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mrs_skeptik [129]
1 year ago
5

The sum of the measures of the angles of a triangle is 180 degrees. angels A,B,C has ratios of 3:5:7. What are their measures ?

Mathematics
2 answers:
Igoryamba1 year ago
4 0

Answer:

<em>36:60:84</em>

<em>A     : B    : C</em>

<em>36° : 60° : 84°</em>

<em>A= 36°</em>

<em>B= 60°</em>

<em>C=84°</em>

Step-by-step explanation:

<em>3:5:7 =15</em>

<em>180÷15=12</em>

<em>3×12=36</em>

<em>36:x:y =180</em>

<em>5×12=60</em>

<em>36:60:y =180</em>

<em>7×12=84</em>

<em>36:60:84 =180</em>

hjlf1 year ago
3 0

Answer: A=45  B=75  C=105

Step-by-step explanation:

3:5:7=180 degrees
   15=180
  15 divided by15=180 divided by 15
so,    1=12

so= 3:5:7 equal to (3*12)  (5*12)  (7*12)

A,B,C angles measures respectively = 36:60:84
  if you add them (A,B,C angles) we can obtain = 36:60:84=180    
  (sum of the measures of the angles of a triangle is 180 degrees)

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Answer:

(a) (x - 3)² + (y + 4)² + (z - 5)² = 25

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Step-by-step explanation:

The equation of a sphere is given by:

(x - x₀)² + (y - y₀)² + (z - z₀)² = r²            ---------------(i)

Where;

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(a) To get the equation of the sphere when it touches the xy-plane, we do the following:

i.  Since the sphere touches the xy-plane, it means the z-component of its centre is 0.

Therefore, we have the sphere now centered at (3, -4, 0).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (3, -4, 0) as follows;

d = \sqrt{(3-3)^2+ (-4 - (-4))^2 + (0-5)^2}

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Now substitute this value r = 5 into the general equation of a sphere given in equation (i) above as follows;

(x - 3)² + (y - (-4))² + (z - 5)² = 5²  

(x - 3)² + (y + 4)² + (z - 5)² = 25  

Therefore, the equation of the sphere when it touches the xy plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 25  

(b) To get the equation of the sphere when it touches the yz-plane, we do the following:

i.  Since the sphere touches the yz-plane, it means the x-component of its centre is 0.

Therefore, we have the sphere now centered at (0, -4, 5).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (0, -4, 5) as follows;

d = \sqrt{(0-3)^2+ (-4 - (-4))^2 + (5-5)^2}

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This distance is the radius of the sphere at that point. i.e r = 3

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(x - 3)² + (y - (-4))² + (z - 5)² = 3²  

(x - 3)² + (y + 4)² + (z - 5)² = 9  

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(x - 3)² + (y + 4)² + (z - 5)² = 9  

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i.  Since the sphere touches the xz-plane, it means the y-component of its centre is 0.

Therefore, we have the sphere now centered at (3, 0, 5).

Using the distance formula, we can get the distance d, between the initial points (3, -4, 5) and the new points (3, 0, 5) as follows;

d = \sqrt{(3-3)^2+ (0 - (-4))^2 + (5-5)^2}

d = \sqrt{(3-3)^2+ (0+4)^2 + (5-5)^2}

d = \sqrt{(0)^2 + (4)^2+ (0)^2}

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This distance is the radius of the sphere at that point. i.e r = 4

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(x - 3)² + (y - (-4))² + (z - 5)² = 4²  

(x - 3)² + (y + 4)² + (z - 5)² = 16  

Therefore, the equation of the sphere when it touches the xz plane is:

(x - 3)² + (y + 4)² + (z - 5)² = 16

 

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