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Crazy boy [7]
3 years ago
10

(05.02 LC) Given triangle ABC, which equation could be used to find the measure of ∠C?

Mathematics
1 answer:
Sidana [21]3 years ago
6 0

Answer:

D. sin m∠C = 2 square root of 5 all over 5

Step-by-step explanation:

The triangle ABC is a right angled triangle. The angle ∠C is where we are asked to find.  The side AB measures 6 , AC measures 3 and the side BC measures 3√5.

The longest side of a right angle triangle is the hypotenuse, that means the side with length  3√5 is the hypotenuse.

Therefore side BC is the hypotenuse side and we are asked to find the angle C. AB is the opposite side while AC is the adjacent side.

Using

sin C = opposite/hypotenuse = 6/3√5 = 2/√5 = 2√5/5

cos C = adjacent/hypotenuse = 3/3√5 = 1/√5

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Answer:

The width is 5m and the length is 10m.

Step-by-step explanation:

Rectangle:

Has two dimensions: Width(w) and length(l).

It's area is:

A = w*l

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This means that l = 3w - 5

The area of the rectangle is 50m^(2)

This means that A = 50. So

A = w*l

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3w^{2} - 5w - 50 = 0

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

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In this question:

3w^{2} - 5w - 50 = 0

So

a = 3, b = -5, c = -50

\bigtriangleup = (-5)^{2} - 4*3*(-50) = 625

w_{1} = \frac{-(-5) + \sqrt{625}}{2*3} = 5

w_{2} = \frac{-(-5) - \sqrt{625}}{2*3} = -3.33

Dimension must be positive result, so

The width is 5m(in meters because the area is in square meters).

Length:

l = 3w - 5 = 3*5 - 5 = 10

The length is 10 meters

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