Asked by lilman
--------------- QUESTION FOLLOWS --------------------------
Can someone Please help I definitely do not know how to do this.
A)Find x, to find x you need to add the arc measures together and set the expression equal to the total degrees of circle then solve for X, must show all steps.
B)Is the triangle equilateral, isosceles, or scalene, explain, plug your x-value from A into each of the arc measures to find what each arc measures.
Apply the Inscribed Angles Theorem to find <P, <R, <S.
Use the angle measure from step 2 above to determine what type of triangle you have.
The picture is a circle with a triangle inside, QP=(8x-10) and PR=(6x), and QR=(10x+10) QP and PR are the sides and QR is the bottom.
Sure hope someone can help.
Can someone Please help I definitely do not know how to do this.
A)Find x, to find x you need to add the arc measures together and set the expression equal to the total degrees of circle then solve for X, must show all steps.
B)Is the triangle equilateral, isosceles, or scalene, explain, plug your x-value from A into each of the arc measures to find what each arc measures.
Apply the Inscribed Angles Theorem to find <P, <R, <S.
Use the angle measure from step 2 above to determine what type of triangle you have.
The picture is a circle with a triangle inside, QP=(8x-10) and PR=(6x), and QR=(10x+10) QP and PR are the sides and QR is the bottom.
Sure hope someone can help.
Answers
Answered by
Reiny
According to the wording of part a) your description of
QP= 8x-10, PR = 6x etc
must have said
arc QP = 8x-10 etc
so 8x-10 + 6x + 10x+10 = 360
24x = 360
x = 15
so arc QP = 8(15) - 10 = 110°
PR = 90°
QR = 160°
By the inscribed angle theorem, the angle opposite the arc must be half the central angle subtended by the arc. Then....
angle R = 55°
angle Q =45°
angle P = 80°
looks like plain old <b>scalene</b> triangle
QP= 8x-10, PR = 6x etc
must have said
arc QP = 8x-10 etc
so 8x-10 + 6x + 10x+10 = 360
24x = 360
x = 15
so arc QP = 8(15) - 10 = 110°
PR = 90°
QR = 160°
By the inscribed angle theorem, the angle opposite the arc must be half the central angle subtended by the arc. Then....
angle R = 55°
angle Q =45°
angle P = 80°
looks like plain old <b>scalene</b> triangle
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