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I need to find the missing angle and arc measures.
please answer
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I need to find the missing angle and arc measures. please answer Step by step-example-1
User Bottleboot
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1 Answer

13 votes
13 votes

9514 1404 393

Answer:

1: 40°; 2: 110°; 3: 70°; 4: 45°; 5: 65°

6: 70°; 7: 70°; 8: 65°; 9: 45°; 10: 17.5°

11: 15°; 12: 37.5°; 13: 70°; 14: 72.5°; 15: 20°

16: 87.5°; 17: 77.5°; 18: 35°; 19: 75°; 20: 20°

21: 50°; 22: 95°

Explanation:

The angles are shown in the attached. The following relationships are used to find them.

  • an inscribed angle is half the measure of the arc it subtends
  • the sum of angles in a triangle is 180°
  • the angles where chords cross are half the sum of the intercepted arcs
  • the total of arcs in a semicircle is 180°
  • an exterior angle of a triangle is equal to the sum of the remote interior angles
  • acute angles in a right triangle are complementary

_____

All of the arcs around the circle are shown except EL and HN. EL is the difference between 180° and the other marked angles in semicircle SLH. It is 90°. HN has a vertical central angle with arc SE, so is the same measure, 40°.

Angle 1 is the central angle for arc HN, so has the same measure: 40°.

Angle 2 is half the sum of arcs SL and HW, so is (130+90)/2 = 110°.

Angle 3 is vertical to the second remote interior angle in triangle BDQ where angle 2 is an exterior angle. That is ∠3 = ∠2 -∠1 = 110° -40° = 70°.

Angle 4 is half arc NL, so is (40° +30° +20°)/2 =45°.

Angle 5 is the remaining interior angle of ∆BNW, so is 180° -70° -45° = 65°.

Angle 6 is half of arc HE, so is (30° +20° +90°)/2 = 70°.

Angle 7 is half of arc SN, so is 140°/2 = 70°.

— note that triangle HNQ is an isosceles triangle with central angle 40°, so the two base angles (∠6 and ∠7) are (180° -40°)/2 = 70°.

Angle 8 is half the measure of arc LS, so is (90° +40°)/2 = 65°.

Angle 9 is half the measure of arc HW, so is (40° +50°)/2 = 45°.

Angle 10 is half the measure of arc ST, so is 35°/2 = 17.5°.

Angle 11 is half the measure of arc OT, so is 30°/2 = 15°.

Angle 12 is half the measure of arc NO, so is (50° +25°)/2 = 37.5°.

Angle 13 is half the measure of arc HE, so is 140°/2 = 70°.

Angle 14 is half the measure of arc HT, so is (180° -35°)/2 = 72.5°.

Angle 15 is half the measure of arc ES, so is 40°/2 = 20°.

Angle 16 is vertical to the remaining interior angle of the triangle containing angles 14 and 15. Its measure is 180° -72.5° -20° = 87.5°.

Angle 17 is half the sum of arcs ES and OH, so is (40° +115°)/2 = 77.5°.

Angle 18 is half the measure of arc RN, so is (30° +40°)/2 = 35°.

Angle 19 is half the sum of arcs LR and EW, so is (20° +130°)/2 = 75°.

Angle 20 is half the measure of arc HN, so is 40°/2 = 20°.

Angle 21 is complementary to angle 1, so is (90° -40°) = 50°.

Angle 22 is the remaining interior angle of triangle ACE, so is (180° -35° -50°) = 95°.

I need to find the missing angle and arc measures. please answer Step by step-example-1
User Jonas Wolff
by
2.9k points