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Kay [80]
3 years ago
15

BCALLE can u help me please

Mathematics
1 answer:
Ksivusya [100]3 years ago
4 0
17. RQ is the same as PS.
PS = -1 + 4x
RQ = 3x + 3
-1 + 4x = 3x + 3
4x = 3x + 4
x = 4
Now plug that into RQ.
3(4) + 3 = RQ
15 = RQ

18. Angles G and E are equal to each other.
G = 5x - 9
E = 3x + 11
5x - 9 = 3x + 11
5x = 3x + 20
2x = 20
x = 10
Plug that x into G.
5(10) - 9
41 = G

19. TE and EV are equal to each other.
TE = 4 + 2x
EV = 4x - 4
4 + 2x = 4x - 4
2x = 4x - 8
-2x = -8
x = 4
Plug that into TE.
4 + 2(4)
12 = TE

20. DB and BF are equal.
DB = 5x - 1
BF = 5 + 3x
5x - 1 = 5 + 3x
5x = 6 + 3x
2x = 6
x = 3
Plug that into DB.
5(3) - 1
14 = DB
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Proof of the Law of Sines

The Law of Sines states that for any triangle ABC, with sides a,b,c (see below)

a

 sin  A

=

b

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For more see Law of Sines.

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c

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Using a similar method it can be shown that in this case

c

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a

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- Q.E.D

Obtuse Triangles

The proof above requires that we draw two altitudes of the triangle. In the case of obtuse triangles, two of the altitudes are outside the triangle, so we need a slightly different proof. It uses one interior altitude as above, but also one exterior altitude.

First the interior altitude. This is the same as the proof for acute triangles above.

Draw the altitude h from the vertex A of the triangle

 sin  B =

h

c

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h

b

or

h = c  sin  B       a n d         h = b  sin  C

Since they are both equal to h

c  sin  B = b  sin  C

Dividing through by sinB and then sinC

c

 sin  C

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b

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Draw the second altitude h from B. This requires extending the side b:

The angles BAC and BAK are supplementary, so the sine of both are the same.

(see Supplementary angles trig identities)

Angle A is BAC, so

 sin  A =

h

c

or

h = c  sin  A

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 sin  C =

h

a

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c  sin  A = a  sin  C

Dividing through by sinA then sinC:

a

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=

c

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Combining (4) and (9):

a

 sin  A

=

b

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=

c

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