{"id":3541,"date":"2023-10-05T11:15:25","date_gmt":"2023-10-05T11:15:25","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=3541"},"modified":"2023-10-05T11:16:19","modified_gmt":"2023-10-05T11:16:19","slug":"week-2-ss2-third-term-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-2-ss2-third-term-mathematics-notes\/","title":{"rendered":"Week 2 – SS2 Third Term Mathematics Notes"},"content":{"rendered":"

WEEK TWO<\/strong>
\n\t\tTopic: Cosine and Sine Rule Relating to Triangle<\/strong>.
\nContent
\n<\/strong>-Sine Rule for Acute and Obtuse Angled Triangle.
\n-Application of Sine Rule to Triangle.
\n-Cosine Rule for Acute and Obtuse Angled Triangle.
\n-Application of Cosine Rule.<\/p>\n

\u00a0Sine Rule for Acute and Obtuse Angled Triangle.
\n<\/strong>Consideration is given to other triangles than a right angled triangle. The angles of any triangle are denoted by capital letters such as; A, B, C, while the sides are represented by small letters; a, b, c, respectively.
\n\"\"\/\"\"\/A<\/p>\n

\u00a0 c b<\/p>\n

\u00a0
\n\u00a0
\n\u00a0\"\"\/ B a C
\nAcute Triangle: This is a type of triangle in which the angles are less than 900<\/sup>.
\nObtuse Triangle: Is a type of triangle in which one of the angles is more than 900<\/sup> but less than 1800<\/sup>.<\/p>\n

\u00a0Deductive Proof of Sine Rule<\/strong>
\n\t\tThe sine rule is the same for acute and obtuse angled triangle.
\nGiven: Any triangle ABC (acute-angled or obtuse-angled triangle).
\n\"\"\/\"\"\/ A<\/p>\n

\u00a0 c b<\/p>\n

\u00a0
\n\u00a0
\n\u00a0 B a C
\n\"\"\/
\n\t\t\"\"\/\"\"\/\"\"\/To prove: a = b = c
\n Sin A SinB Sin C
\n\"\"\/Construction: Draw a perpendicular A to BC ( produced, if necessary)
\n Proof:
\n\"\"\/ Sin B = h \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026(1)
\n c
\n In fig. 7.10 a)
\n\"\"\/ . Sin C = h \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026..(2)
\n b
\nIn fig.7.10 b)
\n\"\"\/ Sin(1800<\/sup> – C ) = h
\n b<\/p>\n

\u00a0Hence, Sin C = h [sin(180 \u2013 \u019f ) = sin \u019f] \u2026\u2026\u2026\u2026\u2026(2)
\n\"\"\/ b
\nFrom (1) h = c SinB
\nFrom(2) h = b Sin C<\/p>\n

\u00a0Hence, cSinB=bSinC
\n b = c
\n\"\"\/\"\"\/ SinB SinC <\/p>\n

\u00a0Example<\/strong>
\n\t\tIn triangle ABC, A= 380<\/sup>, B = 270<\/sup>, and b = 17cm. find a and c.
\nSolution;
\n Using sine rule; Sin A = Sin B
\n\"\"\/\"\"\/\"\"\/\"\"\/ a b
\n\t\t\t<\/sup> A
\n\"\"\/ sin 380<\/sup> = sin 270\u00a0\u00a0\u00a0\u00a0 38<\/sup>
\n\t\t\"\"\/ a 17
\n a = 17 sin 380<\/sup>
\n\t\t\"\"\/ sin 270<\/sup>
\n\t\t\"\"\/
\n\t\ta = 23cm C\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0B
\n To find c; Angle C must be known; A + B + C = 1800<\/sup>
\n\t\t C = 1800<\/sup> \u2013 380<\/sup> \u2013 270<\/sup> , C = 1150<\/sup>
\n\t\t\"\"\/\"\"\/ Sin B = Sin C
\n b c
\n\"\"\/\"\"\/ sin 270<\/sup> = sin 1150<\/sup> , c = 17 x sin 1150<\/sup>
\n\t\t\"\"\/ 17 c sin 270<\/sup>
\n\t\tc= 33.9 approximately, c = 34cm.
\nNB: In any triangle, the longest side correspond to the largest angle while the shortest side corresponds to the smallest angle.
\nEvaluation<\/strong>: 1. Solve the \u2206 completely; A = 390<\/sup>, a = 8.2m and b = 5.6m
\n2.Calculate the values of angles P and R of \u2206 PQR, where q = 14.35cm, p = 7.82cm and Q = 115.60<\/sup><\/p>\n

\u00a0Deductive Proof of Cosine Rule<\/strong>
\n\t\tThe cosine rule is also the same for the acute and obtuse angled triangle.\u00a0\u00a0\u00a0\u00a0A
\n\"\"\/\"\"\/\"\"\/\"\"\/\"\"\/Given: any ABC\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
\n\"\"\/(a)\u00a0\u00a0\u00a0\u00a0 A
\n\"\"\/(b)c
\nbh
\n\u00a0\u00a0\u00a0\u00a0<\/sup>c\u00a0\u00a0\u00a0\u00a0<\/sup>b
\nh
\n\"\"\/\u00a0\u00a0\u00a0\u00a0B\u00a0\u00a0\u00a0\u00a0N
\nB\u00a0\u00a0\u00a0\u00a0a\u00a0\u00a0\u00a0\u00a0C\u00a0\u00a0\u00a0\u00a0x
\n a-x N x\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0C\u00a0\u00a0\u00a0\u00a0a+x
\n\u00a0\u00a0\u00a0\u00a0a<\/p>\n

\u00a0To prove: c2<\/sup> = a2<\/sup> + b2<\/sup> \u2013 2abCos C
\nConstruction: Draw a perpendicular from A to B(produced if necessary).
\nProof: Using the acute triangle; using the obtuse triangle;
\n c2<\/sup> = (a-x)2<\/sup> + h2<\/sup> c2<\/sup> = (a +x)2<\/sup> + h2<\/sup>
\n\t\t c2 = a2<\/sup> -2ax + x2<\/sup> + h2<\/sup> c2<\/sup> =a2<\/sup>+2ax+x2<\/sup>+h2<\/sup>
\n\t\t\"\"\/From \u2206 ACN; b2<\/sup> = x2<\/sup> + h2<\/sup>, and Cos C = x \/ b From ACN,x2<\/sup>+h2<\/sup>=b2<\/sup>
\n\t\t\"\"\/ x= b Cos C =a2<\/sup>+2ax+b2<\/sup>
\n\t\t c2<\/sup> = a2<\/sup> + b2<\/sup> \u2013 2ax From ACN,x\/b=CosACN
\n x =bCosC =Cos(1800<\/sup>-C)
\n = -Cos C ,x= -bCos C
\n c2<\/sup> = a2<\/sup> + b2<\/sup> \u2013 2abCosC c2<\/sup> = a2<\/sup> + b2<\/sup> +2a(-bCos C)
\n c2<\/sup> = a2<\/sup> + b2<\/sup> \u2013 2abCos C
\nSimilarly, for other sides and angles. Therefore the cosine rule can be written as thus:
\n<\/strong>\"\"\/ c2<\/sup> = a2<\/sup> + b2<\/sup> \u2013 2abCosC OR Cos C = a2<\/sup> + b2<\/sup> \u2013 c2<\/sup>
\n\t\t\t\t<\/strong>a2<\/sup>= b2<\/sup> + c2<\/sup> \u2013 2bcCosA 2ab
\n<\/strong>\"\"\/ b2<\/sup> = a2<\/sup> + c2<\/sup> \u2013 2acCosB Cos A = b2<\/sup> + c2<\/sup> \u2013 a2<\/sup>
\n\t\t\t<\/strong> 2bc
\n<\/strong> Cos B = a2<\/sup> + c2<\/sup> \u2013 b2<\/sup>
\n\t\t\t<\/strong>\"\"\/ 2ac
\n<\/strong>Conditions Necessary for Use<\/strong>: The rule is used for solving acute and obtuse angled triangles in which two sides and included angles are given. <\/p>\n

\u00a0Example<\/strong>; Given that A = 1200<\/sup>, b = 7cm, c= 12cm. Solve the triangle completely.
\nSolution
\n<\/strong> Using cosine rule; a2<\/sup> = b2<\/sup> + c2<\/sup> \u2013 2bcCosA
\n a2<\/sup>= 72<\/sup> + 122<\/sup> \u2013 2(7\u00d712)Cos 1200<\/sup>
\n\t\t a2<\/sup>= 49 + 144 \u2013 168 (-0.5)
\n a2<\/sup> = 193 +84<\/p>\n

\u00a0\"\"\/ a =\u221a277
\n a = 16.6cm.
\n To find angle B, Cos B = a2<\/sup> + c2<\/sup> \u2013 b2<\/sup>
\n\t\t\"\"\/ 2ac
\n Cos B = 16.62<\/sup> + 122<\/sup> – 72<\/sup>
\n\t\t\"\"\/ 2x 16.6 x12<\/p>\n

\u00a0 Cos B = 275.56 + 144 \u2013 49
\n\"\"\/ 398.4
\n Cos B = 0.9301, B = cos-1<\/sup>0.9301, B = 21.50<\/sup>.
\n To find < C; < A + < B + <C = 1800<\/sup>,
\n C = 1800<\/sup> \u2013 1200<\/sup> \u2013 21.50<\/sup>, C = 38.50<\/sup>.
\n Hence, a = 16.6cm, B = 21.50<\/sup> and C = 38.50<\/sup>.
\nNB: 1. In any triangle, the longest side corresponds to the largest angle and the shortestside to the smallest angle<\/strong>.
\n 2. It is advisable to always find the smallest angle first , since the angle must be acute<\/strong>.<\/p>\n

\u00a0Evaluation
\n<\/strong>1.Calculate the angles of the \u2206s ABC whose sides are given in centimeters.Give the final answers to the nearest 0.10<\/sup>
\n\t\ta=5.2, b = 6.5cm ,c = 7.8<\/p>\n

\u00a0General Evaluation
\n<\/strong>1.Calculate the smallest angle in the triangle PQR such that p = 7.92m, q= 15.9m and c= 8.44m.
\n2.Calculate the length of the side opposite the given angle in \u2206 XYZ given that x =13.1m, y = 24.2m and Z = 47.80.<\/p>\n

\u00a0
\n\u00a0Revision Questions
\n<\/strong>1 Given that sin \u019f =5\/13 for 0<\u019f<900<\/sup> find
\na sin\u019f -cos\u019f
\nb cos \u019f -3
\n\t\t tan\u019f
\n2 If cos 3y=sin 2y find y for 0<y<900<\/sup><\/p>\n

\u00a0Reading Assignment
\n<\/strong>Essential Mathematics SSS2, page 180-181, exercise13.2, nos 11-15;exercise 13.4,page 185,nos 1a-1f.<\/p>\n

\u00a0Weekend Assignment
\n<\/strong>Objectives
\n<\/strong>Use the information below to answer question 1 \u2013 3. In \u2206ABC, a = 7.8m, b= 8.5m and B = 57.70<\/sup>. correct answers to 1 d.p.
\n1. Find A; A. 50.90<\/sup> B. 510<\/sup> C. 71.40<\/sup> D. 700<\/sup>
\n\t\t2. Find C; A. 510<\/sup> B. 71.40<\/sup> C. 710<\/sup> D. 800<\/sup>
\n\t\t3. What is c? A.10m B. 12m C. 9m D. 9.5m
\n4. In \u2206 ABC, b = 4cm, c= 5cm and A = 1150<\/sup>. Find a to 2 s.f. A. 7.66cm B 7.6cm C.8cm D.7.7cm
\n5. In \u2206PQR, p=1.8cm q = 2.5cm and r = 3.6cm. Calculate P. A. 27.50<\/sup> B. 300<\/sup> C. 320<\/sup> D.280 <\/sup><\/p>\n

\u00a0Theory
\n<\/strong>1.A triangle has sides of length 7cm, 8cm, 9cm. Express the cosine of the smallest angle of the triangle as a fraction in its lowest terms.
\n2.Solve the triangle completely in the \u2206ABC such that B = 34.50<\/sup>, c = 2.8cm, \u2061\u203d
\n\t\ta = 5.1cm. <\/p>\n

\u00a0
\n\t\t\t<\/strong>\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"

WEEK TWO Topic: Cosine and Sine Rule Relating to Triangle. Content -Sine Rule for Acute…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,278],"tags":[],"class_list":["post-3541","post","type-post","status-publish","format-standard","hentry","category-posts","category-third-term-ss2-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3541","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=3541"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3541\/revisions"}],"predecessor-version":[{"id":3542,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3541\/revisions\/3542"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=3541"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=3541"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=3541"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}