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Vera_Pavlovna [14]
3 years ago
5

Can a triangle be formed with the side lenghts of 8cm, 4cm, and 12cm

Mathematics
2 answers:
ioda3 years ago
4 0
No, because the sum of the measures shorter legnths of the sides must be greater than the measure of the greatest angle

4+8=12
so therefor the side legnths if you tried to make into a triangle, it would make a line 12 units long

<span>no is the answer</span>
Morgarella [4.7K]3 years ago
4 0
No, a triangle cannot be formed with side lengths of 8 cm, 4 cm, and 12 cm.  This is because of the triangle inequality which states that the sum of any two of the side lengths of a triangle must be greater than the length of the third side.

This means that 8 + 4 must be greater than 12, however, it is not, it is equal to 12.

Therefore, no a triangle can NOT be formed with the side lengths of 8 cm, 4 cm, and 12 cm.
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The longest side of an acute isosceles triangle is 8 centimeters. Rounded to the nearest tenth , what is the smallest possible l
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Read 2 more answers
I want solution...please help me
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sin(<em>θ</em>) + cos(<em>θ</em>) = 1

Divide both sides by √2:

1/√2 sin(<em>θ</em>) + 1/√2 cos(<em>θ</em>) = 1/√2

We do this because sin(<em>x</em>) = cos(<em>x</em>) = 1/√2 for <em>x</em> = <em>π</em>/4, and this lets us condense the left side using either of the following angle sum identities:

sin(<em>x</em> + <em>y</em>) = sin(<em>x</em>) cos(<em>y</em>) + cos(<em>x</em>) sin(<em>y</em>)

cos(<em>x</em> - <em>y</em>) = cos(<em>x</em>) cos(<em>y</em>) - sin(<em>x</em>) sin(<em>y</em>)

Depending on which identity you choose, we get either

1/√2 sin(<em>θ</em>) + 1/√2 cos(<em>θ</em>) = sin(<em>θ</em> + <em>π</em>/4)

or

1/√2 sin(<em>θ</em>) + 1/√2 cos(<em>θ</em>) = cos(<em>θ</em> - <em>π</em>/4)

Let's stick with the first equation, so that

sin(<em>θ</em> + <em>π</em>/4) = 1/√2

<em>θ</em> + <em>π</em>/4 = <em>π</em>/4 + 2<em>nπ</em>   <u>or</u>   <em>θ</em> + <em>π</em>/4 = 3<em>π</em>/4 + 2<em>nπ</em>

(where <em>n</em> is any integer)

<em>θ</em> = 2<em>nπ</em>   <u>or</u>   <em>θ</em> = <em>π</em>/2 + 2<em>nπ</em>

<em />

We get only one solution from the second solution set in the interval 0 < <em>θ</em> < 2<em>π</em> when <em>n</em> = 0, which gives <em>θ</em> = <em>π</em>/2.

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3 years ago
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