html5-img
1 / 6

# x = (268 – 92) The measure of an angle formed by two lines that intersect outside a circle is half the differe

1 2. x = (268 – 92) The measure of an angle formed by two lines that intersect outside a circle is half the difference of the measures of the intercepted arcs (Theorem 12-11 (2)). Angle Measures and Segment Lengths. LESSON 12-4. Additional Examples.

Télécharger la présentation

## x = (268 – 92) The measure of an angle formed by two lines that intersect outside a circle is half the differe

E N D

### Presentation Transcript

1. 1 2 x = (268 – 92) The measure of an angle formed by two lines that intersect outside a circle is half the difference of the measures of the intercepted arcs (Theorem 12-11 (2)). Angle Measures and Segment Lengths LESSON 12-4 Additional Examples Find the value of the variable. a. x = 88 Simplify.

2. 1 2 94 = (x + 112) The measure of an angle formed by two lines that intersect inside a circle is half the sum of the measures of the intercepted arcs (Theorem 12-11 (1)). b. 1 2 94 = x + 56 Distributive Property 1 2 38 = xSubtract. Angle Measures and Segment Lengths LESSON 12-4 Additional Examples (continued) Quick Check 76 = xMultiply each side by 2.

3. In the diagram, the photographer stands at point T. TX and TY intercept minor arc XY and major arc XAY. Angle Measures and Segment Lengths LESSON 12-4 Additional Examples An advertising agency wants a frontal photo of a “flying saucer” ride at an amusement park. The photographer stands at the vertex of the angle formed by tangents to the “flying saucer.” What is the measure of the arc that will be in the photograph?

4. Then mXAY = 360 – x. 1 2 m T = (mXAY – mXY) The measure of an angle formed by two lines that intersect outside a circle is half the difference of the measures of the intercepted arcs (Theorem 12-11 (2)). 1 2 72 = [(360 – x) – x] Substitute. 1 2 72 = (360 – 2x) Simplify. Angle Measures and Segment Lengths LESSON 12-4 Additional Examples (continued) Let mXY = x. 72 = 180 – xDistributive Property x + 72 = 180 Solve for x. x = 108 Quick Check A 108° arc will be in the advertising agency’s photo.

5. b. Angle Measures and Segment Lengths LESSON 12-4 Additional Examples Find the value of the variable. a. 5 • x = 3 • 7 Along a line, the product of the lengths of two segments from a point to a circle is constant (Theorem 12-12 (1)). 5x = 21 Solve for x. x = 4.2 8(y + 8) = 152Along a line, the product of the lengths of two segments from a point to a circle is constant (Theorem 12-12 (3)). 8y + 64 = 225 Solve for y. 8y = 161 Quick Check y = 20.125

6. The perpendicular bisector of the chord AB contains the center of the circle. Angle Measures and Segment Lengths LESSON 12-4 Additional Examples Quick Check A tram travels from point A to point B along the arc of a circle with a radius of 125 ft. Find the shortest distance from point A to point B. Because the radius is 125 ft, the diameter is 2 • 125 = 250 ft. The length of the other segment along the diameter is 250 ft – 50 ft, or 200 ft. x • x = 50 • 200 Along a line, the product of the lengths of the two segments from a point to a circle is constant (Theorem 12-12 (1)). x2 = 10,000 Solve for x. x = 100 The shortest distance from point A to point B is 200 ft.

More Related