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A regular 9 sided building is 814 ft along one side what is the distance from a Vertex to the center of the building

User MagerValp
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A regular 9 sided building is 814 ft along one side what is the distance from a Vertex to the center of the building ​

we know that

A regular 9 sided polygon, can be divided into 9 isosceles triangle

The equal distances of the isosceles triangle is equal to the radius or the distance from a Vertex to the center of the building ​

the base of the isosceles triangle is given and is equal to 814 ft

The measure of the interior angle of the vertex is equal to

360/9=40 degrees

that means

we have

Applying the law of sines

814/sin(40)=r/sin(70)

solve for r

r=(814/sin(40))*sin(70)

r=1,190 ft

therefore

the distance is 1,190 ft

A regular 9 sided building is 814 ft along one side what is the distance from a Vertex-example-1
User Onion
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