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shtirl [24]
3 years ago
11

In the figure drawn below, two lines are intersecting each other. What is the measure, in degrees, of all the angles adjacent to

or opposite of angle A?
Enter your answers in the following sequence: B, C, D

Mathematics
1 answer:
Roman55 [17]3 years ago
8 0

Answer:

\angle B =155^{\circ} , \angle D =25^{\circ} and \angle D=155^{\circ}

Step-by-step explanation:

\angle A = 25^{\circ}

Vertically opposite angles : the angles opposite each other when two lines interest

So, ∠A is vertically opposite to ∠C

\angle A = \angle C (Vertically opposite angles are equal)

So, \angle C = 25^{\circ}

\angle A+\angle B = 180^{\circ}(\text{Linear pair})\\25^{\circ}+\angle B = 180^{\circ}\\\angle B = 180^{\circ}-25^{\circ}\\\angle B =155^{\circ}

∠B is vertically opposite to ∠D

So,\angle B = \angle D (Vertically opposite angles are equal)

So,\angle D = 155^{\circ}

Hence \angle B =155^{\circ} , \angle D =25^{\circ} and \angle D=155^{\circ}

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kirill [66]

Answer:

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4 0
3 years ago
1) find the equation of the line parallel to
In-s [12.5K]

<u>Answer:</u>

1) The equation of the line parallel to  x-5y=6 and through (4,-2) is 5y = x -14

2) The equation of the line perpendicular to y= -2/5x + 3 and through (2,-1) is  2y = 5x -12

<u>Solution:</u>

<u><em>1) find the equation of the line parallel to  x-5y=6 and through (4,-2).</em></u>

Given, line equation is x – 5y = 6  

We have to find the line equation that is parallel to given line and passing through the point (4, -2)

Now, let us find slope of the given line.

\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-1}{-5}=\frac{1}{5}

Now, we know that, slope of parallel lines are equal.

So, slope of required line is 1/5 and it passes through (4, -2)

Now, using point slope form

y-y_{1}=m\left(x-x_{1}\right) \text { where } m \text { is slope and }\left(x_{1}, y_{1}\right) \text { is point on line. }

y-(-2)=\frac{1}{5}(x-4) \rightarrow 5(y+2)=x-4 \rightarrow 5 y+10=x-4 \rightarrow x-5 y=14

Hence, the line equation is 5y = x -14

<u><em> 2) find the equation of the line perpendicular to y= -2/5x + 3 and through (2,-1)</em></u>

\text { Given, line equation is } y=-\frac{2}{5} x+3 \rightarrow 5 y=-2 x+15 \rightarrow 2 x+5 y=15

We have to find the line equation that is perpendicular to given line and passing through the point (2, -1)

Now, let us find slope of the given line.

\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-2}{5}=-\frac{2}{5}

Now, we know that, product of slopes of perpendicular lines equals to -1.

So, slope of required line \times slope of given line = -1

slope of required line = -1 \times \frac{5}{-2}=\frac{5}{2}

And it passes through (2, -1)

Now, using point slope form

\begin{array}{l}{\text { Line equation is } y-(-1)=\frac{5}{2}(x-2) \rightarrow 2(y+1)=5(x-2)} \\\\ {\rightarrow 2 y+2=5 x-10 \rightarrow 5 x-2 y=12}\end{array}

Hence, the line equation is 2y = 5x -12

4 0
3 years ago
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