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3 years ago

Solve this simultaneous equation m+2n=10 and 2n+3m=18

Mathematics

1 answer:

Ksju [112]3 years ago

3 0

Answer:

m=4

n=3

Step-by-step explanation:

m+2n=10

2n+3m=18

Subtract

-2m=-8

m=4

Solve for n

(4)+2n=10

2n+3(4)=18

2n=10-4

2n=18-12

2n=6

n=3

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Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

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Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

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Considering right angled triangle BOC where O is the point between b-x and x

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In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

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Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

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$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

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Solve this simultaneous equation m+2n=10 and 2n+3m=18​

Solve this simultaneous equation m+2n=10 and 2n+3m=18