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ehidna [41]
3 years ago
14

In ΔRST, r = 530 cm, s = 530 cm and t=950 cm. Find the measure of ∠T to the nearest 10th of a degree.

Mathematics
1 answer:
Pie3 years ago
6 0
The answer to this question is 127.3
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Determine the point of intersection of right bisectors in a triangle ∆ with vertices A (-3, 5), B (1, 1) and (−7, −3). Find the
iris [78.8K]

Answer:

Point of intersection (-11/3 , 1/3)

All distances from the vertices are \frac{10\sqrt{2} }{3}

Step-by-step explanation:

The vertices of the triangle is A(-3,5) , B (1,1) and C (-7,-3)

We need to find the perpendicular bisector of the triangle first

let's take one side connecting A and B

mid point of A and B = (-1,3)

Slope of the line joining A and B = -1

slope of perpendicular to line joining A and B = 1

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slope perpendicular to line joining B and C  =  -2

Equation of perpendicular bisector of line joining B and C =

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On solving 1 and 2

x= -11/3 , y= 1/3

Distances

From A = \frac{10\sqrt{2} }{3}

From B = \frac{10\sqrt{2} }{3}

From C = \frac{10\sqrt{2} }{3}

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