{"id":10593,"date":"2024-03-01T11:07:59","date_gmt":"2024-03-01T05:37:59","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=10593"},"modified":"2024-03-05T12:58:41","modified_gmt":"2024-03-05T07:28:41","slug":"ts-inter-1st-year-maths-1a-properties-of-triangles-important-questions","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-1st-year-maths-1a-properties-of-triangles-important-questions\/","title":{"rendered":"TS Inter 1st Year Maths 1A Properties of Triangles Important Questions"},"content":{"rendered":"

Students must practice these\u00a0TS Inter 1st Year Maths 1A Important Questions<\/a> Chapter 10 Properties of Triangles Important Questions to help strengthen their preparations for exams.<\/p>\n

TS Inter 1st Year Maths 1A Properties of Triangles Important Questions<\/h2>\n

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

Question 1.
\nIn \u0394ABC, If a=3,b=4 and sin A=3\/4, find the angle B.
\nSolution :
\nBy sine rule,\\(\\frac{a}{\\sin A}=\\frac{b}{\\sin B}\\)
\n\u21d2 sinB =\\(\\frac{b \\sin A}{a}=\\frac{4 \\times 3 \/ 4}{3}\\) = 1
\n\u21d2 sinB = 1 = B = 90\u00b0<\/p>\n

Question 2.
\nIf the lengths of the sides of a triangle are 3, 4, 5, find the circumradius of the triangle.
\nSolution:
\nSince 32<\/sup>+42<\/sup>= 52<\/sup> the triangle is right angled and hypotenuse = 5 = circum diameter.
\n\u2234 Circum radius = \\(\\frac{5}{2}\\) cm.<\/p>\n

\"TS<\/p>\n

Question 3.
\nlf a=6,b=5,c=9, then find the angle A.
\nSolution:
\n\"TS<\/p>\n

Question 4.
\nIn a \u0394 ABC, show that \u2211(b +c)cosA = 2s
\nSolution:
\nLH.S. = \u2211 (b + c) cos A
\n= (b + c) cos A+ (c + a) cos B + (a + b) cos C
\n= (bcosA+ acos B)+(ccosA + acosC) + (b cos C + c cos B)
\n= c+b+a
\n= a+b+c=2s=R.H.S.<\/p>\n

Question 5.
\nIn a \u0394 ABC, if
\n(a) (a+b+c)(b+c- a) = 3bc, findA.
\nSolution:
\nGiven (a + b + c)(b + c-a) = 3bc
\n\u21d2 (2s) 2(s – a) = 3 bc
\n\"TS<\/p>\n

(b) If a = 4, b = 5, c = 7, find cos \\(\\frac{B}{2}\\)
\nSolution:
\n2s = a+b+c = 4+5+7=16
\n\u21d2 s = 8
\n\u2234 s – b=8-5=3 and
\n\\(\\cos \\frac{B}{2}=\\sqrt{\\frac{s(s-b)}{a c}}=\\sqrt{\\frac{8 \\times 3}{4 \\times 7}}=\\sqrt{\\frac{6}{7}}\\)<\/p>\n

\"TS<\/p>\n

Question 6.
\nIf \u0394ABC, find \\(b \\cos ^2 \\frac{C}{2}+c \\cos ^2 \\frac{B}{2}\\)
\nSolution:
\n\"TS<\/p>\n

Question 7.
\nIf \\(\\tan \\frac{\\mathrm{A}}{2}=\\frac{5}{6}\\) and \\(\\tan \\frac{\\mathrm{C}}{2}=\\frac{2}{5}\\)
\nSolution:
\n\"TS
\n\u21d2 3s – 3b = s
\n\u21d2 2s = 3b \u21d2 a+b +c = 3b \u21d2 a = 2b
\n\u21d2 a, b, c are in A.P.<\/p>\n

Question 8.
\nIf \\(\\cot \\frac{A}{2}=\\frac{b+c}{a}\\),find angle B
\nSolution:
\n\\(\\cot \\frac{A}{2}=\\frac{b+c}{a}\\)
\n\"TS<\/p>\n

Question 9.
\nIf tan \\(\\left(\\frac{\\mathrm{C}-\\mathrm{A}}{2}\\right)=\\mathrm{k} \\cot \\frac{\\mathrm{B}}{2}\\),find K
\nSolution:
\nUsing Napiers rule,
\n\"TS<\/p>\n

Question 10
\nIn \u0394ABC, Show that \\(\\frac{b^2-c^2}{a^2}=\\frac{\\sin (B-C)}{\\sin (B+C)}\\)
\nSolution:
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 11.
\nShow that \\((b-c)^2 \\cos ^2 \\frac{A}{2}+(b+c)^2 \\sin ^2 \\frac{A}{2}=a^2\\)
\nSolution:
\nL.H.S= \\((b-c)^2 \\cos ^2 \\frac{A}{2}+(b+c)^2 \\sin ^2 \\frac{A}{2}\\)
\n\"TS<\/p>\n

Question 12.
\nIf \u0394ABC, Prove that \\(\\frac{1}{r_1}+\\frac{1}{r_2}+\\frac{1}{r_3}=\\frac{1}{r}\\)
\nSolution:
\n\"TS<\/p>\n

Question 13.
\nShow that r r1<\/sub> r2<\/sub> r3<\/sub> = \u03942<\/sup>
\nSolution:
\nL.H.S. r. r1<\/sub> . r2<\/sub> . r3
\n\"TS
\n<\/sub><\/p>\n

\"TS<\/p>\n

Question 14.
\nIn an equilateral triangle, find the value of\u00a0 \\(\\frac{\\mathbf{r}}{\\mathbf{R}}\\)
\nSolution:
\n\"TS<\/p>\n

Question 15.
\nThe perimeter of \u0394ABC is 12 cm and it’s in radius is 1 cm. Then find the area of the triangle.
\nSolution:
\nGiven 2s = 12 \u21d2 s = 6 cm. ; r = 1 cm
\nArea of \u0394 ABC,
\n\u0394 = rs = (1) (6) = 6sq.cm<\/p>\n

Question 16.
\nShow that r r1<\/sub> = (s – b) (s – c).
\nSolution:
\nL.H.S. = r. r1
\n\"TS
\n<\/sub><\/p>\n

Question 17.
\nIn \u0394ABC, is = 6sq. cm and s = 1.5cm find ‘r’
\nSolution:
\n\\(\\mathrm{r}=\\frac{\\Delta}{\\mathrm{s}}=\\frac{6}{1.5}=\\frac{6}{3 \/ 2}=4 \\mathrm{~cm}\\)<\/p>\n

Question 18.
\nShow that \\(r r_1 \\cot \\frac{A}{2}=\\Delta\\)
\nSolution:
\n\"TS<\/p>\n

\"TS<\/p>\n

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

Question 1.
\nIn a \\(\\Delta \\mathrm{ABC}\\),prove that \\(\\tan \\left(\\frac{B-C}{2}\\right)=\\frac{b-c}{b+c} \\cot \\frac{A}{2}\\)
\nSolution:
\n\"TS
\n\"TS<\/p>\n

Question 2.
\nProve that cot A+cot B+ cot C \\(=\\frac{a^2+b^2+c^2}{4 \\Delta}\\)
\nSolution:
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 3.
\nShow that r + r3<\/sub> + r1<\/sub> – r2<\/sub> = 4R cos B.
\nSolution:
\n\"TS<\/p>\n

Question 4.
\nIn a \u0394ABC, if r1<\/sub>= 8, r2<\/sub>=12, r3 <\/sub>= 24, find a, b, c.
\nSolution:
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 5.
\nIf the sides of a triangle are 13, 14, 15, then find the circum diameter.
\nSolution:
\nLet a= 13,b= 14,c= 15
\n\"TS<\/p>\n

Question 6.
\nShow that
\na2<\/sup>cotA+b2<\/sup>cotB+c2<\/sup>cotC = \\(\\frac{\\mathbf{a b c}}{\\mathbf{R}}\\)
\nSolution:
\n\"TS<\/p>\n

Question 7.
\nProve that a(b cos C – c cos B) = b2 <\/sup>– c2<\/sup>.
\nSolution:
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 8.
\nShow that \\(\\frac{c-b \\cos A}{b-c \\cos A}=\\frac{\\cos B}{\\cos C}\\)
\nSolution:
\nWe have c = a cos B + b cos A and
\nb = a cos C + c cos A
\n\\(\\text { L.H.S. }=\\frac{c-b \\cos A}{b-c \\cos A}\\)
\n\"TS<\/p>\n

Question 9.
\nShow that b2<\/sup> sin 2C + c2<\/sup>sin 2B = 2bc sin A.
\nSolution:
\n\"TS<\/p>\n

Question 10.
\nShow that
\n\\(a \\cos ^2 \\frac{A}{2}+b \\cos ^2 \\frac{B}{2}+c \\cos ^2 \\frac{C}{2}=s+\\frac{\\Delta}{R}\\)
\nSolution:
\n\"TS
\n\"TS<\/p>\n

Question 11.
\nIn a \u0394ABC, if acosA=bcosB, prove that the triangle is either isosceles or right angled.
\nSolution:
\na cos A = b cos B
\n\u21d2 2R sin A cos A = 2R sin B cos B
\n\u21d2 sin 2A = sin 2B
\n\u21d2 sin 2A = sin (180\u00b0 – 2B)
\n\u21d2 2A = 2B or 2A = 180\u00b0 – 2B
\n\u21d2 A=B or A=90\u00b0- B
\n\u21d2 A=B or A+B=90\u00b0
\n\u21d2\u00a0 c = 90\u00b0
\n\u2234 The triangle is isosceles or right angled.<\/p>\n

\"TS<\/p>\n

Question 12.
\nIf \\(\\cot \\frac{A}{2}: \\cot \\frac{B}{2}: \\cot \\frac{C}{2}\\) = 3 : 5 : 7, Show that a : b : c =6 : 5 : 4.
\nSolution:
\n\"TS
\nThens – a = 3ks\u00a0 – b = 5ks – c= 7k
\n\u2234 3s – (a + b + c) = 15k
\n= 3s – 2s= 15k=s= 15k
\n\u2234 a = s – 3k = a = 12k
\nb = s – 5k=b= 10k
\nc = s – 7k c = 8k
\n\u2234 a : b : c = 12k : 10k : 8k = 6 : 5 : 4.<\/p>\n

Question 13.
\nProve that a3<\/sup>cos(B – C)+b3<\/sup>cos(C – A) + cos3<\/sup> cos (A – B) = 3 abc.
\nSolution:
\n\u2211a3<\/sup>cos(B-C)
\n= \u2211 a2<\/sup>. a cos (B – C)
\n=\u2211 a2 <\/sup>2R sin A cos(B – C)
\n= \u2211 a2 <\/sup>2sin(B + C) cos (B – C)
\n= \\(R \\Sigma a^2\\left(2 \\frac{b}{2 R} \\cos B+2 \\frac{c}{2 R} \\cos C\\right)\\)
\n= \u2211 a2<\/sup>(bcos B +ccosC)
\n= a2<\/sup> (bcos B + ccos C) + b2<\/sup>(ccos C + acosA) + c2<\/sup> (a cos A + b cos B)
\n= ab (a cos B+ b cos A) + bc (b cos C + c cos B) + ca(ccos A+acos C)
\n= abc + bca + cab
\n= 3 abc = R.H.S.<\/p>\n

Question 14.
\nIf P1<\/sub> P2<\/sub> p3<\/sub> are the altitudes of a \u0394ABC to the opposite sides show that
\n\"TS
\nSolution:
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 15.
\nThe angle of elevation of the top point P of the vertical tower PQ of height h from a point A is 450 and from a point B Is 600, where B is a point at a distance 30 meters from the point A measured along the line AB which makes an angle 30\u00b0 with AQ. Find the height of the tower.
\nSolution:
\n\"TS<\/p>\n

Question 16.
\nTwo trees A and B are on the same side of a tiver. From a point C in the river, the distances of the trees A and B are 250 m and 300 m respectively. If the angle C is 450, find the distance between the trees (use \\(\\sqrt{2}\\) = 1.414).
\nSolution:
\n\"TS
\nGiven AC = 300 m and BC = 250 m and in the \u0394 ABC, using cosine rule
\nAB2<\/sup> = AC2<\/sup> + BC2<\/sup> – 2AC.BC. cos 45\u00b0
\n= (300)2<\/sup> + (250)2<\/sup> -2 (300) (250) – \\(\\frac{1}{\\sqrt{2}}\\)
\n= 46450
\n\u2234 AB = 215.5 m (approximately)<\/p>\n

Question 17.
\nExpress \\(\\frac{a \\cos A+b \\cos B+c \\cos C}{a+b+c}\\) in terms of R and r.
\nSolution:
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 18.
\nIf a = 13, b = 14, c = 15, find r1<\/sub>.
\nSolution:
\n\\(\\frac{\\mathrm{a}+\\mathrm{b}+\\mathrm{c}}{2}=\\mathrm{s} \\Rightarrow \\mathrm{s}=\\frac{13+14+15}{2}=\\frac{42}{2}=21\\)
\ns – a = 21- 13=8
\ns – b = 21 – 14 = 7 and
\ns – c = 21 – 15 = 6
\n\u03942<\/sup> = 21 x 8 x 7 6
\n\u0394 = 7 x 12 = 84sq. units
\nr1 <\/sub>= \\(\\frac{\\Delta}{s-a}=\\frac{84}{8}\\) = 10.5 units<\/p>\n

Question 19.
\nIf r r2<\/sub> = r1<\/sub> r3<\/sub>, then find B.
\nSolution:
\nGiven r r2<\/sub> = r1<\/sub> r3<\/sub>
\n\"TS<\/p>\n

Question 20.
\nIn a \u0394 ABC, show that the sides a, b, and c are in A.P. If and only If r1<\/sub>, r2<\/sub>, r3<\/sub> are in H.P.
\nSolution :
\nr1<\/sub>, r2<\/sub>, r3<\/sub> are in H.P.
\n\"TS
\n\u21d4 s – a, s – b, s – c are in A.P
\n\u21d4 – a, – b, – c are in AP.
\n\u21d4 a, b, c are in A.P.<\/p>\n

Question 21.
\nIf A = 90\u00b0, show that 2 (r + R) = b + c
\nSolution:
\nL.H.S. = 2 (r + R)
\n= 2r + 2R
\n= 2(s – a) tan\\(\\frac{\\mathrm{A}}{2}\\) + 2R . 1
\n= 2 (s – a) tan 45\u00b0 + 2R sinA (\u2235 A = 90\u00b0)
\n= (2s – 2a) + a
\n= 2s – a = a + b + c – a = b + c = R.H.S.<\/p>\n

Question 22.
\nShow that
\n\"TS
\nSolution:
\n\"TS<\/p>\n

Question 23.
\nProve that \u03a3 (r1<\/sub> + r) \\(\\tan \\left(\\frac{B-C}{2}\\right)=0\\)
\nSolution:
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 24.
\nShow that \\(\\left(r_1+r_2\\right) \\sec ^2 \\frac{C}{2}=\\left(r_2+r_3\\right) \\sec ^2 \\frac{A}{2} =\\left(r_3+r_1\\right) \\sec ^2 \\frac{B}{2}\\)
\nSolution:
\n\"TS
\nHence the triangle is right angled at A.<\/p>\n

Question 25.
\nShow that \\(\\frac{a b-r_1 r_2}{r_3}=\\frac{b c-r_2 r_3}{r_1}=\\frac{c a-r_3 r_1}{r_2}\\)
\nSolution:
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

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

Question 1.
\nIf A, A1<\/sub>, A2<\/sub>, A3<\/sub> are the areas of in circle and exercise of a triangle respectively then prove that
\n\\(\\frac{1}{\\sqrt{A_1}}+\\frac{1}{\\sqrt{A_2}}+\\frac{1}{\\sqrt{A_3}}=\\frac{1}{\\sqrt{A^2}}\\)
\nSolution:
\nLet r1<\/sub>, r2<\/sub>, r3<\/sub> and r be the radii of excircies and incircie respectively whose areas are A1<\/sub>, A2<\/sub>, A3<\/sub> and A.
\n\"TS<\/p>\n

Question 2.
\nIn a \u0394 ABC prove that
\n\\(\\frac{r_1}{b c}+\\frac{r_2}{c a}+\\frac{r_3}{a b}=\\frac{1}{r}-\\frac{1}{2 R}\\)
\nSolution:
\n\"TS
\n\"TS
\n\"TS<\/p>\n

Question 3.
\nIf (r2<\/sub> – r1<\/sub>) (r3<\/sub> – r3<\/sub>) = 2r2<\/sub>r3<\/sub>, show that A=90\u00b0
\nSolution:
\nGiven (r2<\/sub> – r1<\/sub>) (r3<\/sub> – r1<\/sub>) = 2r2<\/sub>r3
\n<\/sub>\"TS<\/p>\n

\"TS<\/p>\n

Question 4.
\nProve that \\(\\frac{r_1\\left(r_2+r_3\\right)}{\\sqrt{r_1 r_2+r_2 r_3+r_3 r_1}}=a\\)
\nSolution:
\n\"TS
\n\"TS<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these\u00a0TS Inter 1st Year Maths 1A Important Questions Chapter 10 Properties of Triangles Important Questions to help strengthen their preparations for exams. TS Inter 1st Year Maths 1A Properties of Triangles Important Questions I. Question 1. In \u0394ABC, If a=3,b=4 and sin A=3\/4, find the angle B. Solution : By sine rule, … Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/10593"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=10593"}],"version-history":[{"count":6,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/10593\/revisions"}],"predecessor-version":[{"id":11333,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/10593\/revisions\/11333"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=10593"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=10593"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=10593"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}