{"id":35021,"date":"2022-08-24T17:42:59","date_gmt":"2022-08-24T12:12:59","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35021"},"modified":"2022-08-24T17:42:59","modified_gmt":"2022-08-24T12:12:59","slug":"inter-1st-year-maths-1a-properties-of-triangles-solutions-ex-10b","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-1st-year-maths-1a-properties-of-triangles-solutions-ex-10b\/","title":{"rendered":"Inter 1st Year Maths 1A Properties of Triangles Solutions Ex 10(b)"},"content":{"rendered":"

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions<\/a> Inter 1st Year Maths 1A Properties of Triangles Solutions Exercise 10(b) will help students to clear their doubts quickly.<\/p>\n

Intermediate 1st Year Maths 1A Properties of Triangles Solutions Exercise 10(b)<\/h2>\n

All problems in this exercise have reference to \u0394ABC.<\/span><\/p>\n

I.<\/span><\/p>\n

Question 1.
\nExpress \\(\\Sigma r_{1} \\cot \\frac{A}{2}\\) in terms of s.
\nSolution:
\n\\(\\Sigma r_{1} \\cot \\frac{A}{2}\\) = \\(\\Sigma\\left(s \\tan \\frac{A}{2}\\right) \\cot \\frac{A}{2}\\)
\n= \u03a3s
\n= s + s + s
\n= 3s<\/p>\n

Question 2.
\nShow that \u03a3a cot A = 2(R + r).
\nSolution:
\nL.H.S = \u03a3a . cot A
\n= \u03a32R sin A \\(\\frac{\\cos A}{\\sin A}\\)
\n= 2R \u03a3 cos A
\n= \\(2 R\\left(1+4 \\sin \\frac{A}{2} \\sin \\frac{B}{2} \\sin \\frac{C}{2}\\right)\\) (From transformants)
\n= \\(2\\left(R+4 R \\sin \\frac{A}{2} \\cdot \\sin \\frac{B}{2} \\cdot \\sin \\frac{C}{2}\\right)\\)
\n= 2(R + r)
\n= R.H.S<\/p>\n

\"Inter<\/p>\n

Question 3.
\nIn \u2206ABC, prove that r1<\/sub> + r2<\/sub> + r3<\/sub> – r = 4R.
\nSolution:
\n\"Inter<\/p>\n

Question 4.
\nIn \u2206ABC, prove that r + r1<\/sub> + r2<\/sub> – r3<\/sub> = 4R cos C.
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 5.
\nIf r + r1<\/sub> + r2<\/sub> + r3<\/sub> then show that C = 90\u00b0.
\nSolution:
\n\"Inter<\/p>\n

II.<\/span><\/p>\n

Question 1.
\nProve that 4(r1<\/sub>r2<\/sub> + r2<\/sub>r3<\/sub> + r3<\/sub>r1<\/sub>) = (a + b + c)2<\/sup>
\nSolution:
\n\"Inter<\/p>\n

Question 2.
\nProve that \\(\\left(\\frac{1}{r}-\\frac{1}{r_{1}}\\right)\\left(\\frac{1}{r}-\\frac{1}{r_{2}}\\right)\\left(\\frac{1}{r}-\\frac{1}{r_{3}}\\right)=\\frac{a b c}{\\Delta^{3}}=\\frac{4 R}{r^{2} s^{2}}\\)
\nSolution:
\n\"Inter<\/p>\n

Question 3.
\nProve that r(r1<\/sub> + r2<\/sub> + r3<\/sub>) = ab + bc + ca – s2<\/sup>.
\nSolution:
\n\"Inter<\/p>\n

Question 4.
\nShow that \\(\\sum \\frac{r_{1}}{(s-b)(s-c)}=\\frac{3}{r}\\)
\nSolution:
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 5.
\nShow that \\(\\left(r_{1}+r_{2}\\right) \\tan \\frac{C}{2}=\\left(r_{3}-r\\right) \\cot \\frac{C}{2}=c\\)
\nSolution:
\n\"Inter<\/p>\n

Question 6.
\nShow that r1<\/sub>r2<\/sub>r3<\/sub> = \\(r^{3} \\cot ^{2} \\frac{A}{2} \\cdot \\cot ^{2} \\frac{B}{2} \\cdot \\cot ^{2} \\frac{C}{2}\\)
\nSolution:
\n\"Inter<\/p>\n

III.<\/span><\/p>\n

Question 1.
\nShow that cos A + cos B + cos C = 1 + \\(\\frac{r}{R}\\)
\nSolution:
\nL.H.S = cos A + cos B + cos C
\n= 2 cos(\\(\\frac{A+B}{2}\\)) cos(\\(\\frac{A-B}{2}\\)) + cos C
\n\"Inter<\/p>\n

Question 2.
\nShow that \\(\\cos ^{2} \\frac{A}{2}+\\cos ^{2} \\frac{B}{2}+\\cos ^{2} \\frac{C}{2}=2+\\frac{r}{2 R}\\)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 3.
\nShow that \\(\\sin ^{2} \\frac{A}{2}+\\sin ^{2} \\frac{B}{2}+\\sin ^{2} \\frac{C}{2}=1-\\frac{r}{2 R}\\)
\nSolution:
\n\"Inter<\/p>\n

Question 4.
\nShow that
\n(i) a = (r2<\/sub> + r3<\/sub>) \\(\\sqrt{\\frac{r r_{1}}{r_{2} r_{3}}}\\)
\n(ii) \u2206 = r1<\/sub>r2<\/sub> \\(\\sqrt{\\frac{4 R-r_{1}-r_{2}}{r_{1}+r_{2}}}\\)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 5.
\nProve that \\(r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+r^{2}\\) = 16R2<\/sup> – (a2<\/sup> + b2<\/sup> + c2<\/sup>).
\nSolution:
\n\"Inter<\/p>\n

Question 6.
\nIf p1<\/sub>, p2<\/sub>, p3<\/sub> are altitudes drawn from vertices A, B, C to the opposite sides of a triangle respectively, then show that
\n(i) \\(\\frac{1}{p_{1}}+\\frac{1}{p_{2}}+\\frac{1}{p_{3}}=\\frac{1}{r}\\)
\n(ii) \\(\\frac{1}{p_{1}}+\\frac{1}{p_{2}}-\\frac{1}{p_{3}}=\\frac{1}{r_{3}}\\)
\n(iii) p1<\/sub> . p2<\/sub> . p3<\/sub> = \\(\\frac{(a b c)^{2}}{8 R^{3}}=\\frac{8 \\Delta^{3}}{a b c}\\)
\nSolution:
\n\"Inter<\/p>\n

Question 7.
\nIf a = 13, b = 14, c = 15, show that R = \\(\\frac{65}{8}\\), r = 4, r1<\/sub> = \\(\\frac{21}{2}\\), r2<\/sub> = 12 and r3<\/sub> = 14.
\nSolution:
\na = 13, b = 14, c = 15
\ns = \\(\\frac{a+b+c}{2}\\)
\n= \\(\\frac{13+14+15}{2}\\)
\n= 21
\ns – a = 21 – 13 = 8
\ns – b = 21 – 14 = 7
\ns – c = 21 – 15 = 6
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 8.
\nIf r1<\/sub> = 2, r2<\/sub> = 3, r3<\/sub> = 6 and r = 1, prove that a = 3, b = 4 and c = 5.
\nSolution:
\n\"Inter<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Properties of Triangles Solutions Exercise 10(b) will help students to clear their doubts quickly. Intermediate 1st Year Maths 1A Properties of Triangles Solutions Exercise 10(b) All problems in this exercise have reference to \u0394ABC. I. Question 1. Express in terms of …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35021"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35021"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35021\/revisions"}],"predecessor-version":[{"id":35022,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35021\/revisions\/35022"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35021"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35021"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35021"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}