Skipton Girls’ High School

Skipton Girls’ High School
KS3/KS4: Angles Skipton Girls’ High School Objectives: Understand notation for angles. Know basic rules of angles (angles in triangle, on straight line). Recognise alternate, corresponding and vertically opposite angles. Find angles in isosceles triangles. Deal with and introduce algebraic angles. Construct diagrams from written information and form angle proofs.

For Teacher Use: Recommended lesson structure:
Lesson 1: Angle notation. Angle basics. Lesson 2: Z/F/C/v. opposite angles. Angle properties of quadrilaterals. Lesson 3: Isosceles Triangles Lesson 4: Algebraic Angles Lesson 5/6: Constructing diagrams from information. Proof Lesson 7: Consolidation/Mini-assessment

Angle Notation ? ? ? ? ? ? 𝐴 𝐵 𝐶 𝑄 ∠𝑆𝑄𝑅 or ∠𝐶𝑄𝑅 𝑃 ∠𝑆𝐶𝑅 𝐶 ∠𝑆𝑃𝑄 𝑅 𝑆
We use capital letters for points. ! We can refer to this angle using: 𝐴 𝐵 𝐶 (with a ‘hat’ on the middle letter) ∠𝐴𝐵𝐶 Or in words: “Angle 𝐴𝐵𝐶” ? 𝐵 ? 𝐶 ? Quick Starter: How would we refer to each of the following angles? (Use ∠𝐴𝐵𝐶 notation) 𝑄 1 ∠𝑆𝑄𝑅 or ∠𝐶𝑄𝑅 1 ? 𝑃 ∠𝑆𝐶𝑅 2 ? 𝐶 3 2 ∠𝑆𝑃𝑄 3 ? 𝑅 𝑆

Angle Basics 30° 128° 𝟓𝟐° ? ! “Angles on a straight line sum to 180°.” (These wordings will be important for angle proofs later) ? 𝟐𝟒𝟎° ? ! “Angles around a point sum to 360°.” ? 85° 𝟒𝟓° ? reflex 50° acute obtuse ! “Angles in a triangle sum to 180°.” ? Name of angle if: Less than 90: Acute Between 90 and 180: Obtuse Over 180: Reflex angle ? ? ?

Test Your Understanding
[JMC 2008 Q4] In this diagram, what is the value of 𝑥? A B 36 C 64 D E 144 Solution: C ? [JMC 2003 Q12] What is the size of the angle marked 𝑥? A 42° B 67° C 69° D 71° E 111° Solution: A ii ?

Exercise 1 ? ? ? ? ? 𝒂=𝟔𝟒°, 𝒃=𝟕𝟎°, 𝒄=𝟔𝟎°, 𝒅=𝟏𝟏𝟐°, 𝒆=𝟑𝟓°
(on provided worksheet) 1 Find the angles marked with letters. 71° 𝑏 150° 45° 𝑎 𝑏 𝑐 30° 𝑏 105° 𝑑 𝑒 52° 60° 110° ? ? ? ? ? 𝒂=𝟔𝟒°, 𝒃=𝟕𝟎°, 𝒄=𝟔𝟎°, 𝒅=𝟏𝟏𝟐°, 𝒆=𝟑𝟓°

Exercise 1 ? ? ? ? ? ? (on provided worksheet) 2 3 4 5 6 7
[JMC 2000 Q3] What is the value of 𝑥? Solution: 28 [JMC 2015 Q6] What is the value of 𝑥 in this triangle? Solution: 50 [JMC 2011 Q6] What is the sum of the marked angles in the diagram? Solution: 𝟑𝟔𝟎° [JMC 1997 Q17] How big is angle 𝑥? Solution: 𝟏𝟎𝟒° [IMC 2014 Q3] An equilateral triangle is placed inside a larger equilateral triangle so that the diagram has three lines of symmetry. What is the value of 𝑥? Solution: 𝟏𝟓𝟎° [IMC 2011 Q9] In the diagram, 𝑋𝑌 is a straight line. What is the value of 𝑥? Solution: 170 2 ? 3 ? 4 ? 5 ? 6 ? 7 ?

Exercise 1 ? ? ? ? ? (on provided worksheet) 8 9 10 11 12
[JMC 2010 Q8] In a triangle with angles 𝑥°, 𝑦°, 𝑧° the mean of 𝑦 and 𝑧 is 𝑥. What is the value of 𝑥? Solution: 60 [JMC 1999 Q13] In the diagram, ∠𝑅𝑃𝑀=20° and ∠𝑄𝑀𝑃=70°. What is ∠𝑃𝑅𝑆? Solution: 𝟏𝟑𝟎° [JMC 2001 Q18] Triangle 𝑃𝑄𝑅 is equilateral. Angle 𝑆𝑃𝑅=40°, angle 𝑇𝑄𝑅=35°. What is the size of the marked angle 𝑆𝑋𝑇? Solution: 𝟏𝟑𝟓° [JMC 2007 Q16] What is the sum of the six marked angles? Solution: 𝟏𝟒𝟒𝟎°. The sum of the angles at the 7 points is 𝟓×𝟑𝟔𝟎=𝟏𝟖𝟎𝟎. We’ve excluded the angles in the 3 triangles, so 𝟏𝟖𝟎𝟎−𝟑×𝟏𝟖𝟎=𝟏𝟒𝟒𝟎. [JMC 2015 Q16] The diagram shows a square inside an equilateral triangle. What is the value of 𝑥+𝑦? Solution: 150 ? 9 ? 10 ? 11 ? 12 ?

Exercise 1 13 [JMO 2001 A3] What is the value of 𝑥 in the diagram alongside? Solution: 15 [Kangaroo Pink 2010 Q9] In the diagram, angle 𝑃𝑄𝑅 is 20°, and the reflex angle at 𝑃 is 330°.The line segments 𝑄𝑇 and 𝑆𝑈 are perpendicular. What is the size of angle 𝑅𝑆𝑃? Solution: 𝟒𝟎° [Kangaroo Pink 2006 Q8] The circle shown in the diagram is divided into four arcs of length 2, 5, 6 and 𝑥 units. The sector with arc length 2 has an angle of 30° at the centre. Determine the value of 𝑥. (Hint: An arc is a part of the line that makes up the circle, i.e. part of the circumference. The arc length grows in proportion to the angle at the centre) Solution: 11 [SMC 2010 Q3] The diagram shows an equilateral triangle touching two straight lines. What is the sum of the four marked angles? Solution: 𝟐𝟒𝟎°. Angles at two points some to 𝟑𝟔𝟎°, but we exclude two angles of 𝟔𝟎°. ? 14 ? 15 ? 16 ?

Angles involving parallel lines
There are three more laws of angles that you need to know, use and quote: (These arrows indicate the lines are parallel) Alternate angles are equal. (Sometimes known as ‘Z’ angles) ? Corresponding angles are equal. (Sometimes known as ‘F’ angles) ? ? Vertically opposite angles are equal. Bro Note: The word ‘vertically’ here is the adjective form of ‘vertex’, which means a point. So “vertically opposite” means “opposite with respect to a point”.

How to spot them To identify alternate angles:
Step 1: Identify a line connecting two parallel lines. Click >> Step 2: At each end on your parallel lines, shoot out along the parallel lines in opposite directions. Your angles are the ones wedged between the lines. Click >> To identify corresponding angles: Step 1: Identify an angle between one line of a parallel pair, and a connecting line. Click >> Step 2: This can be shifted along to the other parallel line of the pair. Click >>

Examples (For extra practise outside of class) Identify the alternate and corresponding angle for each indicated angle (use ∠ notation). 𝑃 𝑊 𝑋 𝐴 𝑄 𝐷 𝐶 𝐵 Double arrows allows us to match another pair of parallel lines. 4 𝐹 𝑆 𝐾 𝐸 2 3 𝐽 𝐺 𝐻 𝐼 5 1 𝑂 𝑁 # Angle Alternate Corresponding 1 ∠𝐽𝑁𝑂 ∠𝐶𝐷𝐽 ∠𝑃𝐷𝑄 or ∠𝐺𝐿𝑀 2 ∠𝑆𝐹𝐻 ∠𝐹𝑀𝑁 ∠𝐿𝑀𝑅 3 ∠DJC ∠BAC ∠𝑊𝐴𝑋 4 ∠𝐼𝐶𝐷 ∠𝑇𝑉𝐼 ∠𝑈𝐿𝑀 5 ∠𝐻𝐺𝐿 ∠𝐷𝐽𝐼 ∠𝑁𝐽𝐾 𝑀 ? ? 𝐿 ? ? ? 𝑇 ? ? ? 𝑅 ? ? ? 𝑈 𝑉 ? ? ?

Check Your Understanding
ANB is parallel to CMD. LNM is a straight line. Angle LMD =68° (i) Work out the size of the angle marked 𝑦. 𝟏𝟏𝟐° (ii) Give reasons for your answer. “Alternate angles are equal” OR “Corresponding angles are equal”, “Angles on straight line sum to 180” i ? ? ii [JMC 2011 Q11] The diagram shows an equilateral triangle inside a rectangle. What is the value of 𝑥+𝑦? A 30 B 45 C 60 D 75 E Solution: C 𝑦 𝑥 ? Hint: Is there a line in the diagram we could add so we can then use alternate angles?

Cointerior angles We’ve seen ‘Z’ and ‘F’ angles. There’s also ‘C’ angles! ! Cointerior angles sum to 𝟏𝟖𝟎° ? 180−𝑥 ? We can identify them in a similar way to alternate angles: identify a line connecting two parallel lines, but this time go in the same direction at each end rather than opposite directions. 𝑥 Application to parallelograms So we can say: 110° ? ? 70° ! Cointerior angles in a parallelogram add to 𝟏𝟖𝟎° ! Opposite angles in a parallelogram are equal. ? ? 70°

Example Can you come up with an alternative explanation for why 𝑦=112°? ∠𝑴𝑵𝑩=𝟏𝟏𝟐° (cointerior angles sum to 180) 𝒚=𝟏𝟏𝟐° (vertically opposite angles are equal) ?

One further property of quadrilaterals
What are the sum of the angles in a quadrilateral? Draw Hint >> ! Angles in a quadrilateral sum to 𝟑𝟔𝟎° ? 50° 𝑥 𝑥=𝟏𝟑𝟎° ? 45° 35°

Exercise 2 ? ? ? ? For each diagram Find the missing angles and
List reasons for each answer. 1 a 𝟓𝟓° Corresponding angles are equal. ? ? b 𝒙=𝟏𝟓𝟎°, 𝒚=𝟗𝟓° For x: Corresponding angles are equal. For y: Angles on straight line add to 180. Alternate angles are equal. ? ?

Exercise 2 ? ? ? ? 2 a b 𝑦 𝑥 115° 122° 80° 𝒙=𝟓𝟖° 𝒚=𝟑𝟓° c d 30° 𝑏 57° 𝑥
𝑐 𝑎 82° 41° 𝒙=𝟕𝟒° 𝒂=𝟑𝟎°, 𝒃=𝟖𝟐°, 𝒄=𝟔𝟖° ? ?

Exercise 2 e f 85° 𝑥 110° 112° 75° 𝑦 𝒙=𝟔𝟖° ? 𝒙=𝟗𝟎° ?

Exercise 2 3 For rhombuses are evenly spaced around a point. Given the angle shown, find 𝑥. Solution: 𝟏𝟐𝟎° ? 30° 𝑥

Exercise 2 ? ? ? ? ? 𝒂=𝟒𝟏° 𝒃=𝟗𝟔° 𝒄=𝟏𝟑𝟔° 𝒅=𝟖𝟒° 𝒆=𝟗𝟓° 4
Find the angles indicated. 100° 41° 𝑐 𝑑 𝑏 𝑒 96° 𝒂=𝟒𝟏° 𝒃=𝟗𝟔° 𝒄=𝟏𝟑𝟔° 𝒅=𝟖𝟒° 𝒆=𝟗𝟓° ? ? ? ? 𝑎 136° ?

Exercise 2 ? ? 5 42° Find the angle 𝛼 Solution: 𝟕𝟏° 𝛼 29° 𝑃 6 𝐷 𝐶
The line 𝑃𝐵 bisects the angle 𝐴𝐵𝐶 (this means it cuts it exactly in half). Determine the angle 𝐵𝑃𝐶. Solution: 𝟓𝟓° ? 70° 𝐴 𝐵

Exercise 2 ? ? ? ? [JMC 2006 Q7] What is the value of 𝑥? 7
Solution: 𝟕𝟓° [JMC 2007 Q9] In the diagram on the right, 𝑆𝑇 is parallel to 𝑈𝑉. What is the value of 𝑥? Solution: 86 [JMC 2009 Q19] The diagram on the right shows a rhombus 𝐹𝐺𝐻𝐼 and an isosceles triangle 𝐹𝐺𝐽 in which 𝐺𝐹=𝐺𝐽. Angle 𝐹𝐽𝐼=111°. What is the size of angle 𝐽𝐹𝐼? Solution: 𝟐𝟕° [IMC 2005 Q7] In the diagram, what is the sum of the marked angles? Solution: 𝟑𝟔𝟎°. Sum of angles in 4 triangles is 𝟏𝟖𝟎°×𝟒=𝟕𝟐𝟎°. The four angles in the quadrilateral sum to 360, thus because of vertically opposite angles, the innermost angles of the triangles sum to 360. Thus sum of angles is 𝟕𝟐𝟎−𝟑𝟔𝟎=𝟑𝟔𝟎°. ? 8 ? 9 ? 10 ?

Exercise 2 ? ? ? [JMO 2003 A9] Find the value of 𝑎+𝑏+𝑐. Solution: 210
[SMC 2006 Q3] The diagram shows overlapping squares. What is the value of 𝑥+𝑦? Solution: 270 [Cayley 2012 Q2] In the diagram, 𝑃𝑄 and 𝑇𝑆 are parallel. Prove that 𝑎+𝑏+𝑐=360. Suppose we add a line horizontal with 𝑹 as shown, where 𝒃 is split into 𝒃 𝟏 and 𝒃 𝟐 . 𝒂 and 𝒃 𝟏 are cointerior thus 𝒂+ 𝒃 𝟏 =𝟏𝟖𝟎°. 𝒃 𝟐 and 𝒄 are cointerior thus 𝒃 𝟐 +𝒄=𝟏𝟖𝟎° Thus 𝒂+𝒃+𝒄=𝒂+ 𝒃 𝟏 + 𝒃 𝟐 +𝒄 =𝟏𝟖𝟎°+𝟏𝟖𝟎°=𝟑𝟔𝟎° 11 ? 12 ? 𝒚 13 ? 𝒃 𝟏 𝒃 𝟐

Exercise 2 [Kangaroo Grey 2005 Q20] Five straight lines intersect at a common point and five triangles are constructed as shown. What is the total of the 10 angles marked in the diagram? A 300° B 450° C 360° D 600° E 720° Answer: 𝟕𝟐𝟎°. Sum of angles in 5 triangles is 𝟏𝟖𝟎°×𝟓=𝟗𝟎𝟎°. Since angles at the central point add to 𝟑𝟔𝟎°, but each angle in a triangle at the centre can be paired with a vertically opposite angle not in a triangle, the central angles in the triangles sum to 𝟏𝟖𝟎°. 𝟗𝟎𝟎°−𝟏𝟖𝟎°=𝟕𝟐𝟎° 14 ?

Isosceles Triangles ? ? ? ? 𝟒𝟎° 𝟔𝟓° 50° 70° 𝟕𝟎°
What do these marks mean? The lines are of the same length. ? 𝟒𝟎° ? 𝟔𝟓° ? 50° ? 70° 𝟕𝟎° ! “Base angles of an isosceles triangle are equal.”

Click for animation >
Warning! Sometimes diagrams are drawn in such a way that it’s not visually obvious what the two angles the same are. You can use the ‘finger slide method’ to identify these. Diagram not drawn accurately. Click for animation > Put your fingers on the two marks. Slide your fingers in the same direction but away from each other. These two corners are where the angles are the same.

[JMC 2013 Q3] What is the value of 𝑥? A 25 B 35 C 40 D 65 E Solution: C [IMC 2012 Q11] In the diagram, 𝑃𝑄𝑅𝑆 is a parallelogram; ∠𝑄𝑅𝑆=50°; ∠𝑆𝑃𝑇=62° and 𝑃𝑄=𝑃𝑇. What is the size of ∠𝑇𝑄𝑅? A 84° B 90° C 96° D 112° E 124° Solution: C ? ii ?

Using angles to give you information about sides
You’ve so far used sides which are equal to find angles. But we can do the opposite too! If two angles are equal, then two sides are equal. [Kangaroo Pink 2004 Q6] In the diagram 𝑄𝑅=𝑃𝑆. What is the size of ∠𝑃𝑆𝑅? 75° 65° Use information provided. Work out some initial angles. Use angles to give us information about side lengths. Use new knowledge of side lengths to work out more angles… Go > Go > Go > Go >

Angle Wall Challenge How far can you get down the angle challenge wall? (do in order, and draw the diagram first) Hint: You may want to add extra lines. 𝟔𝟎° 60° ∠𝐵𝐴𝐶=60° ∠𝐷𝐴𝐶=30° ∠𝐴𝐷𝐵=45° (Δ𝐴𝐷𝐶 is isosceles) ∠𝐴𝐷𝐶=75° (Δ𝐴𝐷𝐶 is isosceles) ∠𝐵𝐷𝐶=30° (75°−45°) ∠𝐷𝐵𝐶=15° (angles within Δ𝐷𝐶𝐵) ?  𝟏𝟓° 𝟑𝟎° ? 𝟏 ?  ? 𝟔𝟎° ? 𝟕𝟓° 𝟕𝟓° 𝟒𝟓° ?  𝟑𝟎° This was a Junior Maths Olympiad problem! (to prove that ∠𝐵𝐷𝐶=2×∠𝐷𝐵𝐶)

Exercise 3 ? ? ? ? ? 1 2 3 4 5 Find the value of each variable.
Solution: 𝒙=𝟒𝟔°, 𝒚=𝟕𝟒°, 𝒛=𝟐𝟎° [JMC 2014 Q10] An equilateral triangle is surrounded by three squares, as shown. What is the value of 𝑥? Solution: 30 [JMC 2005 Q13] The diagram shows two equal squares. What is the value of 𝑥? Solution: 140 [IMC 2010 Q6] In triangle 𝑃𝑄𝑅, 𝑆 is a point on 𝑄𝑅 such that 𝑄𝑆=𝑆𝑃=𝑃𝑅 and ∠𝑄𝑃𝑆=20°. What is the size of ∠𝑃𝑅𝑆? Solution: 𝟒𝟎° [IMC 1999 Q8] In the diagram 𝑃𝑄=𝑃𝑅=𝑅𝑆. What is the size of angle 𝑥? Solution: 𝟏𝟎𝟖° 1 ? 2 ? 3 ? 4 ? 5 ?

Exercise 3 ? ? ? ? ? (on provided worksheet) 6 7 8 9 10
[JMC 2008 Q19] In the diagram on the right, 𝑃𝑇=𝑄𝑇=𝑇𝑆, 𝑄𝑆=𝑆𝑅, ∠𝑃𝑄𝑇=20°. What is the value of 𝑥? Solution: 35 [Kangaroo Grey 2010 Q9] The diagram shows a quadrilateral 𝐴𝐵𝐶𝐷, in which 𝐴𝐷=𝐵𝐶, ∠𝐶𝐴𝐷=50°, ∠𝐴𝐶𝐷=65° and ∠𝐴𝐶𝐵=70°. What is the size of ∠𝐴𝐵𝐶? Solution: 𝟓𝟓° [Cayley 2004 Q1] The “star” octagon shown in the diagram is beautifully symmetrical and the centre of the star is at the centre of the circle. If angle 𝑁𝐴𝐸 = 110° (as indicated), how big is the angle 𝐷𝑁𝐴? (i.e. 𝑥) Solution: 𝟐𝟎° [IMC 2003 Q8] Lines 𝐴𝐵 and 𝐶𝐷 are parallel and 𝐵𝐶=𝐵𝐷. Given that 𝑥 is an acute angle not equal to 60°, how many other angles in this diagram are equal to 𝑥? Solution: 4 [IMC 1997 Q11] In the quadrilateral 𝐴𝐵𝐶𝐷, ∠𝐴𝐵𝐶=90°, ∠𝐵𝐴𝐷=70°, and 𝐴𝐵=𝐵𝐷=𝐵𝐶. What is the size of ∠𝐵𝐷𝐶? Solution: 𝟔𝟓° ? 7 ? 𝑥 110° 𝑁 𝐴 𝐸 𝐷 𝐵 𝐶 𝑆 𝑊 8 ? 9 ? 10 ?

Exercise 3 ? ? ? (on provided worksheet)
[Kangaroo Pink 2013 Q9] The diagram shows an equilateral triangle 𝑅𝑆𝑇 and also the triangle 𝑇𝑈𝑉 obtained by rotating triangle 𝑅𝑆𝑇 about the point 𝑇. Angle 𝑅𝑇𝑉=70°. What is angle 𝑅𝑆𝑉? Solution: 𝟑𝟓° [IMC 2000 Q8] In the triangle 𝐴𝐵𝐶, 𝐴𝐷=𝐵𝐷=𝐶𝐷. What is the size of angle 𝐵𝐴𝐶? Solution: 𝟗𝟎° In the diagram 𝐴𝐵 is parallel to 𝐶𝐷 and 𝐴𝐸=𝐴𝐶. Determine ∠𝐶𝐴𝐸. 11 ? 12 ? 13 𝐴 𝐵 30° Solution: 𝟏𝟐° ? 𝐸 54° 𝐶 𝐷

Exercise 3 ? (on provided worksheet)
[TMC Regional 2008 Q9] If 𝐴𝐵=𝐵𝐶=𝐶𝐷=𝐷𝐸=𝐸𝐹 and angle 𝐴𝐸𝐹=75°, what is the size of angle 𝐸𝐴𝐹? Solution: 𝟐𝟏° 14 ?

STARTER : Algebraic Angles
What are the angles in each case, in terms of the variables given? (Hint: Just think what you’d do usually – it’s no different here!) i 180−𝑥 ? 𝑥 ii iii 180−2𝑧 ? 𝑧 90−𝑦 ? 𝑦

Overview ! There are two types of problems you’ll have which involve algebraic angles: 1. Angles given 2. No angles given …and you have to find an expression for a given angle. …and you have to introduce variables yourself, either so that you can prove two angles have some relationship, or so you can form an equation and hence find an angle. Example: Example: Given that 𝐴𝐵=𝐴𝐶 and 𝑧<90, which of the following expressions must equal 𝑧? A 𝑥−𝑦 B 𝑥+𝑦 C 𝑥+𝑦−180 D 𝑥−𝑦 E 180−𝑥+𝑦 [JMO 2010 A10] In the diagram, 𝐽𝐾 and 𝑀𝐿 are parallel. 𝐽𝐾=𝐾𝑂=𝑂𝐽=𝑂𝑀 and 𝐿𝑀=𝐿𝑂=𝐿𝐾. Find the size of angle 𝐽𝑀𝑂.

Finding remaining angle in triangle/on line
? If Angle 1 is 90+𝑥, what is angle 2? 180− 90+𝑥 =180−90−𝑥 =90−𝑥 Quickfire Questions: ? Angle 1 Angle 2 𝑥 180−𝑥 𝑥+20 160−𝑥 𝑥−10 190−𝑥 2𝑥+60 120−2𝑥 𝑥−𝑦 180−𝑥+𝑦 180−2𝑥 2𝑥 20+𝑥 ? ? (Expand brackets) ? ? (Simplify) ? ? Bro Tip: An easier way to do the subtraction is to think what we have to do to 90+𝑥 to get to 180. Adding 90 gets us from 90 to 180, and subtracting 𝑥 cancels out the 𝑥. ? ? ?

Finding remaining angle in triangle/on line
? Angle 1 Angle 2 Remaining 𝑥 180−2𝑥 𝑥+20 𝑥+50 110−2𝑥 2𝑥 90−𝑥 90−3𝑥 90−2𝑥 5𝑥 𝑥−𝑦 160+𝑥 20−2𝑥+𝑦 30+2𝑥−𝑦 50+𝑥+𝑦 100−3𝑥 ? ? ? ? ? ?

Full example [JMC 2002 Q21] Given that 𝐴𝐵=𝐴𝐶 and 𝑧<90, which of the following expressions must equal 𝑧? A 𝑥−𝑦 B 𝑥+𝑦 C 𝑥+𝑦−180 D 𝑥−𝑦 E 180−𝑥+𝑦 Try to gradually work out angles in the diagram (in terms of 𝑥 and 𝑦) 𝐷 Two possible ways: ∠𝐴𝐵𝐶=𝑦 (Δ𝐴𝐵𝐶 is isosceles) ∠𝐵𝐴𝐶=180−2𝑦 (angles in triangle 𝐴𝐵𝐶 sum to 180) ∠𝐷𝐴𝐶=180−𝑥−𝑦 (angles in triangle 𝐴𝐷𝐶 sum to 180) Therefore 𝑧=∠𝐵𝐴𝐶−∠𝐷𝐴𝐶 =180−2𝑦− 180−𝑥−𝑦 =𝑥−𝑦 ? ∠𝐴𝐵𝐶=𝑦 (Δ𝐴𝐵𝐶 is isosceles) ∠𝐴𝐷𝐵=180−𝑥 (angles on straight line sum to 180) Therefore using angles in Δ𝐴𝐵𝐷: 𝑧=180−𝑦− 180−𝑥 =180−𝑦−180+𝑥 =𝑥−𝑦 ?

Check Your Understanding
How far can you get down the angle challenge wall? (do in order, and draw the diagram first) ? 𝐶 ∠𝐴𝐶𝐵=180−2𝑥° ∠𝐵𝐶𝐷=2𝑥−90° ∠𝐴𝐵𝐷=180−𝑥  𝐷 ?   ? 𝑥 𝐴 𝐵

Modelling restrictions on angles
We have already seen we can represent the two base angles of an isosceles triangle both as 𝑥 to ensure they have the same value. But how can we model other descriptions? 𝐶 “Angle 𝐶𝐵𝐴 is twice the value of angle 𝐶𝐴𝐵.” ? 2𝑥 ? 𝑥 𝐵 𝐴 𝐷 𝐶 “The line 𝐴𝐶 bisects the angle 𝐷𝐴𝐵.” ? 𝑥 Bro Note: To ‘bisect’ means to cut in half. ? 𝑥 𝐴 𝐵

Modelling restrictions on angles
𝐴 In a triangle 𝐴𝐵𝐶, ∠𝐶𝐴𝐵=2×∠𝐵𝐶𝐴 and ∠𝐴𝐵𝐶=7×∠𝐵𝐶𝐴 Suggest suitable expressions for each of the angles. 𝒙, 𝟐𝒙, 𝟕𝒙 Hence determine ∠𝐵𝐶𝐴. 𝒙+𝟐𝒙+𝟕𝒙=𝟏𝟎𝒙 𝟏𝟎𝒙=𝟏𝟖𝟎 So 𝒙=𝟏𝟖° 𝐵 ? ? 𝐶

Exercise 4 𝐷 ? ? ? 𝐴 𝐵 𝐶 ? ? 𝐴 ? ? ? 𝐵 ? 𝐶 ? (on provided worksheet) 1
Given the value of ∠𝐷𝐵𝐴 indicated, work out the value of ∠𝐷𝐵𝐶. ∠𝑫𝑩𝑨 ∠𝑫𝑩𝑪 𝑤 180−𝑤 𝑥+10 170−𝑥 2𝑥−30 160−2𝑥 90+2𝑥 90−2𝑥 130+𝑥−2𝑦 50−𝑥+2𝑦 𝐷 ? ? ? 𝐴 𝐵 𝐶 ? ? 2 Given the values of ∠𝐴𝐵𝐶 and ∠𝐵𝐴𝐶 indicated, work out the value of ∠𝐴𝐶𝐵. 𝐴 ∠𝑨𝑩𝑪 ∠𝑩𝑨𝑪 ∠𝑨𝑪𝑩 𝑥 180−2𝑥 10 170−𝑥 𝑥+10 150 20−𝑥 2𝑥+30 20−5𝑥 130+3𝑥 4𝑥−20 7𝑥−30 230−11𝑥 ? ? ? 𝐵 ? 𝐶 ?

Exercise 4 ? ? ? ? ? ? (on provided worksheet)
Determine all the angles in each diagram in terms of 𝑥 and/or 𝑦. 3 a c 𝑦 180−2𝑦 ? ? 𝑥 𝟏𝟖𝟎−𝒙−𝒚 ? 𝒙+𝒚 ? 𝑦 𝑦 b d 90−𝑥 ? ? 180−2𝑥 𝟐𝒙 𝑥

Exercise 4 ? ? ? ? ? ? ? ? ? (on provided worksheet) h e 150−𝑦 20+𝑥
𝟏𝟔𝟎−𝒙 𝑥 𝟑𝟎−𝒙+𝒚 ? f i 𝑥 ? 𝒙−𝟏𝟖𝟎 𝟏𝟖𝟎−𝟐𝒙 ? 𝟏𝟖𝟎−𝟐𝒙 ? ? 𝟐𝟕𝟎−𝒙 𝟏𝟖𝟎−𝒙 ? ? 𝒙 𝑥 g 𝑦−10 ? 𝑥+20 𝟏𝟕𝟎−𝒙−𝒚

Exercise 4 Write suitable expressions in terms of 𝑥 for each missing angle. Hence determine the smallest angle in the diagram in each case. 4 a Angle 𝐴𝐵𝐷 is 3 times bigger than angle 𝐷𝐵𝐶. 𝐷 4𝒙=𝟏𝟖𝟎 ∴𝒙=𝟒𝟓 ? ? ? 3𝑥 𝑥 𝐴 𝐵 𝐶 b Angle 𝐵𝐶𝐴 is twice as large as angle 𝐵𝐴𝐶. 𝐴 𝑥 ? ? 𝟗𝟎°+𝟑𝒙=𝟏𝟖𝟎° 𝒙=𝟑𝟎° ? 2𝑥 𝐵 𝐶 The angles 𝐴𝐶𝐵, 𝐴𝐵𝐶 and 𝐵𝐴𝐶 are in the ratio 3:4:5. c 𝐴 𝟏𝟐𝒙=𝟏𝟖𝟎 ∴𝒙=𝟏𝟓° Smallest angle =𝟒𝟓° ? 5𝑥 ? 4𝑥 ? ? 3𝑥 𝐵 𝐶

Exercise 4 [Based on JMC 2011 Q23] The points 𝑆, 𝑇, 𝑈 lie on the sides of the triangle 𝑃𝑄𝑅, as shown, so that 𝑄𝑆=𝑄𝑈 and 𝑅𝑆=𝑅𝑇. ∠𝑇𝑆𝑈=40°. a) Let ∠𝑇𝑆𝑅=𝑥° and ∠𝑈𝑆𝑄=𝑦°. What is the value of 𝑥+𝑦? Solution: 𝟏𝟒𝟎° b) Using the information provided, find expressions for ∠𝑇𝑅𝑆 and ∠𝑈𝑄𝑆. (See diagram) c) Hence find an expression for ∠𝑅𝑃𝑄. (See diagram) d) Using your answer to part (a), hence find the value of ∠𝑅𝑃𝑄. 𝟐𝒙+𝟐𝒚−𝟏𝟖𝟎=𝟐𝟖𝟎−𝟏𝟖𝟎=𝟏𝟎𝟎° 5 [Based on JMO 2005 B4] In this figure 𝐴𝐷𝐶 is a straight line and 𝐴𝐵=𝐵𝐶=𝐶𝐷. Also, 𝐷𝐴=𝐷𝐵. We wish to find the size of ∠𝐵𝐴𝐶. Let ∠𝐷𝐴𝐵=𝑥. a) By using the information provided, determine all the remaining angles in the diagram in terms of 𝑥. b) By considering the three angles in Δ𝐷𝐶𝐵, hence determine ∠𝐵𝐴𝐶. Solution: 𝟑𝟔° 6 ? 𝑥 ? ? 2𝑥 or 180−3𝑥 180−2𝑥 ? 180−3𝑥 ? 𝑥 𝑥 ? ? 2𝑥+2𝑦−180 ? ? 180−2𝑥 ? 𝑥 𝑦 180−2𝑦

Exercise 4 [Based on TMC Final 2014 Q4] The triangle 𝐴𝐵𝐶 is isosceles with 𝐴𝐶=𝐵𝐶 as shown. The point 𝐷 lies on the line 𝐵𝐶 such that the triangle 𝐴𝐵𝐷 is isosceles with 𝐴𝐵=𝐵𝐷 as shown. ∠𝐵𝐴𝐶=2×∠𝐵𝐶𝐴 and we wish to work out the value of ∠𝐴𝐷𝐶. Suppose we let ∠𝐵𝐶𝐴=𝑥. Hence determine: a) ∠𝐵𝐴𝐶 𝟐𝒙 (as specified) b) ∠𝐶𝐵𝐴 𝟐𝒙 c) Hence by considering the triangle 𝐶𝐴𝐷, determine 𝑥. 𝟓𝒙=𝟏𝟖𝟎° therefore 𝒙=𝟑𝟔° d) Use the remaining information to determine ∠𝐴𝐷𝐶 𝟏𝟐𝟔° 7 8 [JMO 2004 B1] In the rectangle ABCD, M and N are the midpoints of AB and CD respectively; AB has length 2 and AD has length 1. Given that ∠𝐴𝐵𝐷=𝑥°, calculate ∠𝐷𝑍𝑁 in terms of 𝑥. ? ? 45° ? ? (because Δ𝑀𝐵𝐶 is isosceles and right-angled) ? 𝑥 Solution: 𝟏𝟑𝟓−𝒙 ?

Exercise 4 [Based on JMO 2006 B3] In this diagram, Y lies on the line AC, triangles ABC and AXY are right angled and in triangle ABX, AX = BX. The line segment AX bisects angle BAC and angle AXY is seven times the size of angle XBC. Let ∠𝑋𝐵𝐶=𝑥. In terms of 𝑥, find expressions for: a) ∠𝐴𝑋𝑌 (see diagram) b) ∠𝐵𝑋𝑌 (hint: the lines 𝑋𝑌 and 𝐵𝐶 are parallel) (see diagram) c) ∠𝑋𝐴𝑌 (see diagram) d) ∠𝐵𝐴𝑋 (see diagram: AX bisects ∠𝑩𝑨𝑪) e) ∠𝐵𝑋𝐴 (using the fact that Δ𝐵𝑋𝐴 is isosceles) (see diagram) f) Optional: By considering angles around a suitable point, hence find ∠𝐴𝐵𝐶. 𝟏𝟒𝒙+𝟕𝒙+𝟏𝟖𝟎−𝒙=𝟑𝟔𝟎 𝟐𝟎𝒙+𝟏𝟖𝟎=𝟑𝟔𝟎 𝟐𝟎𝒙=𝟏𝟖𝟎 𝒙=𝟗° ∠𝑨𝑩𝑪=𝟗+𝟐𝟕=𝟑𝟕° 9 ? ? ? ? ? ? 90−7𝑥 90−7𝑥 14𝑥 7𝑥 180−𝑥 90−7𝑥 𝑥

Exercise 4 [JMC 2003 Q23] In the diagram alongside, 𝐴𝐵=𝐴𝐶 and 𝐴𝐷=𝐶𝐷. How many of the following statements are true for the angles marked? 𝑤=𝑥 𝑥+𝑦+𝑧= 𝑧=2𝑥 A all of them B two C one D none of them E it depends on 𝑥 Solution: A [JMC 2005 Q23] In the diagram, triangle 𝑋𝑌𝑍 is isosceles, with 𝑋𝑌=𝑋𝑍. What is the value of 𝑟 in terms of 𝑝 and 𝑞? A 𝑝−𝑞 B 𝑝+𝑞 C 𝑝−𝑞 D 𝑝+𝑞 E Impossible to determine Solution: B [JMC 2015 Q25] The four straight lines in the diagram are such that 𝑉𝑈=𝑉𝑊. The sizes of ∠𝑈𝑋𝑍, ∠𝑉𝑌𝑍 and ∠𝑉𝑍𝑋 are 𝑥°, 𝑦° and 𝑧°. Which of the following equations gives 𝑥 in terms of 𝑦 and 𝑧? A 𝑥=𝑦−𝑧 B 𝑥=180−𝑦−𝑧 C 𝑥=𝑦− 𝑧 2 D 𝑥=𝑦+𝑧−90 E 𝑥= 𝑦−𝑧 2 Solution: E 10 ? 11 ? 12 ?

Exercise 4 13 [JMO 2008 B3] In the diagram ABCD and APQR are congruent rectangles. The side PQ passes through the point D and ∠𝑃𝐷𝐴=𝑥°. Find an expression for ∠𝐷𝑅𝑄 in terms of x. Solution: 𝟏 𝟐 𝒙 ? 1 2 𝑥 90− 1 2 𝑥 𝑥° 90− 1 2 𝑥

Proof Now that we have a number of angle skills, including introducing algebraic angles, we now have all the skills to form a ‘proof’. A ‘proof’ is a sequence of justified statements that results in the desired conclusion. ! Examples (which we’ll do later) 𝐴 Using the diagram, prove that ∠𝐴𝐶𝐷=∠𝐴𝐵𝐶+∠𝐵𝐴𝐶 𝐵 𝐶 𝐷 [JMO 2015 B2] The diagram shows triangle 𝐴𝐵𝐶, in which ∠𝐴𝐵𝐶=72° and ∠𝐶𝐴𝐵=84°. The point 𝐸 lies on 𝐴𝐵 so that 𝐸𝐶 bisects ∠𝐵𝐶𝐴. The point 𝐹 lies on 𝐶𝐴 extended. The point 𝐷 lies on 𝐶𝐵 extended so that 𝐷𝐴 bisects ∠𝐵𝐴𝐹. Prove that 𝐴𝐷=𝐶𝐸.

STARTER: Constructing diagrams
Sometimes you are not given a diagram, but have to construct one given information. Can you form a suitable diagram given each of the descriptions? Make sure that you use marks/arrows to indicate when sides are the same length or parallel, or where angles are equal. 𝐴 “𝐴𝐵𝐶 is an equilateral triangle.” Bro Tip: We name ‘vertices’ (i.e. corners) using capital letters, and go either clockwise or anticlockwise around the shape. e.g. If “𝐴𝐵𝐶𝐷 is a square”, then the first two diagrams are OK, but the last is not. 𝐵 𝐶 “𝐴𝐵𝐶 is a triangle such that 𝐴𝐵=𝐵𝐶. A point 𝑃 lies outside the triangle such that 𝑃𝐴=𝑃𝐵. 𝑃 ? 𝐵 𝐴 𝐶 𝐴 “𝐴𝐵𝐶 is a triangle where ∠𝐴𝐵𝐶=90°. A point 𝐷 lies on 𝐴𝐶 such that ∠𝐴𝐷𝐵=90°.” ? 𝐷 𝐵 𝐶 𝐴 𝐵 𝐶 𝐵 ? 𝐴 Bro Note: For parallelograms/ rhombuses, we need not indicate parallel sides because it is implied by the lengths.   “The point 𝐹 lies inside the regular pentagon 𝐴𝐵𝐶𝐷𝐸 so that 𝐴𝐵𝐹𝐸 is a rhombus.” 𝐸 𝐵 𝐷 𝐶 𝐷 𝐴 𝐴 𝐵 𝐹  𝐷 𝐶 𝐶 𝐷

A harder one for discussion…
Two of the angles of triangle 𝐴𝐵𝐶 are given by ∠𝐶𝐴𝐵=2𝛼 and ∠𝐴𝐵𝐶=𝛼, where 𝛼<45°. The bisector of angle 𝐶𝐴𝐵 meets 𝐵𝐶 at 𝐷. The point 𝐸 lies on the bisector, but outside the triangle, so that ∠𝐵𝐸𝐴=90°. When produced, 𝐴𝐶 and 𝐵𝐸 meet at 𝑃. Construct the diagram. Note: To ‘bisect’ an angle is to cut it in half. So a ‘bisector’ of an angle is a line which bisects the angle. ?

Simple Proofs The simplest proofs just require you to find an angle, but you need to give a reason for each step. 𝐴𝐵 is parallel to 𝐶𝐷 and 𝐿𝑁𝑀 is a straight line. Prove that 𝑦=112°. ? ∠𝐿𝑁𝐵=68° (corresponding angles are equal) ∠𝐿𝑁𝐴=112° (angles on a straight line sum to 180°) Tip: A helpful way to write out each step of the proof (when referring to angles) is in the form: ∠𝒂𝒏𝒈𝒍𝒆=𝒗𝒂𝒍𝒖𝒆 (reason)

𝑇𝑃=𝑄𝑃 and 𝑃𝑄𝑅 is a straight line. Prove that ∠𝑇𝑃𝑄=40° ∠𝑇𝑄𝑃=70° (angles on straight line sum to 180°) ∠𝑄𝑇𝑃=70° (base angles of isosceles triangle are equal) ∠𝑇𝑃𝑄=40° (angles in triangle sum to 180°) ? ii (if you finish) 𝐴 𝑦 𝐵𝐶𝐷 is a straight line. Let ∠𝐴𝐵𝐶=𝑥 and ∠𝐵𝐴𝐶=𝑦. Prove that ∠𝐴𝐶𝐷=𝑥+𝑦. 𝑥 𝐵 𝐶 𝐷 ? ∠𝐴𝐶𝐷=180−𝑥−𝑦° (angles in triangle sum to 180°) ∠𝐴𝐶𝐷=𝑥+𝑦 (angles on a straight line sum to 180°)

Exercise 5a ? ? ? (on provided worksheet) 1
Prove ∠𝑅𝑄𝐶=55°. Your proof should only require one line and be in the form “∠𝑎𝑛𝑔𝑙𝑒=𝑣𝑎𝑙𝑢𝑒 𝑟𝑒𝑎𝑠𝑜𝑛 ” ∠𝑹𝑸𝑪=𝟓𝟓° (corresponding angles are equal) a ? 𝐴 b 85° 𝐵 𝐷 Prove that ∠𝐸𝐹𝐻=85°. (Your proof should consist of two lines) (Lots of possible proofs, but here’s one…) ∠𝑪𝑭𝑮=𝟖𝟓° (corresponding angles are equal) ∠𝑬𝑭𝑮=𝟖𝟓° (vertically opposite angles are equal) 𝐶 ? 𝐸 𝐹 𝐺 ? 𝐻 c 𝐵 Prove that ∠𝐵𝐶𝐷=2𝑥. (Your proof should consist of three lines) ∠𝑨𝑩𝑪=𝒙 (base angles of isosceles triangle 𝑨𝑩𝑪 are equal) ∠𝑩𝑪𝑨=𝟏𝟖𝟎−𝟐𝒙 (angles in triangle sum to 𝟏𝟖𝟎°) ∠𝑩𝑪𝑫=𝟐𝒙 (angles on straight line sum to 𝟏𝟖𝟎°) ? 𝑥 ? 𝐴 𝐶 𝐷

Exercise 5a ? ? ? ? (on provided worksheet) 2 a 𝐶 𝐷 𝐵 𝐵 𝐴 b 𝐴 𝐷 𝐶 𝐷 c
Draw diagrams which satisfy the following criteria, ensuring you note where lines are of equal length or parallel. You do NOT need to find any angles. 2 a ? 𝐶 𝐴𝐵𝐶 is a right-angled triangle such that ∠𝐶𝐴𝐵=90°. 𝐷 is a point on 𝐵𝐶 such that 𝐴𝐷=𝐵𝐷. 𝐷 ? 𝐵 𝐴 𝐵 𝐴𝐵𝐶 is an isosceles triangle where 𝐴𝐵=𝐴𝐶. 𝐷 is a point that lies inside the triangle such that 𝐵𝐶𝐷 is equilateral. b 𝐴 𝐷 𝐶 𝐷 ? c 25° Points 𝐴, 𝐵 and 𝐶 lie in that order on a straight line. A point 𝐷, not on the line, is placed such that 𝐴𝐷=𝐴𝐵, 𝐵𝐷=𝐵𝐶, and ∠𝐵𝐷𝐶=25°. 𝐶 𝐵 𝐴 ? 𝑄 𝑃 Points 𝑇 and 𝑈 lie outside a parallelogram 𝑃𝑄𝑅𝑆, and are such that triangles 𝑅𝑄𝑇 and 𝑆𝑅𝑈 are equilateral and lie wholly outside the parallelogram. d 𝑇 𝑆 𝑅 𝑈

Exercise 5a ? ? (on provided worksheet) 3 𝐵 𝐶 𝐴 a Prove that ∠𝐵𝐶𝐴=20°
∠𝑩𝑨𝑫=𝟒𝟎° (base angles of isosceles 𝚫 are equal) ∠𝑨𝑩𝑫=𝟏𝟎𝟎° (angles in triangle sum to 180) ∠𝑪𝑩𝑫=𝟖𝟎° (angles on straight line sum to 180) ∠𝑩𝑫𝑪=𝟖𝟎° (base angles of isosceles 𝚫 are equal) ∠𝑩𝑪𝑫=𝟐𝟎° (angles in triangle sum to 180) ? 40° 𝐷 b 𝐴 𝐵 70° In the diagram, 𝐴𝐵 and 𝐶𝐸 are parallel, and 𝐷𝐹=𝐷𝐸. Prove that ∠𝐷𝐸𝐹=35° ∠𝑩𝑫𝑬=𝟕𝟎° (alternate angles are equal) ∠𝑭𝑫𝑬=𝟏𝟏𝟎° (angles on a straight line sum to 180) ∠𝑫𝑬𝑭=𝟑𝟓° (as base angles of isosceles 𝚫𝑫𝑬𝑭 are equal and angles in triangle sum to 180) 𝐶 𝐷 𝐸 ? 𝐹

Other Types of Proof ? Proof ? ? ? 𝐵 𝐶 𝐴 𝐷 𝐷 𝐴 𝐵 𝐶 𝐵 𝐴 𝐶
𝐵𝐶𝐷 is a straight line and 𝐴𝐵=𝐴𝐶. Prove that 𝐴𝐶 bisects the angle 𝐵𝐴𝐷. 70° 𝐶 Before we start this proof, what specifically are we trying to show? That ∠𝑩𝑨𝑪=∠𝑪𝑨𝑫 ? ∠𝐵𝐶𝐴=70° (base angles of isosceles triangle are equal) ∠𝐴𝐶𝐷=110° (angles on straight line sum to 180°) ∠𝐵𝐴𝐶=40° (angles in triangle sum to 180°) ∠𝐶𝐴𝐷=40° (angles in triangle sum to 180°) ∠𝐵𝐴𝐶=∠𝐶𝐴𝐷 therefore 𝐴𝐶 bisects ∠𝐵𝐴𝐷. 𝐴 Proof ? 30° 𝐷 Bro Note: You need a conclusion so it’s clear that your proof is complete.. Refer to the provided ‘cheat sheet’. 𝐵 Similarly, what would we need to do in order to: …prove that 𝐴𝐵𝐶 is isosceles? 𝐷 …prove that 𝐴𝐵𝐶 is a straight line? ? Show that ∠𝐷𝐵𝐴+∠𝐷𝐵𝐶=180° Show that ∠𝐵𝐴𝐶=∠𝐵𝐶𝐴 ? 𝐴 𝐵 𝐶 𝐴 𝐶

𝐶𝐷𝐵 is a straight line and 𝐴𝐶=𝐴𝐷=𝐷𝐵. ∠𝐷𝐴𝐵=72° as shown. Prove that triangle 𝐴𝐵𝐶 is isosceles. 𝐶 𝐷 ∠𝑫𝑩𝑨=𝟑𝟔° (base angles of isosceles triangle 𝑨𝑫𝑩 are equal) ∠𝑪𝑫𝑨=𝟕𝟐° (exterior angle of triangle is sum of two other interior angles) ∠𝑫𝑪𝑨=𝟕𝟐° (base angles of isosceles triangle 𝑨𝑪𝑫 are equal) ∠𝑪𝑨𝑫=𝟑𝟔° (angles in triangle sum to 𝟏𝟖𝟎°) ∠𝑪𝑨𝑩=𝟑𝟔°+𝟑𝟔°=𝟕𝟐° ∠𝑩𝑪𝑨=∠𝑪𝑨𝑩=𝟕𝟐° ∴ triangle 𝑨𝑩𝑪 is isosceles. Proof ? 𝐵 36° 𝐴 ii (if you finish) 𝐷 In the diagram, 𝐴𝐵=𝐵𝐶=𝐵𝐷 and ∠𝐴𝐷𝐶=90°. Prove that 𝐴𝐵𝐶 is a straight line. Let ∠𝑫𝑨𝑩=𝒙. ∠𝑨𝑩𝑫=𝒙 (base angles of isosceles triangle are equal) ∠𝑨𝑩𝑫=𝟏𝟖𝟎−𝟐𝒙 (angles in triangle sum to 𝟏𝟖𝟎°) ∠𝑩𝑫𝑪=𝟗𝟎−𝒙 (as ∠𝑨𝑫𝑪=𝟗𝟎°) ∠𝑩𝑪𝑫=𝟗𝟎−𝒙 (base angles of isosceles triangle are equal) ∠𝑫𝑩𝑪=𝟐𝒙 (angles in triangle 𝑩𝑪𝑫 sum to 𝟏𝟖𝟎°) ∠𝑫𝑩𝑨+∠𝑫𝑩𝑪=𝟏𝟖𝟎−𝟐𝒙+𝟐𝒙=𝟏𝟖𝟎° therefore 𝑨𝑩𝑪 is a straight line. Proof ? 𝐶 𝐵 𝐴 Hint: You should choose a suitable angle to be 𝑥 and start by writing “Let ∠𝑚𝑦𝑎𝑛𝑔𝑙𝑒=𝑥”

Exercise 5b ? ? (on provided worksheet) 1 𝐴 𝐷 𝐵 a 𝐶 𝐸 b 𝐵 𝐴 𝐶 𝐷
𝐴𝐷𝐶 and 𝐴𝐵𝐶 are right-angled triangles where ∠𝐴𝐶𝐷=10° and ∠𝐷𝐶𝐵=70°. Prove that triangle 𝐴𝐵𝐶 is isosceles. 10° ∠𝑪𝑨𝑫=𝟖𝟎° (angles in triangle 𝑨𝑪𝑫 add to 𝟏𝟖𝟎°). ∠𝑨𝑪𝑩=∠𝑨𝑪𝑫+∠𝑫𝑪𝑩=𝟖𝟎° ∠𝑫𝑨𝑪=∠𝑨𝑪𝑩 therefore 𝚫𝑨𝑩𝑪 is isosceles. ? 70° 𝐶 𝐸 b 𝐵 𝐴𝐶=𝐵𝐶, 𝐴𝐶𝐷 is a straight line, ∠𝐵𝐶𝐴=40° and ∠𝐵𝐶𝐸=70° as per the diagram. Prove that 𝐴𝐵 and 𝐶𝐸 are parallel. ∠𝑩𝑨𝑪=𝟕𝟎° (base angles in isosceles triangle 𝑨𝑩𝑪 are equal) ∠𝑬𝑪𝑫=𝟕𝟎° (angles on straight line at 𝑪 sum to 𝟏𝟖𝟎°) ∠𝑩𝑨𝑪=∠𝑬𝑪𝑫 which are corresponding angles, therefore 𝑨𝑩 and 𝑪𝑬 are parallel. ? 70° 40° 𝐴 𝐶 𝐷

Exercise 5b 2 𝐸 As before, 𝐴𝐶=𝐵𝐶. Suppose also that 𝐶𝐸 bisects the angle 𝐵𝐶𝐷, but unlike before, no angles are known. By introducing a suitable variable (you may wish to start your proof “Let ∠𝑎𝑛𝑔𝑙𝑒=𝑥”) prove that 𝐴𝐵 and 𝐶𝐸 are parallel. 𝐵 ? Let ∠𝑬𝑪𝑫=𝒙. ∠𝑩𝑪𝑬=𝒙 (as 𝑪𝑬 bisects ∠𝑩𝑪𝑫). ∠𝑩𝑪𝑨=𝟏𝟖𝟎−𝟐𝒙 (angles on straight line at 𝑪 sum to 𝟏𝟖𝟎°) ∠𝑩𝑨𝑪= 𝟏𝟖𝟎− 𝟏𝟖𝟎−𝟐𝒙 𝟐 =𝒙 (base angles of isosceles triangle 𝑨𝑩𝑪 are equal) ∠𝑩𝑨𝑪=∠𝑬𝑪𝑫=𝒙, and 𝑨𝑪𝑫 is a straight line, therefore 𝑨𝑩 and 𝑪𝑬 are parallel. 𝐴 𝐶 𝐷

Exercise 5b 3 [JMO 2001 B3] In the diagram, B is the midpoint of AC and the lines AP, BQ and CR are parallel. The bisector of ∠𝑃𝐴𝐵 meets BQ at Z. Draw a diagram to show this, and join Z to C (i) Given that ∠𝑃𝐴𝑍=𝑥°, find ∠𝑍𝐵𝐶 in terms of x (ii) Show that CZ bisects ∠𝐵𝐶𝑅. ?

Exercise 5b 4 [JMO 2015 B2] The diagram shows triangle 𝐴𝐵𝐶, in which ∠𝐴𝐵𝐶=72° and ∠𝐶𝐴𝐵=84°. The point 𝐸 lies on 𝐴𝐵 so that 𝐸𝐶 bisects ∠𝐵𝐶𝐴. The point 𝐹 lies on 𝐶𝐴 extended. The point 𝐷 lies on 𝐶𝐵 extended so that 𝐷𝐴 bisects ∠𝐵𝐴𝐹. Prove that 𝐴𝐷=𝐶𝐸. ?

Skipton Girls’ High School
KS3/4: Polygons Skipton Girls’ High School Objectives: Be able to classify shapes, reason about sides and angles, and find interior/exterior angles of polygons.

STARTER: Identifying 2D polygons
! A polygon is a 2D shape with straight sides. ? Sides: 3 Triangle Equilateral ? Isosceles ? Scalene ? 4 ? Square Rectangle ? Rhombus ? Quadrilateral ? Parallelogram ? Trapezium ? Kite ? Arrowhead ? 5 Pentagon ? 8 12 Octagon ? Dodecagon ? 6 Hexagon ? 9 20 Nonagon ? Icosagon ? 7 Heptagon ? 10 Decagon ?

Key Properties of Polygons
Shape Description 2D Shape Polygon A 2D shape with straight edges. Ellipse (Also known as an oval) Circle Quadrilateral Polygon with four edges. Square Regular polygon with four edges. Rectangle Two pairs of parallel sides, all angles equal. Oblong All rectangles which are not squares. Trapezium Quadrilateral with at least one pair of parallel sides. Parallelogram Quadrilateral with two pairs of parallel sides. Rhombus Quadrilateral with all edges the same length. Kite Quadrilateral where sides in adjacent pairs are equal in length. Arrowhead Kite with a reflex angle.

Properties of quadrilaterals
Diagonals always equal in length? Shape Name Lines of symmetry Num pairs of parallel sides Diagonals perpendicular? Square Rectangle Kite Rhombus Parallelogram Arrowhead ? 4 2 1 ? 2 Yes No ? Yes No ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?

RECAP: Interior angles of quadrilateral
The interior angles of a quadrilateral add up to 360. ? Parallelogram 1 2 y 100° x x 50° x = 130° ? x = 100° y = 80° ? ? 3 4 Trapezium Kite x x = 120° ? x 55° 95° 60° x = 105° ?

Sum of interior angles n = 3 n = 4 Total of interior angles = 360°
Can you guess what the angles add up to in a pentagon? How would you prove it?

! For an n-sided shape, the sum of the interior angles is:
Sum of interior angles Click to animate We can cut a pentagon into three triangles. The sum of the interior angles of the triangles is: 3 x 180° = 540° ! For an n-sided shape, the sum of the interior angles is: 180(n-2) ?

A regular decagon (10 sides). 160° x x 80° 130° 120° x = 140° ? x = 144° ? 120° x 40° 100° x = 240° ? 40°

Exercise 1 ? ? ? ? ? ? 1a b c b = 260 x = 75 x = 25 d e f a = 100

Exercise 1 ? ? ? ? ? g h i x = 120 x = 252 x = 54
The total of the interior angles of a polygon is 1260°. How many sides does it have? The interior angle of a regular polygon is 179°. How many sides does it have? 2 ? N1 ?

Exercise 1 If a n-sided polygon has exactly 3 obtuse angles (i.e. 90 <  < 180), then determine the possible values of (Hint: determine the possible range for the sum of the interior angles, and use these inequalities to solve). N2 ?

Interior Angles NO YES NO ? ? ?
An exterior angle of a polygon is an angle between the line extended from one side, and an adjacent side. Which of these are exterior angles of the polygon? ? ? NO YES ? NO

Click to Start animation
Interior Angles To defeat Kim Jon Il, Matt Damon must encircle his pentagonal palace. What angle does Matt Damon turn in total? 360° ? ! The sum of the exterior angles of any polygon is 360°. Click to Start animation

Interior Angles ? ? Exterior angle of pentagon = 360 / 5 = 72°
If the pentagon is regular, then all the exterior angles are clearly the same. Therefore: Exterior angle of pentagon = 360 / 5 = 72° Interior angle of pentagon = 180 – 72 = 108° ? ?

Angles in Regular Polygons
Num Sides Name of Regular Polygon Exterior Angle Interior Angle 3 Triangle 120° 60° 4 Quadrilateral 90° 5 Pentagon 72° 108° 6 Hexagon 7 Heptagon 51.4° 128.6° 8 Octagon 45° 135° 9 Nonagon 40° 140° 10 Decagon 36° 144° ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Bonus Question: What is the largest number of sides a shape can have such that its interior angle is an integer? 360 sides. The interior angle will be 179°. ?

GCSE question The diagram shows a regular hexagon and a regular octagon. Calculate the size of the angle marked x. You must show all your working. x = 105° ?

GCSE Question Hint: Fill in what angles you do know. You can work out what the interior angle of Tile A will be. Question: The pattern is made from two types of tiles, tile A and tile B. Both tile A and tile B are regular polygons. Work out the number of sides tile A has. Sides = 12 ?

Q1 Q2 50 85 75 a 80 80 80 c a = 110° ? c = 70° ? Q3 What is the exterior angle of a 180-sided regular polygon? 360 ÷ 180 = 2 Q4 The interior angle of a regular polygon is 165. How many sides does it have? Interior angle = 180 – 165 = 15 n = 360 ÷ 15 = 24 Alternative method: Total interior angle = 165n Then solve 180(n – 2) = 165n ? ?

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