Presentation on theme: "GCSE: Non-right angled triangles"— Presentation transcript:
1GCSE: Non-right angled triangles Dr J FrostLast modified: 16th November 2014
2RECAP: Right-Angled Triangles We’ve previously been able to deal with right-angled triangles, to find the area, or missing sides and angles.5634Area = 15?30.96°?553?Using Pythagoras:𝑥= − =3Using 𝑨= 𝟏 𝟐 𝒃𝒉:𝐴𝑟𝑒𝑎= 1 2 ×6× =15Using trigonometry:tan 𝜃 = 3 5 𝜃= tan − =30.96°
3We’ll revisit this later... Learning ObjectivesThere’s 2 things you’ll need to be able to do with non-right angle triangles:3cm??59°7cmHow would I find the missing side and angle?How do I find the area of this non-right angled triangle?We’ll revisit this later...
4Labelling Sides of Non-Right Angle Triangles Right-Angled Triangles:Non-Right-Angled Triangles:𝑎ℎ𝐶?𝑜𝑏𝐵??𝐴𝑎𝑐We label the sides 𝑎, 𝑏, 𝑐 and their corresponding OPPOSITE angles 𝐴, 𝐵, 𝐶
5OVERVIEW: Finding missing sides and angles You haveYou wantUse#1: Two angle-side opposite pairsMissing angle or side in one pairSine rule#2 Two sides known and a missing side opposite a known angleRemaining sideCosine rule#3 All three sidesAn angle#4 Two sides known and a missing side not opposite known angleSine rule twice
6The Sine Rule 5.02 10 9.10 ! Sine Rule: 𝑎 sin 𝐴 = 𝑏 sin 𝐵 = 𝑐 sin 𝐶 ? cCbBaAFor this triangle, try calculating each side divided by the sin of its opposite angle. What do you notice in all three cases?65°5.021085°! Sine Rule:𝑎 sin 𝐴 = 𝑏 sin 𝐵 = 𝑐 sin 𝐶?30°9.10You haveYou wantUse#1: Two angle-side opposite pairsMissing angle or side in one pairSine rule
7Examples8Q2Q1850°85°100°15.76?45°30°11.27?𝒙 𝐬𝐢𝐧 𝟖𝟓 = 𝟖 𝐬𝐢𝐧 𝟒𝟓𝒙= 𝟖 𝐬𝐢𝐧 𝟖𝟓 𝐬𝐢𝐧 𝟒𝟓 =𝟏𝟏.𝟐𝟕𝒙 𝐬𝐢𝐧 𝟏𝟎𝟎 = 𝟖 𝐬𝐢𝐧 𝟑𝟎𝒙= 𝟖 𝐬𝐢𝐧 𝟏𝟎𝟎 𝐬𝐢𝐧 𝟑𝟎 =𝟏𝟓.𝟕𝟔You haveYou wantUse#1: Two angle-side opposite pairsMissing angle or side in one pairSine rule
8ExamplesWhen you have a missing angle, it’s better to ‘flip’ your formula to get𝐬𝐢𝐧 𝑨 𝒂 = 𝐬𝐢𝐧 𝑩 𝒃i.e. in general put the missing value in the numerator.5Q3Q48126°85°40.33°?10?56.11°6sin 𝜃 5 = sin sin 𝜃 = 5 sin 𝜃= sin − sin =56.11°sin 𝜃 8 = sin 126° sin 𝜃 = 8 sin 𝜃= sin − sin =40.33°
9Test Your Understanding 𝑄82°𝑃20°10𝑚𝜃85°5𝑐𝑚12𝑚𝑅Determine the angle 𝜃.sin 𝜽 𝟏𝟎 = sin 𝟖𝟐 𝟏𝟐𝜽=𝒔𝒊 𝒏 −𝟏 𝟏𝟎 sin 𝟖𝟐 𝟏𝟐 =𝟓𝟓.𝟔°Determine the length 𝑃𝑅.𝑷𝑹 𝐬𝐢𝐧 𝟐𝟎 = 𝟓 𝐬𝐢𝐧 𝟖𝟓𝑷𝑹= 𝟓 𝐬𝐢𝐧 𝟐𝟎 𝐬𝐢𝐧 𝟖𝟓 =𝟏.𝟕𝟐𝒄𝒎??
10Exercise 1Find the missing angle or side. Please copy the diagram first! Give answers to 3sf.Q1Q2Q315101685°12𝑦30°𝑥30°40°𝑥20𝑥=53.1°?𝑦=56.4°?𝑥=23.2?𝑥Q4Q6Q570°35°1040°510𝛼20𝛼=16.7°?𝑥=5.32?𝑥𝑥=6.84?
11Cosine Rule How are sides labelled ? Calculation? Cosine Rule: 15 The sine rule could be used whenever we had two pairs of sides and opposite angles involved.However, sometimes there may only be one angle involved. We then use something called the cosine rule.𝑎𝐴𝑏𝑐Cosine Rule:𝑎 2 = 𝑏 2 + 𝑐 2 −2𝑏𝑐 cos 𝐴15The only angle in formula is 𝐴, so label angle in diagram 𝐴, label opposite side 𝑎, and so on (𝑏 and 𝑐 can go either way).𝑥 2 = − 2×15×12× cos 𝑥 2 = … 𝑥=22.83How are sides labelled ?115°𝑥Calculation?12
12Sin or Cosine Rule? Sine Cosine Sine Cosine Sine Cosine If you were given these exam questions, which would you use?𝑥10𝑥1070°70°1515SineCosineSineCosine1010𝛼7𝛼70°1512SineCosineSineCosine
13Test Your Understanding e.g. 1e.g. 2𝑥𝑥47847°106.4°7𝑥=6.05?𝑥=8.99?You haveYou wantUseTwo sides known and a missing side opposite a known angleRemaining sideCosine rule
14Exercise 2Use the cosine rule to determine the missing angle/side. Quickly copy out the diagram first.Q1Q2Q3135°58558𝑥100°60°70𝑥7𝑦𝑥=6.24?𝑦=10.14?𝑥=50.22?Q6Q5𝑥Q464575°𝑥1043°65°863𝑥=4.398?𝑥3𝑥=9.513?𝑥=6.2966?
15Dealing with Missing Angles You haveYou wantUseAll three sidesAn angleCosine rule𝒂 𝟐 = 𝒃 𝟐 + 𝒄 𝟐 −𝟐𝒃𝒄 𝐜𝐨𝐬 𝑨7𝛼Label sides then substitute into formula.4𝟒 𝟐 = 𝟕 𝟐 + 𝟗 𝟐 − 𝟐×𝟕×𝟗× 𝐜𝐨𝐬 𝜶 𝟏𝟔=𝟏𝟑𝟎−𝟏𝟐𝟔 𝐜𝐨𝐬 𝜶 𝟏𝟐𝟔 𝐜𝐨𝐬 𝜶 =𝟏𝟑𝟎−𝟏𝟔 𝐜𝐨𝐬 𝜶 = 𝟏𝟏𝟒 𝟏𝟐𝟔 𝜶= 𝐜𝐨𝐬 −𝟏 𝟏𝟏𝟒 𝟏𝟐𝟔 =𝟐𝟓.𝟐°?9?Simplify each bit of formula.?Rearrange (I use ‘swapsie’ trick to swap thing you’re subtracting and result)?𝛼=25.2°?
16Test Your Understanding 4𝑐𝑚87𝑐𝑚5𝜃𝜃9𝑐𝑚7?4 2 = − 2×7×9× cos 𝜃 16=130−126 cos 𝜃 114=126 cos 𝜃 cos 𝜃 = 𝜃= cos − =𝟐𝟓.𝟐𝟏°?8 2 = − 2×7×5× cos 𝜃 64=74−70 cos 𝜃 10=70 cos 𝜃 cos 𝜃 = 𝜃= cos − =𝟖𝟏.𝟕𝟗°
18Using sine rule twice ? 4 32° 3 𝑥 You haveYou wantUse#4 Two sides known and a missing side not opposite known angleRemaining sideSine rule twiceGiven there is just one angle involved, you might attempt to use the cosine rule:𝟑 𝟐 = 𝒙 𝟐 + 𝟒 𝟐 − 𝟐×𝒙×𝟒× 𝐜𝐨𝐬 𝟑𝟐𝟗= 𝒙 𝟐 +𝟏𝟔−𝟖𝒙 𝐜𝐨𝐬 𝟑𝟐4?32°3𝑥This is a quadratic equation!It’s possible to solve this using the quadratic formula (using 𝒂=𝟏, 𝒃=−𝟖 𝐜𝐨𝐬 𝟑𝟐 , 𝒄=𝟕). However, this is a bit fiddly and not the primary method expected in the exam…
19! Using sine rule twice ? ? ? 4 32° 3 𝑥 You haveYou wantUse#4 Two sides known and a missing side not opposite known angleRemaining sideSine rule twice!2: Which means we would then know this angle.4𝟏𝟖𝟎−𝟑𝟐−𝟒𝟒.𝟗𝟓𝟓𝟔=𝟏𝟎𝟑.𝟎𝟒𝟒𝟒?1: We could use the sine rule to find this angle.32°33: Using the sine rule a second time allows us to find 𝑥?sin 𝐴 4 = sin 32 3𝐴= °𝑥𝑥 sin = 3 sin 32 𝑥=5.52 𝑡𝑜 3𝑠𝑓?
20Test Your Understanding 9?𝑦=6.97𝑦61°104353°𝑦=5.01?𝑦
21Where C is the angle wedged between two sides a and b. Area of Non Right-Angled Triangles3cmArea = 0.5 x 3 x 7 x sin(59)= 9.00cm2?59°7cm!Area = 1 2 𝑎 𝑏 sin 𝐶Where C is the angle wedged between two sides a and b.
225 5 5 Test Your Understanding ? ? 𝐴= 1 2 ×6.97×10×𝑠𝑖𝑛61 =30.48 9 6.97 𝐴= 1 2 ×6.97×10×𝑠𝑖𝑛 =30.48?96.9761°1055𝐴= 1 2 ×5×5× sin =?5
23Harder Examples 6 7 8 ? ? Q1 (Edexcel June 2014) Q2 Finding angle ∠𝐴𝐶𝐷:sin 𝐷 9 = sin 𝐷= ° ∠𝐴𝐶𝐷=180−100− = °𝐴𝑟𝑒𝑎 𝑜𝑓 Δ= 1 2 ×9×11× sin =21.945Area of 𝐴𝐵𝐶𝐷=2×21.945=43.9 𝑡𝑜 3𝑠𝑓?Using cosine rule to find angle opposite 8:cos 𝜃 = 𝜃= °𝐴𝑟𝑒𝑎= 1 2 ×6×7× sin 75.5…=𝟐𝟎.𝟑
24Exercise 4Q3Q1Q2Q453.63115100°3.875°815.2Area = 7.39?70°𝐴𝑟𝑒𝑎= =0.433?Area = 9.04?Area = 8.03?2cmQ5Q7Q6110°8.7𝑐𝑚3cmArea = 3.11𝑐 𝑚 2?49°64°Q84.2m3m𝐴𝑟𝑒𝑎=29.25𝑐 𝑚 2?𝑃 is the midpoint of 𝐴𝐵 and 𝑄 the midpoint of 𝐴𝐶. 𝐴𝑃𝑄 is a sector of a circle. Find the shaded area.5.3m?1 2 × 6 2 × sin 60 − 1 6 𝜋 =10.9𝑐 𝑚 2Area = 6.29 𝑚 2?
25Segment Area𝐴𝑂𝐴𝐵 is a sector of a circle, centred at 𝑂.Determine the area of the shaded segment.10𝑐𝑚𝑂70°𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑒𝑐𝑡𝑜𝑟= × 𝜋× = 𝑐 𝑚 2 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒= 1 2 × 10 2 × sin 70 = 𝑐 𝑚 2𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑒𝑔𝑚𝑒𝑛𝑡= − =14.1 𝑡𝑜 3𝑠𝑓???𝐵