Let the angles be `/_A=80^@,/_B=90^@,/_C=94^@,/_D=96^@` , and let the points of tangency be x(between A and B), y,z,w in order.

The measure of an angle formed by two tangents is equal to 1/2 the difference of the two arcs. Thus:

`m/_A=1/2(arcxyz-arcxw)`

`m/_B=1/2(arcyzx-arcyx)`

`m/_C=1/2(arczwy-arczy)`

`m/_D=1/2(arcwxz-arcwz)`

Let a=arc(xy),b=arc(yz),c=arc(wz) and d=arc(wx)

Then substituting...

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Let the angles be `/_A=80^@,/_B=90^@,/_C=94^@,/_D=96^@` , and let the points of tangency be x(between A and B), y,z,w in order.

The measure of an angle formed by two tangents is equal to 1/2 the difference of the two arcs. Thus:

`m/_A=1/2(arcxyz-arcxw)`

`m/_B=1/2(arcyzx-arcyx)`

`m/_C=1/2(arczwy-arczy)`

`m/_D=1/2(arcwxz-arcwz)`

Let a=arc(xy),b=arc(yz),c=arc(wz) and d=arc(wx)

Then substituting we get:

`80=1/2(a+b+c-d)`

`90=1/2(b+c+d-a)`

`94=1/2(c+d+a-b)`

`96=1/2(d+a+b-c)`

We know that a=90. (A radius drawn to a point of tangency is perpendicular to the tangent; thus since angle B is 90 and the two radii to the points of tangent are 90 then the central angle is 90 and thus the arc is 90)

Then we have the following system:

160=90+b+c-d 70=b+c-d

180=-90+b+c+d 270=b+c+d

188=90-b+c+d 98=-b+c+d

192=90+b-c+d 102=b-c+d

Adding the first to the second yields 340=2b+2c or 170=b+c

Adding the third to the fourth yields 200=2d or d=100

Adding the second to the third yields 368=2c+2d or 2c=168; c=84

Then b=170-84=86.

**Thus a=90,b=86,c=84, and d=100. These are the four arcs.**