9.6 Secants, Tangents and Angle Measures Geometry
Objectives Use angles formed by tangents and chords to solve problems in geometry. Use angles formed by lines that intersect a circle to solve problems.
Using Tangents and Chords Measure of an angle inscribed in a circle is half the measure of its intercepted arc. m ADB = ½m AB
Theorem 9.11 If a tangent and a chord intersect at a point on a circle, then the measure of each angle formed is one half the measure of its intercepted arc. m1= ½m AB m2= ½m ABC
Finding Angle and Arc Measures Line m is tangent to the circle. Find the measure of the red angle or arc. Solution: m1= ½ AB m1= ½ (150°) m1= 75° 150°
Finding Angle and Arc Measures Line m is tangent to the circle. Find the measure of the red angle or arc. Solution: m RSP = 2(130°) m RSP = 260° 130°
Finding an Angle Measure is tangent to the circle. Find m CBD Solution: m CBD = ½ m DAB 5x = ½(9x + 20) 10x = 9x +20 x = 20 mCBD = 5(20°) = 100° (9x + 20)° 5x° D
m1 = ½ ( m CD + m AB) m2 = ½ ( m BC+ m AD)
Finding the Measure of an Angle Formed by Two Chords 106° Find the value of x Solution: x° = ½ (m QR +m PS) x° = ½ (106° + 174°) x = 140 x° 174°
Using Theorem 9.13 Find the value of x 72° = ½ (200° - x°) m GHF = ½ (m EGD - m GF ) x° 72°
Using Theorem 9.13 Find the value of x = ½ (268 - 92) = ½ (176) = 88 Because MN and MLN make a whole circle, m MLN =360°-92°=268° x° 92° Find the value of x = ½ (268 - 92) = ½ (176) = 88 m GHF = ½ (m MLN - m MN)
Practice
Practice m1 = ½ ( 40 + 52) =46 m2 = ½ ( 134) = 67
Practice 100 = ½ ( 130 + x) 200 = 130+ x X = 70 50 = ½ ( (360 – x) -x)
CD = CQD = 120 E = ½ ( AD -BC) 25 = ½ (x -30) 50 = x – 30 X = 80 AB = 360-30 – 120 – 80 = AB = 130 QDC = (180- 120) / 2 = 30
360 = 140+ 2y + y +2y 360= 140 +5y 220 = 5y Y = 44 Y = 44 2 * 44 = 88 Y = 44 2 * 44 = 88 BCD = ½( AE – BD) BCD = ½( 140-44) BCD = 48
A = FB = 50 BCA = ½ * FB = 25 ABC = 180- 50 -25 = 105 GBC =180-105 =75 360 = 4x – 50 +x + x + 25+ x – 15 + 50 360=7x +10 350 = 7x X = 50 X = 50 CFD = ½*50 = 25 FHE = ½( 35 + 50) FHE = 42.5