{"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 I. <\/span><\/p>\n Question 1. Question 2. <\/p>\n Question 3. Question 4. Question 5. (b) If a = 4, b = 5, c = 7, find cos \\(\\frac{B}{2}\\) <\/p>\n Question 6. Question 7. Question 8. Question 9. Question 10 <\/p>\n Question 11. Question 12. Question 13. <\/p>\n Question 14. Question 15. Question 16. Question 17. Question 18. <\/p>\n II.<\/span><\/p>\n Question 1. Question 2. <\/p>\n Question 3. Question 4. <\/p>\n Question 5. Question 6. Question 7. <\/p>\n Question 8. Question 9. Question 10. Question 11. <\/p>\n Question 12. Question 13. Question 14. <\/p>\n Question 15. Question 16. Question 17. <\/p>\n Question 18. Question 19. Question 20. Question 21. Question 22. Question 23. <\/p>\n Question 24. Question 25. <\/p>\n III.<\/span><\/p>\n Question 1. Question 2. Question 3. <\/p>\n Question 4. 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}]}}TS Inter 1st Year Maths 1A Properties of Triangles Important Questions<\/h2>\n
\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
\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
\nlf a=6,b=5,c=9, then find the angle A.
\nSolution:
\n<\/p>\n
\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
\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<\/p>\n
\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
\nIf \u0394ABC, find \\(b \\cos ^2 \\frac{C}{2}+c \\cos ^2 \\frac{B}{2}\\)
\nSolution:
\n<\/p>\n
\nIf \\(\\tan \\frac{\\mathrm{A}}{2}=\\frac{5}{6}\\) and \\(\\tan \\frac{\\mathrm{C}}{2}=\\frac{2}{5}\\)
\nSolution:
\n
\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
\nIf \\(\\cot \\frac{A}{2}=\\frac{b+c}{a}\\),find angle B
\nSolution:
\n\\(\\cot \\frac{A}{2}=\\frac{b+c}{a}\\)
\n<\/p>\n
\nIf tan \\(\\left(\\frac{\\mathrm{C}-\\mathrm{A}}{2}\\right)=\\mathrm{k} \\cot \\frac{\\mathrm{B}}{2}\\),find K
\nSolution:
\nUsing Napiers rule,
\n<\/p>\n
\nIn \u0394ABC, Show that \\(\\frac{b^2-c^2}{a^2}=\\frac{\\sin (B-C)}{\\sin (B+C)}\\)
\nSolution:
\n<\/p>\n
\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<\/p>\n
\nIf \u0394ABC, Prove that \\(\\frac{1}{r_1}+\\frac{1}{r_2}+\\frac{1}{r_3}=\\frac{1}{r}\\)
\nSolution:
\n<\/p>\n
\nShow that r r1<\/sub> r2<\/sub> r3<\/sub> = \u03942<\/sup>
\nSolution:
\nL.H.S. r. r1<\/sub> . r2<\/sub> . r3
\n
\n<\/sub><\/p>\n
\nIn an equilateral triangle, find the value of\u00a0 \\(\\frac{\\mathbf{r}}{\\mathbf{R}}\\)
\nSolution:
\n<\/p>\n
\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
\nShow that r r1<\/sub> = (s – b) (s – c).
\nSolution:
\nL.H.S. = r. r1
\n
\n<\/sub><\/p>\n
\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
\nShow that \\(r r_1 \\cot \\frac{A}{2}=\\Delta\\)
\nSolution:
\n<\/p>\n
\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
\n<\/p>\n
\nProve that cot A+cot B+ cot C \\(=\\frac{a^2+b^2+c^2}{4 \\Delta}\\)
\nSolution:
\n
\n<\/p>\n
\nShow that r + r3<\/sub> + r1<\/sub> – r2<\/sub> = 4R cos B.
\nSolution:
\n<\/p>\n
\nIn a \u0394ABC, if r1<\/sub>= 8, r2<\/sub>=12, r3 <\/sub>= 24, find a, b, c.
\nSolution:
\n<\/p>\n
\nIf the sides of a triangle are 13, 14, 15, then find the circum diameter.
\nSolution:
\nLet a= 13,b= 14,c= 15
\n<\/p>\n
\nShow that
\na2<\/sup>cotA+b2<\/sup>cotB+c2<\/sup>cotC = \\(\\frac{\\mathbf{a b c}}{\\mathbf{R}}\\)
\nSolution:
\n<\/p>\n
\nProve that a(b cos C – c cos B) = b2 <\/sup>– c2<\/sup>.
\nSolution:
\n<\/p>\n
\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<\/p>\n
\nShow that b2<\/sup> sin 2C + c2<\/sup>sin 2B = 2bc sin A.
\nSolution:
\n<\/p>\n
\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
\n<\/p>\n
\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
\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
\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
\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
\nIf P1<\/sub> P2<\/sub> p3<\/sub> are the altitudes of a \u0394ABC to the opposite sides show that
\n
\nSolution:
\n
\n<\/p>\n
\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<\/p>\n
\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
\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
\nExpress \\(\\frac{a \\cos A+b \\cos B+c \\cos C}{a+b+c}\\) in terms of R and r.
\nSolution:
\n
\n<\/p>\n
\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
\nIf r r2<\/sub> = r1<\/sub> r3<\/sub>, then find B.
\nSolution:
\nGiven r r2<\/sub> = r1<\/sub> r3<\/sub>
\n<\/p>\n
\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
\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
\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
\nShow that
\n
\nSolution:
\n<\/p>\n
\nProve that \u03a3 (r1<\/sub> + r) \\(\\tan \\left(\\frac{B-C}{2}\\right)=0\\)
\nSolution:
\n<\/p>\n
\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
\nHence the triangle is right angled at A.<\/p>\n
\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
\n<\/p>\n
\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<\/p>\n
\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
\n
\n<\/p>\n
\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><\/p>\n
\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
\n<\/p>\n","protected":false},"excerpt":{"rendered":"