AoPS Chapter 2 Angles

When 2 rays share an origin, they form an angle. Vertex: O Sides: OA, OB, OC Angle AOB, < BOA Angle BOC, < COB Angle AOC, < COA

Problem 2.3 Given that < AOC = 60° and < AOB = 20°, find < BOC.

Measure of central angles in a circle What is the measure of a central angle that cuts off ¼ of a circle?

Measure of central angles in a circle What is the measure of a central angle that cuts off ¼ of a circle? What is the measure of a central angle that cuts off ⅓ of a circle?

Measure of central angles in a circle What is the measure of a central angle that cuts off ¼ of a circle? What is the measure of a central angle that cuts off ⅓ of a circle? What is the measure of a central angle that cuts off ½ of a circle?

Measure of central angles in a circle What is the measure of a central angle that cuts off ¼ of a circle? What is the measure of a central angle that cuts off ⅓ of a circle? What is the measure of a central angle that cuts off ½ of a circle? What is the measure of a central angle that cuts off ⅛ of a circle?

Right Angles An angle that measures exactly 90°. We usually mark right angles with a little box as shown. Two lines, rays, or segments that form a right angle are said to be perpendicular. JK KL J KL

Adjacent Angles We call angles that share a side adjacent angles. Ray OB is the common side and O is the common vertex.

Straight & Vertical Angles A straight angle measures 180 °.

Straight & Vertical Angles A straight angle measures 180 °. Vertical angles are formed by the intersection of 2 lines.

Straight & Vertical Angles Vertical angles are formed by the intersection of 2 lines. They are opposite angles and are always congruent. What are the measures of angles 1, 2, and 3?

Supplementary Angles

Linear Pair

Complementary Angles

Problem 2.7 If the measure of < 2 = 55°, what is the measure of < 1, <3, <4?

Parallel Lines Two lines on the same plane that never intersect are parallel. XY || m m

Corresponding Angles When lines are parallel, many special relation- ships exist. Corresponding angles are congruent.

Alternate Interior Angles When lines are parallel, many special relation- ships exist. Alternate interior angles are congruent.

Alternate Exterior Angles When lines are parallel, many special relation- ships exist. Alternate exterior angles are congruent.

Same-side Interior Angles When lines are parallel, many special relation- ships exist. Same-side interior angles are supplemen- tary.

Angles in a Triangle The sum of the measures of a triangle is 180°.

Problem 2.18 One angle in a triangle is twice another angle, and the third angle is 54°. What is the measure of the smallest angle?

Problem 2.18 One angle in a triangle is twice another angle, and the third angle is 54°. What is the measure of the smallest angle? 2x + x + 54 = 180 3x = 126 x = 42

Problem 2.19 In the diagram, m || n, AB | m, <ADC = <BCE = 3x, <CEB = 50°, and <BCD = x. Find x. m n A D B C x 3x 50°E

Problem 2.19 In the diagram, m || n, AB | m, <ADC = <BCE = 3x, <CEB = 50°, and <BCD = x. Find x. m n A D B C x 3x 50° Start with the parallel lines See that < ABE = 90° because m || n and AB | m E

Problem 2.19 In the diagram, m || n, AB | m, <ADC = <BCE = 3x, <CEB = 50°, and <BCD = x. Find x. m n A D B C x 3x 50° Start with the parallel lines See that < ABE = 90° because m || n and AB | m Then from ABE we have <EAB + <ABE + <AEB = 180°, so <EAB = 180° - <ABE - <AEB = 40°. E

Problem 2.19 In the diagram, m || n, AB | m, <ADC = <BCE = 3x, <CEB = 50°, and <BCD = x. Find x. m n A D B C x 3x 50° We can then use ACD to find <ACD = 180° - <DAC - <ADC = 180° - 40° - 3x = 140° - 3x E

Problem 2.19 In the diagram, m || n, AB | m, <ADC = <BCE = 3x, <CEB = 50°, and <BCD = x. Find x. m n A D B C x 3x 50° We can then use ACD to find <ACD = 180° - <DAC - <ADC = 180° - 40° - 3x = 140° - 3x When angle-chasing, it’s best to write the values you find for angles on your diagram as you find them, even when these values include variables. E 140° - 3x 40°

Problem 2.19 In the diagram, m || n, AB | m, <ADC = <BCE = 3x, <CEB = 50°, and <BCD = x. Find x. m n A D B C x 3x 50° We can then use ACD to find <ACD = 180° - <DAC - <ADC = 180° - 40° - 3x = 140° - 3x This diagram now suggests a way to finish the problem. We have three angles with vertices at C that together make a straight line, so E 140° - 3x 40°

Problem 2.19 In the diagram, m || n, AB | m, <ADC = <BCE = 3x, <CEB = 50°, and <BCD = x. Find x. m n A D B C x 3x 50° so we have <ACD + <DCB + <BCE = Substitution gives 140° - 3x + x + 3x = 180° so x = 40°. E 140° - 3x 40°

Problem 2.12 Given AB || CD and AD || BC and given the measures of the four angles as shown in terms of x and y, find x and y. 3y + 15° 3x - 15° AB D C y x

Problem 2.12 There’s no obvious way to make an equation for x and y so we start off by using our parallel lines and vertical angles to write the measures of all the angles we know in terms of x and y. 3y + 15° 3x - 15° AB D C y x

Problem 2.12 There’s no obvious way to make an equation for x and y so we start off by using our parallel lines and vertical angles to write the measures of all the angles we know in terms of x and y. 3y + 15° 3x - 15° AB D C y x

Problem 2.12 There’s no obvious way to make an equation for x and y so we start off by using our parallel lines and vertical angles to write the measures of all the angles we know in terms of x and y. 3y + 15° 3x - 15° AB DC y x 3y + 15° x x y y y x 3x - 15°

Problem 2.12 There’s no obvious way to make an equation for x and y so we start off by using our parallel lines and vertical angles to write the measures of all the angles we know in terms of x and y. 3y + 15° 3x - 15° AB DC y x 3y + 15° x x y y y x 3x - 15° N H EF G KL M J I          

Problem 2.12 After labeling the angles we know in terms of x and y we look for ways to build equations. We can use angles that together form straight angles at A and B: 3y + 15° 3x - 15° AB DC y x 3y + 15° x x y y y x 3x - 15° N H EF G KL M J I          

Problem 2.12 After labeling the angles we know in terms of x and y we look for ways to build equations. We can use angles that together form straight angles at A and B: <EAN + <EAB = (3y + 15) + (3x – 15) =180° 3y + 15° 3x - 15° AB DC y x 3y + 15° x x y y y x 3x - 15° N H EF G KL M J I          

Problem 2.12 After labeling the angles we know in terms of x and y we look for ways to build equations. We can use angles that together form straight angles at A and B: <EAN + <EAB = (3y + 15) + (3x – 15) =180° <FBA + <FBG +< GBH (3y + 15) + x + y =180° 3y + 15° 3x - 15° AB DC y x 3y + 15° x x y y y x 3x - 15° N H EF G KL M J I          

Problem 2.12 After labeling the angles we know in terms of x and y we look for ways to build equations. We can use angles that together form straight angles at A and B: <EAN + <EAB = (3y + 15) + (3x – 15) =180° <FBA + <FBG +< GBH (3y + 15) + x + y =180° Rearranging these gives x + y = 60° x + 4y = 165° 3y + 15° 3x - 15° AB DC y x 3y + 15° x x y y y x 3x - 15° N H EF G KL M J I          

Problem 2.12 Subtracting the 1 st from the 2 nd gives us 3y = 105°, so y = 35°. Of course we didn’t have to label every angle – we could have stopped when we had enough information to set up a pair of equations to solve for x and y. 3y + 15° 3x - 15° AB DC y x 3y + 15° x x y y y x 3x - 15° N H EF G KL M J I          

Problem 2.12 Subtracting the 1 st from the 2 nd gives us 3y = 105°, so y = 35°. Of course we didn’t have to label every angle – we could have stopped when we had enough information to set up a pair of equations to solve for x and y. Since AB || CD, we have <DAN=<KDM so 3x – 15 = x + y. 3y + 15° 3x - 15° AB DC y x 3y + 15° x x y y y x 3x - 15° N H EF G KL M J I          

Problem 2.12 Subtracting the 1 st from the 2 nd gives us 3y = 105°, so y = 35°, then x = 25°. Of course we didn’t have to label every angle - we could have stopped when we had enough information to setup a pair of equations to solve for x and y. Since AB || CD, we have <DAN=<KDM so 3x – 15 = x + y. Also <HBC & <BCI are supplementary so (3y + 15) + (3x – 15) = 180°. 3y + 15° 3x - 15° AB DC y x 3y + 15° x x y y y x 3x - 15° N H EF G KL M J I          

Problem 2.12 Solving these 2 equations gives us the same answer as before. Solving a problem with 2 different methods is an excellent way to check your answer. Since AB || CD, we have <DAN=<KDM so 3x – 15 = x + y. Also <HBC & <BCI are supplementary so (3y + 15) + (3x – 15) = 180°. 3y + 15° 3x - 15° AB DC y x 3y + 15° x x y y y x 3x - 15° N H EF G KL M J I          

Problem 2.13 Given that WV || YZ and WZ || VY, find x. Z V Y W P 140° 3x x

Problem 2.13 Given that WV || YZ and WZ || VY, find x. W e start by using what we know about parallel lines to find as many angles as we can. We find <V = 180 – 140 = 40° since WZ || VY. Z V Y W P 140° 3x x

Problem 2.13 Given that WV || YZ and WZ || VY, find x. W e start by using what we know about parallel lines to find as many angles as we can. We find <V = 180 – 140 = 40° since WZ || VY. Similarly, <Z = 40° and <ZYV = 180° - <Z = 140°. Z V Y W P 140° 3x x

Problem 2.13 Given that WV || YZ and WZ || VY, find x. W e start by using what we know about parallel lines to find as many angles as we can. We find <V = 180 – 140 = 40° since WZ || VY. Similarly, <Z = 40° and <ZYV = 180° - <Z = 140°. Since <PYV = 3x, we have <PYZ = <VYZ - <PYV = 140° - 3x. Since this angle, the 90° angle, & the angle with measure x together give us straight line PY, we have Z V Y W P 140° 3x x

Problem 2.13 < PYZ + 90° + x = 180° W e then substitute <PYZ = 140° – 3x into this equation, and we have 140° - 3x + 90° + x = 180°. We then solve this equation for x to find that x = 25°. Using information about angles to find information about other angles has 3 important tools: straight angles, vertical angles,and parallel lines. Z V Y W P 140° 3x x

Reflex Angle Angle that measures more than 180° is called a reflex angle. Suppose instead of measuring an angle in the ‘regular’ way, we go the ‘long’ way around. The ‘regular’ angle has a measure of 40°. What is the measure of the reflex angle?

Reflex Angle Angle that measures more than 180° is called a reflex angle. Suppose instead of measuring an angle in the ‘regular’ way, we go the ‘long’ way around. The ‘regular’ angle has a measure of 40°. What is the measure of the reflex angle? 360° - 40° = 320°.

Exterior Angles Find x given the angle measures shown. x is known as an exterior angle and the 35° and the 79° angles are known as its remote interior angles. A ACB 79° 35° x°x°

Exterior Angles Find x given the angle measures shown. x is known as an exterior angle and the 35° and the 79° angles are known as its remote interior angles. Important: Any exterior angle of a triangle is equal to the sum of its remote interior angles. A ACB 79° 35° x°x°

Exterior Angles Find x given the angle measures shown. So now it is easy to find x: x = 79° + 35° x = 114° Important: Any exterior angle of a triangle is equal to the sum of its remote interior angles. A ACB 79° 35° x°x°

Problem 2.22 In the diagram AB || DE, <BAC = 2x - 20°, <ACB = 30°, and <DEF = x + 55°. Find <CED.  A B C E D F 30° x + 55° 2x - 20°  G

Problem 2.22 In the diagram AB || DE, <BAC = 2x - 20°, <ACB = 30°, and <DEF = x + 55°. Find <CED. We know we’ll probably need to find x to answer the problem. We could label every angle we know in terms of x, but 1 st we take a minute to look for a faster way to get x.  A B C E D F 30° x + 55° 2x - 20° G 

Problem 2.22 In the diagram AB || DE, <BAC = 2x - 20°, <ACB = 30°, and <DEF = x + 55°. Find <CED. We have <BAC and <ACB of ΔABC so we know the exterior angle <GBC = <BAC + <ACB = 2x  A B C E D F 30° x + 55° 2x - 20° G  2x + 10°

Problem 2.22 In the diagram AB || DE, <BAC = 2x - 20°, <ACB = 30°, and <DEF = x + 55°. Find <CED. From AB || DE, we know that <GBC = <DEF = = x + 55°. Hence we know that 2x + 10 = x + 55, so x = 45°.  A B C E D F 30° x + 55° 2x - 20° G  2x + 10°

Problem 2.22 In the diagram AB || DE, <BAC = 2x - 20°, <ACB = 30°, and <DEF = x + 55°. Find <CED. Our desired angle is the supplement of <DEF, so our answer is <CED = 180° - <DEF = 180° - (x + 55) = 80°.  A B C E D F 30° x + 55° 2x - 20° G  2x + 10°

Problem 2.22 There are many other ways we could have approached this problem. This is almost always true when we have problems involving exterior angles. <CED = 80°.  A B C E D F 30° x + 55° 2x - 20° G  2x + 10°

Review Problem 2.36 Let Δ ABC have (interior) angles in the ratio of 3:4:5. What is the measure of its smallest exterior angle?

Review Problems Let Δ ABC have (interior) angles in the ratio of 3:4:5. What is the measure of its smallest exterior angle? Suppose that ΔABC is labeled in such a way that <A ≤ <B ≤ <C. Then since the angles are in the ratio of 3:4:5, there exists some number x such that <A = 3x, <B = 4x, and <C = 5x. Using the fact that the 3 angles of a triangle add up to 180°, we obtain 12x = <A + <B + <C = 180°, so x = 15°.

Review Problems Let Δ ABC have (interior) angles in the ratio of 3:4:5. What is the measure of its smallest exterior angle? x = 15° Since the exterior angles of the triangle are equal to the sums of their remote interior angles, the smallest exterior angle has measure equal to the least of 3x + 4x = 7x, 3x + 5x = 8x, and 4x + 5x = 9x. Since the smallest of these is 7x, the measure of the smallest exterior angle of the triangle is 7x = 105°.

Review Problem 2.40 Three straight lines intersect at O and <COD = <DOE in the diagram. The ratio of <COB to <BOF is 7 : 2. What is the number of degrees in <COD? (Source: MATHCOUNTS) B A F O C D E

Review Problem 2.40 Three straight lines intersect at O and <COD = <DOE in the diagram. The ratio of <COB to <BOF is 7 : 2. What is the number of degrees in <COD? (Source: MATHCOUNTS) B A F O C D E Since the ratio of <COB = <BOF is 7 : 2, there is some x such that <COB = 7x and <BOF = 2x. Since <COF is a straight angle, 9x = <COB + <BOF = <COF = 180 so x = 20°.

Review Problem 2.40 Three straight lines intersect at O and <COD = <DOE in the diagram. The ratio of <COB to <BOF is 7 : 2. What is the number of degrees in <COD? (Source: MATHCOUNTS) B A F O C D E Since the ratio of <COB = <BOF is 7 : 2, there is some x such that <COB = 7x and <BOF = 2x. Since <COF is a straight angle, 9x = <COB + <BOF = <COF = 180 so x = 20°. Since <BOF & <COE are vertical angles, <COE = <BOF = 2x = 40° Finally, since <COD = <DOE, we have <COD = ½ <COE = 20°

Review Problem 2.43 The measures of the angles in the diagram are marked. Find <C. AC || BE. C A B E D 4x4x3x3x 2x2x

Review Problem 2.43 The measures of the angles in the diagram are marked. Find <C. AC || BE. C A B E D 4x4x3x3x 2x2x Since AD cuts the parallel segments AC and EB, we have 3x = <CAB = <ABE. Angle ABD is a straight line, so 5x = <ABE + <EBD = 180°, giving x = 36°.

Review Problem 2.43 The measures of the angles in the diagram are marked. Find <C. AC || BE. C A B E D 4x4x3x3x 2x2x Since AD cuts the parallel segments AC and EB, we have 3x = <CAB = <ABE. Angle ABD is a straight line, so 5x = <ABE + <EBD = 180°, giving x = 36°. Also, <CBD is an exterior angle of ΔABC, so 4x = <CBD = <C + <CAB = <C + 3x. Therefore, <C = x = 36°.

Review Problem 2.44 Three angles of a triangle have measures <A = x – 2y, <B = 3x + 5y, <C = 5x – 3y. Find x.

Review Problem 2.44 Three angles of a triangle have measures <A = x – 2y, <B = 3x + 5y, <C = 5x – 3y. Find x. Use the fact that the sum of the measures of the angles of a triangle is 180° to determine 180° = <A + <B + <C = (x – 2y) + (3x + 5y) + (5x – 3y) = 9x. Solving for x, we find that x = 20°.

Challenge 2.48 What is the number of degrees of the angle formed by the minute and hour hands of a clock at 11:10 PM? (Source: MATHCOUNTS)

Challenge 2.48 What is the number of degrees of the angle formed by the minute and hour hands of a clock at 11:10 PM? (Source: MATHCOUNTS) 1 st, find the position of the minute hand. The minute hand makes a 0° angle with 12 o’clock on the hour, and it travels a total of 360° in an hour, so in the 10 minutes after 11 o’clock, it moves (10/60) * 360° = 60° clockwise from the 12.

Challenge 2.48 What is the number of degrees of the angle formed by the minute and hour hands of a clock at 11:10 PM? (Source: MATHCOUNTS) Next, we consider the hour hand. It travels all the way around the clock in 12 hours, so it travels at a rate of (1/12) * 360° = 30° per hour. Since it will be pointing at the 12 at 12 o’clock, and it will travel (5/6 hour)(30° per hour) = 25° in the time between 11:10 PM and 12:00 midnight, it currently rests 25° counterclockwise from the 12. Therefore since the angle made by the 2 hands is the sum of the 2 angles made between the hands & the ray pointing from the center of the clock towards the 12, the angle between the hands is 60° + 25° = 85°

These are just a few examples of what can be found in Chapter 2, Introduction to Geometry. If you are interested in the series of books, go to