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Nata [24]
4 years ago
9

1.)Mario and Hank are arguing. Mario says it is impossible to draw a right triangle with sides measuring 8 inches, 12 inches, an

d 18 inches. Hank says it is possible. Who is correct?
2.) Hank says it is possible to draw a right triangle with the measurements shown in the diagram below right triangle with sides of 10 in, and 26 in. What is the length of the third side of this right triangle?
Mathematics
1 answer:
erastovalidia [21]4 years ago
3 0
Problem One
Test it.
a^2 + b^2 = c^2
a = 8
b = 12 What will c calculate to be.

8^2 + 12^12 = c^2
c^2 = 64 + 144
c^2 = 208 
If 18 is to be the hypotenuse of the triangle c^ must equal 324. 
208 is nowhere's near that amount. 
Mario is right.

Question Two
The only way Hank will be right is if 26 is the hypotenuse. If that is so, the third side is 24. This is how you do it.
a^2 + b^2 = c^2
c = 26
a = 10

10^2 + b^2 = 26^2
100 + b^2 = 676 Subtract 100
b^2 = 676 - 100
b^2 = 576
b = sqrt(576)
b = 24

If 26 is not the hypotenuse, the right triangle cannot be drawn.
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It’s worth 10 points if you answer!
Maksim231197 [3]

Answer:

m∠x = 10°

Step-by-step explanation:

Since we are dealing with a <u>right triangle</u> and its two sides and one angle, we can use <u>trigonometry ratios</u>. Remember them all with the acronym SohCahToa.

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The ratios are:

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In this triangle, θ = x. The sides we know are hypotenuse and opposite. Therefore, we will use the sinθ ratio.

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sinx = CB/AB      Substitute the labels in the diagram

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x = sin⁻¹(4/23)      Isolate 'x'. Use calculator to solve.

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