12.4 Other Angle Relationships in Circles Geometry

Objectives/DFA/HW Objectives: DFA HW You will use angles formed by tangents and chords to solve problems in geometry. You will find the length of segments associated with circles. DFA pp.794 #12 HW Assignment: pp. 794-796 (2-34 even, 38)

Using Tangents and Chords You know that measure of an angle inscribed in a circle is half the measure of its intercepted arc. This is true even if one side of the angle is tangent to the circle. mADB = ½m

Theorem 10.12 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. m1= ½m m2= ½m

Ex. 1: Finding Angle and Arc Measures Line m is tangent to the circle. Find the measure of the red angle or arc. Solution: m1= ½ m1= ½ (150°) m1= 75° 150°

Ex. 1: Finding Angle and Arc Measures Line m is tangent to the circle. Find the measure of the red angle or arc. Solution: m = 2(130°) m = 260° 130°

Ex. 2: Finding an Angle Measure In the diagram below, is tangent to the circle. Find mCBD Solution: mCBD = ½ m 5x = ½(9x + 20) 10x = 9x +20 x = 20  mCBD = 5(20°) = 100° (9x + 20)° 5x° D

Lines Intersecting Inside or Outside a Circle If two lines intersect a circle, there are three (3) places where the lines can intersect. on the circle

Inside the circle

Outside the circle

Lines Intersecting You know how to find angle and arc measures when lines intersect ON THE CIRCLE. You can use the following theorems to find the measures when the lines intersect INSIDE or OUTSIDE the circle.

Theorem 10.13 If two chords intersect in the interior of a circle, then the measure of each angle is one half the sum of the measures of the arcs intercepted by the angle and its vertical angle. m1 = ½ m + m m2 = ½ m + m

Theorem 10.14 If a tangent and a secant, two tangents or two secants intercept in the EXTERIOR of a circle, then the measure of the angle formed is one half the difference of the measures of the intercepted arcs. m1 = ½ m( - m )

Theorem 10.14 If a tangent and a secant, two tangents or two secants intercept in the EXTERIOR of a circle, then the measure of the angle formed is one half the difference of the measures of the intercepted arcs. m2 = ½ m( - m )

Theorem 10.14 If a tangent and a secant, two tangents or two secants intercept in the EXTERIOR of a circle, then the measure of the angle formed is one half the difference of the measures of the intercepted arcs. 3 m3 = ½ m( - m )

Ex. 3: Finding the Measure of an Angle Formed by Two Chords 106° Find the value of x Solution: x° = ½ (m +m x° = ½ (106° + 174°) x = 140 x° 174° Apply Theorem 10.13 Substitute values Simplify

Ex. 4: Using Theorem 10.14 Find the value of x Solution: 200° Find the value of x Solution: 72° = ½ (200° - x°) 144 = 200 - x° - 56 = -x 56 = x x° mGHF = ½ m( - m ) 72° Apply Theorem 10.14 Substitute values. Multiply each side by 2. Subtract 200 from both sides. Divide by -1 to eliminate negatives.

Ex. 4: Using Theorem 10.14 Find the value of x Solution: Because and make a whole circle, m =360°-92°=268° x° 92° Find the value of x Solution: = ½ (268 - 92) = ½ (176) = 88 mGHF = ½ m( - m ) Apply Theorem 10.14 Substitute values. Subtract Multiply

Ex. 5: Describing the View from Mount Rainier You are on top of Mount Rainier on a clear day. You are about 2.73 miles above sea level. Find the measure of the arc that represents the part of Earth you can see.

Ex. 5: Describing the View from Mount Rainier You are on top of Mount Rainier on a clear day. You are about 2.73 miles above sea level. Find the measure of the arc that represents the part of Earth you can see.

Ex. 5: Describing the View from Mount Rainier and are tangent to the Earth. You can solve right ∆BCA to see that mCBA  87.9°. So, mCBD  175.8°. Let m = x° using Trig Ratios

From the peak, you can see an arc about 4°. 175.8  ½[(360 – x) – x] 175.8  ½(360 – 2x) 175.8  180 – x x  4.2 Apply Theorem 10.14. Simplify. Distributive Property. Solve for x. From the peak, you can see an arc about 4°.

Finding the Lengths of Chords When two chords intersect in the interior of a circle, each chord is divided into two segments which are called segments of a chord. The following theorem gives a relationship between the lengths of the four segments that are formed.

Theorem 12.15 If two chords intersect in the interior of a circle, then the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord. EA • EB = EC • ED

Proving Theorem 12.15 You can use similar triangles to prove Theorem 12.15. Given: , are chords that intersect at E. Prove: EA • EB = EC • ED

Proving Theorem 12.15 Paragraph proof: Draw and . Because C and B intercept the same arc, C  B. Likewise, A  D. By the AA Similarity Postulate, ∆AEC  ∆DEB. So the lengths of corresponding sides are proportional. = Lengths of sides are proportional. EA • EB = EC • ED Cross Product Property

Ex. 1: Finding Segment Lengths Chords ST and PQ intersect inside the circle. Find the value of x. RQ • RP = RS • RT Use Theorem 12.15 Substitute values. 9 • x = 3 • 6 9x = 18 x = 2 Simplify. Divide each side by 9.

Using Segments of Tangents and Secants In the figure shown, PS is called a tangent segment because it is tangent to the circle at an end point. Similarly, PR is a secant segment and PQ is the external segment of PR.

Theorem 12.16 If two secant segments share the same endpoint outside a circle, then the product of the length of one secant segment and the length of its external segment equals the product of the length of the other secant segment and the length of its external segment. EA • EB = EC • ED

Theorem 12.17 If a secant segment and a tangent segment share an endpoint outside a circle, then the product of the length of the secant segment and the length of its external segment equal the square of the length of the tangent segment. (EA)2 = EC • ED

Ex. 2: Finding Segment Lengths Find the value of x. RP • RQ = RS • RT Use Theorem 12.16 9•(11 + 9)=10•(x + 10) Substitute values. Simplify. 180 = 10x + 100 Subtract 100 from each side. 80 = 10x Divide each side by 10. 8 = x

Ex. 3: Estimating the radius of a circle Aquarium Tank. You are standing at point C, about 8 feet from a circular aquarium tank. The distance from you to a point of tangency is about 20 feet. Estimate the radius of the tank.

Solution (CB)2 = CE • CD Use Theorem 12.17 (20)2  8 • (2r + 8) Substitute values. Simplify. 400  16r + 64 Subtract 64 from each side. 336  16r Divide each side by 16. 21  r So, the radius of the tank is about 21 feet.

Ex. 4: Finding Segment Lengths (BA)2 = BC • BD Use Theorem 12.17 (5)2 = x • (x + 4) Substitute values. Simplify. 25 = x2 + 4x Write in standard form. 0 = x2 + 4x - 25 Use Quadratic Formula. x = x = Simplify. Use the positive solution because lengths cannot be negative. So, x = -2 +  3.39.