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olga nikolaevna [1]

3 years ago

5

The circle shown below has AB and BC as its tangents: AB and BC are two tangents to a circle which intersect outside the circle

at a point B. If the measure of arc AC is 130°, what is the measure of angle ABC? (1 point) 55° 50° 60° 65°

2 answers:

Nadusha1986 [10]3 years ago

3 0

We are given with
arc AC = 130
The measure of the arc on the other side of the circle is
360 - 130 = 230
Therefore, according to the theorem on circles, the measure of angle ABC is
(1/2) ( 230 - 130) = 50

Anit [1.1K]3 years ago

3 0

he picture in the attached figure

we know that

The measure of the external angle is the semidifference of the arcs that it covers.

so

the arcs that it covers are

arc AC and arc ABC

We have

AB and BC as its tangents

the measure of arc AC is $130$ °

In addition, a circle has $360$ degrees by definition

so

$360$ °= arc AC + arc ABC

$360$ °= $130$ ° + arc ABC

arc ABC= $360$ °- $130$ °= $230$ °

Then

Angle ABC = $\frac{1}{2}*(230-130)$

Angle ABC= $50$ °

therefore

the answer is

Angle ABC= $50$ °

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Read 2 more answers

Lines k and n intersect on the y-axis

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a) The equation of line k is:

$y = -\frac{202}{167}x + \frac{598}{167}$

b) The equation of line j is:

$y = \frac{167}{202}x + \frac{1546}{202}$

The equation of a line, in <u>slope-intercept formula</u>, is given by:

$y = mx + b$

In which:

m is the slope, which is the rate of change.
b is the y-intercept, which is the value of y when x = 0.

Item a:

Line k intersects line m with an angle of 109º, thus:

$\tan{109^{\circ}} = \frac{m_2 - m_1}{1 + m_1m_2}$

In which $m_1$ and $m_2$ are the slopes of <u>k and m.</u>

Line k goes through points (-3,-1) and (5,2), thus, it's slope is:

$m_1 = \frac{2 - (-1)}{5 - (-3)} = \frac{3}{8}$

The tangent of 109 degrees is $\tan{109^{\circ}} = -\frac{29}{10}$
Thus, the slope of line m is found solving the following equation:

$\tan{109^{\circ}} = \frac{m_2 - m_1}{1 + m_1m_2}$

$-\frac{29}{10} = \frac{m_2 - \frac{3}{8}}{1 + \frac{3}{8}m_2}$

$m_2 - \frac{3}{8} = -\frac{29}{10} - \frac{87}{80}m_2$

$m_2 + \frac{87}{80}m_2 = -\frac{29}{10} + \frac{3}{8}$

$\frac{167m_2}{80} = \frac{-202}{80}$

$m_2 = -\frac{202}{167}$

Thus:

$y = -\frac{202}{167}x + b$

It goes through point (-2,6), that is, when $x = -2, y = 6$ , and this is used to find b.

$y = -\frac{202}{167}x + b$

$6 = -\frac{202}{167}(-2) + b$

$b = 6 - \frac{404}{167}$

$b = \frac{6(167)-404}{167}$

$b = \frac{598}{167}$

Thus. the equation of line k, in slope-intercept formula, is:

$y = -\frac{202}{167}x + \frac{598}{167}$

Item b:

Lines j and k intersect at an angle of 90º, thus they are perpendicular, which means that the multiplication of their slopes is -1.

Thus, the slope of line j is:

$-\frac{202}{167}m = -1$

$m = \frac{167}{202}$

Then

$y = \frac{167}{202}x + b$

Also goes through point (-2,6), thus:

$6 = \frac{167}{202}(-2) + b$

$b = \frac{(2)167 + 202(6)}{202}$

$b = \frac{1546}{202}$

The equation of line j is:

$y = \frac{167}{202}x + \frac{1546}{202}$

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