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Alborosie
2 years ago
5

Experimentally verify sum of three angles of an triangle is 180⁰.​

Mathematics
1 answer:
svetlana [45]2 years ago
4 0

Draw line a through points A and B. Draw line b through point C and parallel to line a.

Since lines a and b are parallel, <)BAC = <)B'CA and <)ABC = <)BCA'.

It is obvious that <)B'CA + <)ACB + <)BCA' = 180 degrees.

Thus <)ABC + <)BCA + <)CAB = 180 degrees.

Lemma

If ABCD is a quadrilateral and <)CAB = <)DCA then AB and DC are parallel.

Proof

Assume to the contrary that AB and DC are not parallel.

Draw a line trough A and B and draw a line trough D and C.

These lines are not parallel so they cross at one point. Call this point E.

Notice that <)AEC is greater than 0.

Since <)CAB = <)DCA, <)CAE + <)ACE = 180 degrees.

Hence <)AEC + <)CAE + <)ACE is greater than 180 degrees.

Contradiction. This completes the proof.

Definition

Two Triangles ABC and A'B'C' are congruent if and only if

|AB| = |A'B'|, |AC| = |A'C'|, |BC| = |B'C'| and,

<)ABC = <)A'B'C', <)BCA = <)B'C'A', <)CAB = <)C'A'B'.

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Answer C,D,E,F<br><br><br>Snsjma
Ulleksa [173]

Answer:

f) 97° = m∠6

e) 83° = m∠5

d) 83° = m∠4

c) 97° = m∠3

b) 97° = m∠2

a) 83° = m∠1

Step-by-step explanation:

  • m∠5 ≅ m∠4; Angle Correspondence Theorem
  • m∠5 ≅ m∠1; Alternate Interior Angles Theorem
  • m∠6 ≅ m∠2; Alternate Exterior Angles Theorem
  • 180° = m∠2 + m∠1; Definition of Linear Pair [Supplementary Angles]
  • 180° = m∠5 + m∠3; Definition of Linear Pair [Supplementary Angles]

I am joyous to assist you anytime.

6 0
3 years ago
Fred's sporting goods is providing a $5 discount on a bicycle that costs $80. What is the percentage of the discount?
Sonja [21]

Answer:

6.25%

Step-by-step explanation:

discount/ normal price

5/80

And we are trying to find out the percentage, so

x/100

Set the 2 equations equal to each other

5/80=x/100

Cross multiply

80x=500

Divide by 80 on both sides

x= 6.25

8 0
3 years ago
What is the radius for the circle given by the equation? 
alisha [4.7K]

Use:

(a-b)^2=a^2-2ab+b^2\qquad(*)

x^2+y^2-6y-12=0\ \ \ \ |+12\\\\x^2+y^2-6y=12\\\\x^2+y^2-2y\cdot3=12\ \ \ \ |+3^2\\\\x^2+\underbrace{y^2-2\cdot y\cdot3+3^2}_{(*)}=12+3^2\\\\x^2+(y-3)^2=21\\\\\boxed{k=3}

7 0
3 years ago
What is an equation that goes through the point (-7, 10) and is perpendicular to the line -2x-4y=9
algol13

Step-by-step explanation:

m1.m2= -1

(4y = -2x -9)= (y = -1x/2 -9/4)

-1/2.m2= -1

m2 = 2

formula = y - y1 = m2 ( x - x1)

y - 10 = 2 ( x - (-7))

y - 10 = 2x + 14

y = 2x + 24

7 0
3 years ago
Pls help me!!!!!!!!!!!!!
solmaris [256]
The answer is the third option
7 0
3 years ago
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