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bezimeni [28]
2 years ago
5

In the figure, if m∠ABD = 120º, then m∠ADC =? º.

Mathematics
2 answers:
Anna11 [10]2 years ago
4 0
The figure that this question is referring to is attached. We must use the Law of Sines to solve this question, which is as follows:

a/sinA = b/sinB = c/sinC

This applies to any triangle. We are told that ∠ABD = 120º. We are asked to solve for ∠ADC. We know that ∠ADC + ∠ADB = 180º. If we assign ∠ADC = x, then ∠ADB = 180<span>º - x. We can now apply the law of sines to this data.

35/sin120 = 30/sin(180-x)
sin(180-x) = (30/35)(sin120)
sin(180-x) = 0.742
sin-1(sin(180-x)) = sin-1(0.742)
180 - x = 48</span><span>º
x = 132</span><span>º
</span>
We have already assigned x = ∠ADC; therefore, ∠ADC = 132<span>º.</span>

Andru [333]2 years ago
4 0

The correct answer is

132

:)

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What are the coordinates of the midpoint of the segment whose endpoints are (-4,6) and (-8,-2)
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6 0
3 years ago
What is a necessary step for constructing perpendicular lines through a point off the line?
Nostrana [21]

Answer:

Find another point on the perpendicular line.

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Most often, this is accomplished by exploiting the fact that "p" is the set of all points that are equidistant from any pair of points that are symmetric about "p".

Since the symmetry must be about "p", and we don't even know where "p" is, one often finds two points on "m" that are equidistant from "Q".

This can be accomplished by adjusting a compass to a fixed radius (larger than the distance from "Q" to "m"), and making an arc that intersects "m" in two places.  Those two places will be equidistant from "Q", and are simultaneously on line "m".  Thus, these two points, "A" & "B" are symmetric about "p".

Since "A" & "B" are symmetric about "p", they are equidistant from "p", and are on "m".  One could try to find the point of intersection between "p" and "m" through construction, but this is unnecessary.  We need only find a second point (besides "Q") that is equidistant from "A" & "B", which will necessarily be a point on "p", to form the line perpendicular to "m".

To do this, fix the compass with any radius, and from "A" make a large arc generally in the direction of "B", and make the same radius arc from "B" in the direction of "A" such that the two arcs intersect at some point that isn't "Q".  This point of intersection we can call point "T", and the line QT is line "p", the line perpendicular to the original line, necessarily containing "Q".

8 0
2 years ago
I need help i don’t understand this
lys-0071 [83]
Answer:
Angle BAC: 56
ANGLE BCA: 42

Why? First find angle BCA because that angle is on a straight line with another angle on the other side you can do
180-138=42 which is the measurement of angle BCA.

For angle BAC you have to know that the sun of all triangles interior angles is 180 so you do
180-82-42=56
So angle BAC is 56

Hope this helps :D
6 0
3 years ago
Read 2 more answers
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