lines is -1.

(½)(gradient of perpendicular bisector)= -1

Gradient of perpendicular bisector

= -1 ÷(½)

= -1(2)

= -2

Substitute m= -2 into the equation:

y= -2x +c

To find the value of c, we need to substitute a pair of coordinates that the line passes through into the equation. Since the perpendicular bisector passes through the midpoint of the given line, let's find the coordinates of the midpoint.

\boxed{midpoint = ( \frac{x1 + x2}{2} , \frac{y1 + y2}{2})  }

Midpoint of given line

= ( \frac{3  -  9}{2} , \frac{1 - 5}{2} )

= ( \frac{ - 6}{2}  , \frac{ - 4}{2} )

= ( - 3 , - 2)

Substituting (-3, -2) into the equation:

-2= -2(-3) +c

-2= 6 +c

c= -2 -6 <em>(</em><em>-</em><em>6</em><em> </em><em>on both</em><em> </em><em>sides</em><em>)</em>

c= -8

Thus, the equation of the perpendicular bisector is y= -2x -8.

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3 years ago
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