25. angle EFG and angle GFH are a linear pair,
1 answer:
Answer:
m<EFG = , and m<GFH =
Step-by-step explanation:
Linear pair angles are two supplementary angles.
Thus,
m<EFG + m<GFH =
2n + 21 + 4n + 15 =
collecting like terms, we have:
6n + 36 =
6n = - 36
6n =
divide both both sides by 6,
n =
Therefore,
m<EFG = 2n + 21
= 2 x + 21
= 48 + 21
=
m<GFH = 4n + 15
= 4 x + 15
= 96 + 15
=
Thus m<EFG = , and m<GFH =
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<h3>Answer: 80 degrees</h3>
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Explanation:
I'm assuming that segments AD and CD are tangents to the circle.
We'll need to add a point E at the center of the circle. Inscribed angle ABC subtends the minor arc AC, and this minor arc has the central angle AEC.
By the inscribed angle theorem, inscribed angle ABC = 50 doubles to 2*50 = 100 which is the measure of arc AC and also central angle AEC.
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Focus on quadrilateral DAEC. In other words, ignore point B and any segments connected to this point.
Since AD and CD are tangents, this makes the radii EA and EC to be perpendicular to the tangent segments. So angles A and C are 90 degrees each for quadrilateral DAEC.
We just found angle AEC = 100 at the conclusion of the last section. So this is angle E of quadrilateral DAEC.
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Here's what we have so far for quadrilateral DAEC
- angle A = 90
- angle E = 100
- angle C = 90
- angle D = unknown
Now we'll use the idea that all four angles of any quadrilateral always add to 360 degrees
A+E+C+D = 360
90+100+90+D = 360
D+280 = 360
D = 360-280
D = 80
Or a shortcut you can take is to realize that angles E and D are supplementary
E+D = 180
100+D = 180
D = 180-100
D = 80
This only works if AD and CD are tangents.
Side note: you can use the hypotenuse leg (HL) theorem to prove that triangle EAD is congruent to triangle ECD; consequently it means that AD = CD.