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In-s [12.5K]
3 years ago
10

There are two non congruent triangles where B=55 degrees a=15 and b=13. Find the measures of the angles of the triangle with the

greater perimeter. Round to the nearest tenth if necessary.
Mathematics
1 answer:
liq [111]3 years ago
5 0

Answer:

The side c needs to be 12.85 for the triangle to have greatest perimeter

Step-by-step explanation:

We are given;

b = 13

a = 15

Angle at b: B = 55°

Let's find side c.

Using the law of cosines,we have;

b² = a² + c² - 2ac•cos(B)

13² = 15² + c² - 2•15•c•cos(55)

169 = 225 + c² - 30c•cos(55)

c² - 30c•cos(55) + 225 - 169 = 0

c² - 30c•cos(55) + 56 = 0

c² - 30c•(0.5736) + 56 = 0

c² - 30c•(0.5736) + 56 = 0

c² - 17.208c + 56 = 0

Using quadratic formula;

c = [-(-17.208) ± √((-17.208)² - (4•1•56)]/2(1)

c = [17.208 ± √(296.115 - 224)]/2

c = 8.604 ± 4.246

To have the greater perimeter, we need the larger value of c, thus we will use the positive sign and ignore the negative one ;

Thus,

c = 8.604 + 4.246 = 12.85

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The graph below represents the solution set of which inequality?
natulia [17]

Answer:

option: B (x^2+2x-8) is correct.

Step-by-step explanation:

We are given the solution set as seen from the graph as:

(-4,2)

1)

On solving the first inequality we have:

x^2-2x-8

On using the method of splitting the middle term we have:

x^2-4x+2x-8

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And we know that the product of two quantities are negative if either one of them is negative so we have two cases:

case 1:

x+2>0 and x-4

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so we have the region as:

(-2,4)

Case 2:

x+2 and x-4>0

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Hence, we did not get a common region.

Hence from both the cases we did not get the required region.

Hence, option 1 is incorrect.

2)

We are given the second inequality as:

x^2+2x-8

On using the method of splitting the middle term we have:

x^2+4x-2x-8

⇒ x(x+4)-2(x+4)

⇒ (x-2)(x+4)

And we know that the product of two quantities are negative if either one of them is negative so we have two cases:

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Hence, we do not get a common region.

Case 2:

x-2 and x+4>0

i.e. x<2 and x>-4

Hence the common region is (-4,2) which is same as the given option.

Hence, option B is correct.

3)

x^2-2x-8>0

On using the method of splitting the middle term we have:

x^2-4x+2x-8>0

⇒ x(x-4)+2(x-4)>0

⇒ (x-4)(x+2)>0

And we know that the product of two quantities are positive if either both of them are negative or both of them are positive so we have two cases:

Case 1:

x+2>0 and x-4>0

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Hence, the common region is (4,∞)

Case 2:

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Hence, from both the cases we did not get the desired answer.

Hence, option C is incorrect.

4)

x^2+2x-8>0

On using the method of splitting the middle term we have:

x^2+4x-2x-8>0

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Case 2:

x-2>0 and x+4>0

i.e. x>2 and x>-4.

Hence, the common region is: (2,∞)

Hence from both the case we do not have the desired region.

Hence, option D is incorrect.




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