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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
If `a^2, b^2,c^2`are in A.P., then prove that `tanA ,tanB ,tanC`are in H.P. |
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Answer» `a^(2), b^(2), c^(2)` are in A.P. `rArr b^(2) - a^(2) = c^(2) - b^(2)` `rArr sin^(2)B - sin^(2) A = sin^(2) C - sin^(2) B` (Using sine Rule) or `sin(B + A) sin (B - A) = sin (C + B) sin (C - B)` or `sin C (sin B cos A - cos B sin A)` `= sin A (sin C cos B - cos C sin B)` Dividing both sides by sin A sin B sin C, we get `cot A - cot B = cot B - cot C` `rArr cot A, cot B, cot C` are in A.P. `rArr tan A, tan B, tan C` are in H.P. |
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| 52. |
If A is the area and 2s is the sum of the sides of a triangle, thenA. `A le (s^(2))/(4)`B. `A le (s^(2))/(3sqrt3)`C. `A lt (s^(2))/(sqrt3)`D. none of these |
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Answer» Correct Answer - A::B We have `2s = a + b + c, A^(2) = s (s-a) (s-b) (s-c)` Now, `A.M. ge G.M.` `rArr (s+(s-a) +(s-b)+(s-c))/(4)` `ge [s (s-a) (s-b) (s-c)]^(1//4)` `rArr (4s-2s)/(4) ge [A^(2)]^(1//4)` `rArr ((s)/(2)) ge A^(1//2) or A le ((s^(2))/(4))` Also, `((s-a) + (s-b) + (s-c))/(3)` `ge [(s-a) (s-b) (s-c)]^(1//3)` or `(s)/(3) ge [(A^(2))/(s)]^(1//3) or (A^(2))/(s) le (s^(3))/(27) or A le (s^(2))/(3sqrt3)` |
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| 53. |
In a triangle ABC, the altitude from A is not less than BC and the altitude from B is not less than AC. The triangleA. right angledB. isoscelesC. obtuse angledD. equilateral |
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Answer» Correct Answer - A Given `c sin B ge a` `rArr sin C sin B ge sin A` and `a sin C ge b` `rArr sin C sin A ge sin B` `rArr sin C sin A ge sin B ge (sinA)/(sinC)` `:. Sin^(2) C ge 1` `rArr sin C = 1` `rArr angle C " is " (pi)/(2)` |
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| 54. |
In any triangle `ABC, (a^(2) + b^(2) + c^(2))/(R^(2))` has the maximum value ofA. 3B. 6C. 9D. none of these |
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Answer» Correct Answer - C In triangle ABC, `(a^(2) + b^(2) + c^(2))/(R^(2)) = 4(sin^(2) A + sin^(2) B + sin^(2) C)` `= 8(1 + cos A cos B cos C) le 8(1 + (1)/(8)) le 9` |
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| 55. |
In `DeltaABC, " if " b^(2) + c^(2) = 2a^(2)`, then value of `(cotA)/(cot B + cot C)` isA. `(1)/(2)`B. `(3)/(2)`C. `(5)/(2)`D. `(5)/(3)` |
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Answer» Correct Answer - A `(cotA)/(cotB cotC) = ((R(b^(2) + c^(2) -a^(2)))/(abc))/((R(a^(2) + c^(2) -b^(2)))/(abc) + (R(a^(2) + b^(2) -c^(2)))/(abc))` `= (b^(2) + c^(2) -a^(2))/(2a^(2))` `= (2a^(2) -a^(2))/(2a^(2))` `= (1)/(2)` |
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| 56. |
Lengths of the tangents from A,B and C to the incircle are in A.P., thenA. `r_(1) , r_(2) r_(3)` are in H.PB. `r_(1), r_(2), r_(3)` are in APC. a, b, c are in A.PD. `cos A = (4c -3b)/(2c)` |
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Answer» Correct Answer - A::C::D Given that `s -a, s-b, and s-c` are in A.P. `rArr a, b ,c` are in A.P. `rarr (Delta)/(s-a), (Delta)/(s-b),(Delta)/(s-c)` are in H.P `rArr r_(1), r_(2), r_(3)` are in H.P. Also, `cos A = (b^(2) + c^(2) -a^(2))/(2bc)` `= (b^(2) + c^(2) - (2b -c)^(2))/(2bc) = (4c - 3b)/(2c)` |
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| 57. |
The area of any cyclic quadrilateral ABCD is given by `A^(2) = (s -a) (s-b) (s-c) (s-d)`, where `2s = a + b ++ c + d, a, b, c and d` are the sides of the quadrilateral Now consider a cyclic quadrilateral ABCD of area 1 sq. unit and answer the following question The minium perimeter of the quadrilateral isA. 4B. 2C. 1D. none of these |
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Answer» Correct Answer - A Using A.M. `ge` G.M for `s -a, s-b, s-c, s-d`, we have `(s-a + s-b + s-c + s-d)/(4) ge sqrtA` or `(4s -2s)/(4) ge 1` or `2s ge 4` |
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| 58. |
If in `Delta ABC`, sides a, b, c are in A.P. thenA. `B gt 60^(@)`B. `B lt 60^(@)`C. `B le 60^(@)`D. `B = |A - C|` |
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Answer» Correct Answer - C `2b = a + c rArr 2 sin B = sin A + sin C` or `4 sin.(B)/(2) cos.(B)/(2) = 2 sin ((A +C)/(2)) cos ((A -C)/(2))` `rArr sin ((B)/(2)) = (1)/(2) cos (A - C) le (1)/(2) rArr B le 60^(@)` |
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| 59. |
In triangle ABC, `R (b + c) = a sqrt(bc)`, where R is the circumradius of the triangle. Then the triangle isA. isosceles but not rightB. right but not isoscelesC. right isoscelesD. equilateral |
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Answer» Correct Answer - C `R(b +c) = a sqrt(bc) = 2R sin A sqrt(bc)` `:. sin A = (b +c)/(2sqrt(bc))` Now `sin A le 1` `rArr (b + c)/(2sqrt(bc)) le 1` or `(sqrtb - sqrtc)^(2) le 0` or `b = c` `rArr sin A = 1` `rArr A = 90^(@) and b = c` Hence, the triangle is right isosceles. |
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| 60. |
Sides of `DeltaABC` are in A.P. If `a lt "min" {b,c}`, then cos A may be equal toA. `(4b - 3c)/(2b)`B. `(3c -4b)/(2c)`C. `(4c -3b)/(2b)`D. `(4c -3b)/(2c)` |
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Answer» Correct Answer - A::D Sides are in A.P. and `a lt` min {b,c} Therefore, order of A.P. can be b, c, a, or c, b, a Case I : If `2c = a + b`, then `cos A = (b^(2) + c^(2) -a^(2))/(2bc) = (b^(2) + c^(2) -(2c -b)^(2))/(2bc) = (4b - 3c)/(2b)` Case II If `2b = a + c`, then `cos A = (b^(2) + c^(2) -a^(2))/(2bc) = (b^(2) + c^(2) - (2b -c)^(2))/(2bc) = (4c -3b)/(2c)` |
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| 61. |
Let a = 6, b = 3 and `cos (A -B) = (4)/(5)` Value of `sin A` is equal toA. `(1)/(2sqrt5)`B. `(1)/(sqrt3)`C. `(1)/(sqrt5)`D. `(2)/(sqrt5)` |
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Answer» Correct Answer - D `cos(A -B) = (4)/(5)` `rArr (1 - tan^(2).(A-B)/(2))/(1 + tan^(2).(A-B)/(2)) = (4)/(5)` or `"tan"^(2) (A -B)/(2) = (1)/(9)` or `"tan" (A -B)/(2) = (1)/(3)` Now, `tan.(A-B)/(2) = (a-b)/(a+b) "cot"(C)/(2)` or `(1)/(3) = (6-3)/(6+3) "cot"(C)/(2)` or `"cot"(C)/(2) = 1 " or " C = (pi)/(2)` Area of triangle `= (1)/(2) ab sin C = (1)/(2) xx 6 xx 3 xx 1 = 9` `(a)/(sinA) = (sqrt(a^(2) + b^(2)))/(1)` or `(6)/(sinA) = sqrt45` or `sin A = (2)/(sqrt5)` |
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| 62. |
In `DeltaABC`, if `(sin A)/(c sin B) + (sin B)/(c) + (sin C)/(b) = (c)/(ab) + (b)/(ac) + (a)/(bc)`, then the value of angle A isA. `120^(@)`B. `90^(@)`C. `60^(@)`D. `30^(@)` |
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Answer» Correct Answer - B `(sinA)/(c sin B) + (sin B)/(c) + (sin C)/(b) = (c)/(ab) + (b)/(ac) + (a)/(bc)` or `(a)/(bc) + (sinB)/(c) + (sin C)/(b) = (c)/(ab) + (b)/(ab) + (a)/(bc)` or `(sin B)/(c) + (sin C)/(b) = (c)/(ab) + (b)/(ac)` or `(b sin B + c sin C)/(bc) = (c^(2) + b^(2))/(abc)` or `a = (b^(2) + c^(2))/(b sin B + c sin C)` `= (b (2R sin B) + c(2R sin C))/(b sin B + c sin C)` `= 2R` `rArr angleA = (pi)/(2)` |
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| 63. |
If H is the orthocentre of triangle ABC, R = circumradius and `P = AH + BH + CH`, thenA. `P = 2 (R + r)`B. max. of P is 3RC. min. of P is 3RD. `P = 2 (R -r)` |
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Answer» Correct Answer - A::B `AH = 2R cos A, BH = 2R cos B, CH = 2R cos C` `:. P = 2R (cos A + cos B+ cos C)` `= 2R (1+(r)/(R))` `=2 (R + r)` We know that in any triangle, `r le(R)/(2)` `:. P le 2R + R` `rArr P le 3R` |
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| 64. |
In any triangle. `if(a^2-b^2)/(a^2+b^2)=("sin"(A-B))/("sin"(A+B))`, then prove that the triangle is either right angled or isosceles. |
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Answer» `(a^(2) - b^(2))/(a^(2) + b^(2)) = (sin (A - B))/(sin (A + B))` or `(4R^(2) sin^(2) A - 4R^(2) sin^(2) B)/(4R^(2) sin^(2) A + 4R^(2) sin^(2) B) = (sin (A - B))/(sin (A + B))` (Using Sine Rule) or `(sin (A + B) sin (A- B))/(sin^(2) A + sin^(2)B) = (sin(A - B))/(sin(A + B))` `rArr sin(A - B) = 0 " or " (sin (pi C))/(sin^(2) A + sin^(2) B) = (1)/(sin (pi - C))` or `A = B " or " sin^(2) C = sin^(2) A + sin^(2) B` or `A = B " or " c^(2) = a^(2) + b^(2)` [from the sine rule] Therefore, the triangle is isosceles or right angled |
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| 65. |
Let a = 6, b = 3 and `cos (A -B) = (4)/(5)` Angle C is equal toA. `(3pi)/(4)`B. `(pi)/(4)`C. `(pi)/(2)`D. none of these |
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Answer» Correct Answer - C `cos(A -B) = (4)/(5)` `rArr (1 - tan^(2).(A-B)/(2))/(1 + tan^(2).(A-B)/(2)) = (4)/(5)` or `"tan"^(2) (A -B)/(2) = (1)/(9)` or `"tan" (A -B)/(2) = (1)/(3)` Now, `tan.(A-B)/(2) = (a-b)/(a+b) "cot"(C)/(2)` or `(1)/(3) = (6-3)/(6+3) "cot"(C)/(2)` or `"cot"(C)/(2) = 1 " or " C = (pi)/(2)` Area of triangle `= (1)/(2) ab sin C = (1)/(2) xx 6 xx 3 xx 1 = 9` `(a)/(sinA) = (sqrt(a^(2) + b^(2)))/(1)` or `(6)/(sinA) = sqrt45` or `sin A = (2)/(sqrt5)` |
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| 66. |
A circle of radius 4 cm is inscribed in `DeltaABC`, which touches side BC at D. If BD = 6 cm, DC = 8 cm thenA. the triangle is necessarily acute angledB. `tan.(A)/(2) = (4)/(7)`C. perimeter of the triangle ABC is 42 cmD. area of `DeltaABC " is " 84 cm^(2)` |
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Answer» Correct Answer - A::B::C::D `"tan"(B)/(2) = (2)/(3), "tan"(C)/(2) = (1)/(2) rArr tan.(A)/(2) = (4)/(7)`, `rArr "tan"(B)/(2) "tan"(C)/(2) = (s-a)/(s) = (1)/(3)` `rArr 2s = 3a = 42` `:. Delta = rs = 84 cm^^(2)` Thus, `"tan"(A)/(2), "tan"(B)/(2), "tan"(C)/(2)` all are less than 1 Hence, all angles of the triangle are acute |
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| 67. |
Find the value of `(a^2+b^2+c^2)/R^2` in any right-angled triangle. |
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Answer» Correct Answer - 8 Let `angle A = (pi)/(2) rArr a^(2) = b^(2) + c^(2) and 2R = a` `rArr (a^(2) + b^(2) c^(2))/(R^(2)) = (2a^(2))/(R^(2)) = (2a^(2) xx4)/(a^(2)) = 8` |
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| 68. |
In any triangle, the minimum value of `r_(1) r_(2) r_(3) //r^(3)` is equal toA. 1B. 9C. 27D. none of these |
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Answer» Correct Answer - C `((r_(1))/(r)(r_(2))/(r)(r_(3))/(r))^(1//3) ge (3)/((r)/(r_(1)) + (r)/(r_(2)) + (r)/(3))` (using G.M. `ge` H.M.) `ge (3)/((s-a)/(s) + (s-b)/(s) + (s-c)/(s)) ge (3s)/(s-a+s -b + s-c) ge (3s)/(s)` `rArr (r_(1))/(r) (r_(2))/(r) (r_(3))/(r) ge 27` |
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| 69. |
Prove that `(b+c)cosA+(c+a)cosB+(a+b)cosC=2sdot` |
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Answer» `(b + c) cos A + (c + a) cos B + (a + b) cos C` `= (b cos A + a cos B) + (c cos A + a cos C) + (b cos C + c cos B) = c + b + a = 2s` |
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| 70. |
Prove that `cosA+cosB+cosC=1+r/R` |
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Answer» `cosA + cos B + cosC = 1 + 4 sin((A)/(2)) sin((B)/(2)) sin((C)/(2))` `= 1 + ([4R sin (A//2) sin (B//2) sin(C//2)])/(R) = 1 + (r)/(R)` |
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| 71. |
A variable triangle `A B C`is circumscribed about a fixed circle of unit radius. Side `B C`always touches the circle at D and has fixed direction. If B and C vary insuch a way that (BD) (CD)=2, then locus of vertex Awill be a straight line.parallel to side BCperpendicular to side BCmaking an angle `(pi/6)`with BCmaking an angle `sin^(-1)(2/3)`with `B C`A. parallel to side BCB. perpendicular to side BCC. making an angle `(pi//6)` with BCD. making an angle `sin^(-1) (2//3)` with BC |
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Answer» Correct Answer - A `BD = (s -b), CD = (s-c)` `rArr (s-b) (s-c) =2` or `s(s-a) (s-b) (s-c) = 2 s(s -a)` or `Delta^(2) = 2s (s-a)` or `(Dleta^(2))/(s^(2)) = (2(s-a))/(s)` (using `Delta = rs`) or `r^(2) = (2(s-a))/(s)` or `(a)/(s)` = constant Now, `Delta = (1)/(2) aH_(a)`, where `H_(a)` is the distance of A from BC. Thus, `(Delta)/(2) = (1)/(2) (aH_(a))/(s) = 1 " or " H_(a) = (2s)/(a)` = constant Therefore, locus of A will be a straight line parallel to side BC |
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| 72. |
Let a = 6, b = 3 and `cos (A -B) = (4)/(5)` Area (in sq. units) of the triangle is equal toA. 9B. 12C. 11D. 10 |
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Answer» Correct Answer - A `cos(A -B) = (4)/(5)` `rArr (1 - tan^(2).(A-B)/(2))/(1 + tan^(2).(A-B)/(2)) = (4)/(5)` or `"tan"^(2) (A -B)/(2) = (1)/(9)` or `"tan" (A -B)/(2) = (1)/(3)` Now, `tan.(A-B)/(2) = (a-b)/(a+b) "cot"(C)/(2)` or `(1)/(3) = (6-3)/(6+3) "cot"(C)/(2)` or `"cot"(C)/(2) = 1 " or " C = (pi)/(2)` Area of triangle `= (1)/(2) ab sin C = (1)/(2) xx 6 xx 3 xx 1 = 9` `(a)/(sinA) = (sqrt(a^(2) + b^(2)))/(1)` or `(6)/(sinA) = sqrt45` or `sin A = (2)/(sqrt5)` |
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| 73. |
In a triangle `ABC, (a)/(b) = (2)/(3) and sec^(2) A = (8)/(5)`. Find the number of triangle satisfying these conditions |
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Answer» Correct Answer - two We have `(a)/(b) = (b)/(3) = k` and `sec^(2) A = (8)/(5)` `rArr cos^(2) A = (5)/(8)` `rArr (5)/(8) = ((9k^(2) + c^(2) - 4k^(2))/(6kc))^(2) = ((5k^(3) + c^(2))/(6kc))^(2)` `rArr 45k^(2) c^(2) = 50 k^(4) + 20 k^(2) c^(2) + 2c^(4)` `rArr 2c^(4) - 25 k^(2) c^(2) + 50k^(4) = 0` `rArr c^(2) = (25 k^(2) +- sqrt(625 k^(4) - 400 k^(4)))/(4)` `= (25k^(2) +- 15 k^(2))/(4) = 10 k^(2), (5)/(2) k^(2)` There are two possible valid values of `c^(2)`. Hence there exist two triange satisfying the given conditions |
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| 74. |
Let the angles `A , Ba n dC`of triangle `A B C`be in `AdotPdot`and let `b : c`be `sqrt(3):sqrt(2)`. Find angle `Adot` |
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Answer» Correct Answer - `(5pi)/(12)` Angles are in A.P., so, `B = (pi)/(3)` Also, `(b)/(c) = (sqrt3)/(sqrt2) " or " (b)/(sqrt3) = (c)/(sqrt2) " or " (b)/((sqrt3)/(2)) = (c)/((1)/(sqrt2))` But we know that `(b)/(sin B) = (c)/(sin C) = (a)/(sin A)` `:. B = (pi)/(3), C = (pi)/(4) rArr A = (5pi)/(12)` |
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| 75. |
In a triangle, if the angles `A , B ,a n dC`are in A.P. show that `2cos1/2(A-C)=(a+c)/(sqrt(a^2-a c+c^2))` |
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Answer» Since angles A, B, and C are in A.P., we have `A + C = 2B` But, `A + B + C = 180^(@) " or " 3B = 180^(@) " or " B = 60^(@)` Now, `cos B = (1)/(2) = (a^(2) + c^(2) - b^(2))/(2ac)` or `a^(2) + c^(2) - b^(2) = ac` or `a^(2) - ac + c^(2) = b^(2)` `:. (a + c)/(sqrt(a^(2) - ac + c^(2))) = (a + c)/(b)` `= (2R(sin A + sin C))/(2R sin B)` `= (2 sin ((A+C)/(2)) cos((A - C)/(2)))/(sin B)` `= (2 sin 60^(@))/(sin 60^(@)) cos ((A - C)/(2))` `= 2 cos ((A - C)/(2))` |
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| 76. |
Let `alt=blt=c`be the lengths of the sides of a triangle. If `a^2+b^2 |
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Answer» `a^(2) + b^(2) lt c^(2)` `rArr a^(2) + b^(2) lt a^(2) + b^(2) - 2ab cos C` `rArr cos C lt 0` Hence, C is an obtuse angles |
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| 77. |
In a triangle ABC, if `(cosA)/a=(cosB)/b=(cosC)/c` and the side `a =2`, then area of triangle is |
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Answer» Correct Answer - `sqrt3` sq. unit `(cos A)/(a) = (cos B)/(b) = (cos C)/(c)` or `(cos A)/(2R sin A) = (cos B)/(2R sin B) = (cos C)/(2R sin C)` or `tan A = tan B = tan C` Hence, triangle is equilateral. Therefore, Area of triangle `= (sqrt3)/(4) a^(2) = sqrt3` (as a = 2) |
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| 78. |
In `Delta ABC`, if a = 10 and `b cot B + c cot C = 2(r + R)` then the maximum area of `DeltaABC` will beA. 50B. `sqrt50`C. 25D. 5 |
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Answer» Correct Answer - C `b cot B + c cot C = 2 (r + R)` `rARr 2R sin B .(cosB)/(sinB) + 2R sin C.(cos C)/(sinC) = 2(r + R)` `rArr cos B + cos C = 1 + (r)/(R)` `rArr cos B + cos C = 1 + 4 sin.(A)/(2) sin.(B)/(2) sin.(C)/(2)` `rArr cos B + cos C = cos A + cos B + cos C` `:. cos A = 0` `rArr A = (pi)/(2)` `rArr a^(2) = b^(2) + c^(2)` `rArr b^(2) + c^(2) = 100` Using A.M. `ge` G.M. we get `(b^(2) + c^(2))/(2) ge sqrt(b^(2) c^(2))` `rArr bc le 50` Hence, area of `Delta ABC = (1)/(2) bc le 25` |
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| 79. |
In `Delta ABC`, if `r_(1) lt r_(2) lt r_(3)`, then find the order of lengths of the sides |
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Answer» Correct Answer - `a lt b lt c` In `DeltaABC`, we have `r_(1) lt r_(2) lt r_(3)`. Thus, `(1)/(r_(1)) gt (1)/(r_(2)) gt (1)/(r_(3))` `rArr (s -a)/(Delta) gt (s-b)/(Delta) gt (s -c)/(Delta)` `rArr s -a gt s-b gt s-c` `rArr -a gt -b gt -c` or `a lt b lt c` |
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| 80. |
Given that `Delta = 6, r_(1) = 2,r_(2)=3, r_(3) = 6` Difference between the greatest and the least angles isA. `cos^(-1).(4)/(5)`B. `tan^(-1).(3)/(4)`C. `cos^(-1).(3)/(5)`D. none of these |
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Answer» Correct Answer - C We have, `r_(1) = (Delta)/(s-a) = 2, r_(2) = (Delta)/(s-b) = 3, r_(3) = (Delta)/(s-c) = 6` Given `Delta =6`, `:. S-a = 3` ...(i) `s-b =2` ...(ii) `s-c =1` ...(iii) Adding Eqs. (i) and (ii), `2s - a - b =5 " or " a + b + c -a -b = 5 or c = 5` Adding Eqs. (i) and (iii), `2s - a- c = 4, or b = 4` And adding Eqs. (ii) and (iii), `2s -b -c = 3 or a = 3` Hence the sides of the `Delta` are `a =3, b = 4, c = 5` Since the triangle is right angled, the greatest angle is `90^(@)`. Also, the least angle is opposite to side a, which is `sin^(-1).(3)/(5)`. Therefore, `90^(@) - "sin"^(-1)(3)/(5) = "cos"^(-1)(3)/(5)` Also, `R = (abc)/(4Delta) = (60)/(24) = 2.5` `r = (Delta)/(s) = (6)/(6) = 1` |
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| 81. |
In a triangle, the lengths of the two larger sides are 10 and 9,respectively. If the angles are in A.P., then the length of the third sidecan be`5-sqrt(6)`(b) `3sqrt(3)`(c)`5`(d) `5+sqrt(6)` |
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Answer» Correct Answer - `5 + sqrt6 " or " 5 - sqrt6` Given that the angle of triangle are in A.P., Let `angleA = x - d, angle B = x, angle C = x + d` Now, `angle A + angle B + angle C = 180^(@)` `rArr x - d + x + x + d = 180^(@)` `rArr 3x =180^(@)` `rArr x = 60^(@)` `:. Angle B = 60^(@)` Using cosine formula `cos B = (a^(2) + c^(2) -B^(2))/(2AC)`, we get `cos 60^(@) = (100 + c^(2) - 81)/(2 xx 10 xx c)` `rArr (1)/(2) = (19 + c^(2))/(2 xx 10c)` `rArr c^(2) - 10 c + 19 = 0` `rArr c = (10 +- sqrt(100 - 76))/(2)` `rArr c = 5 +- sqrt6` Given that `a = 10 and b = 9` are the longer sides Therefore, `c = 5 + sqrt6 and 5 - sqrt6` both are possible |
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| 82. |
Let `a , ba n dc`be the three sides of a triangle, then prove that the equation `b^2x^2+(b^2=c^2-alpha^2)x+c^2=0`has imaginary roots. |
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Answer» `b^(2) x^(2) + (b^(2) + c^(2) - a^(2)) x + c^(2) = 0` Let `f(x) = b^(2) x^(2) + (2 bc cos A) x + c^(2) = 0` Also in `Delta ABC`, where `A in (0, pi)` in a triangle, we find `cos A in (-1, 1)`. Now, `D = (2bc cos A)^(2) - 4b^(2) c^(2) = 4b^(2) c^(2) (cos^(2) A - 1) lt 0` Hence, the roots are imaginary |
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| 83. |
In triangle ABC, if `r_(1) = 2r_(2) = 3r_(3)`, then `a : b` is equal toA. `(5)/(4)`B. `(4)/(5)`C. `(7)/(4)`D. `(4)/(7)` |
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Answer» Correct Answer - A `r_(1) = 2r_(2) = 3r_(3)` `rArr (Delta)/(s-a) = 2 (Delta)/(s-b) = 3 (Delta)/(s-c)` or `(1)/(s-a) = (2)/(s-b) = (3)/(s-c) =k` (say) or `s-a = (1)/(k), s-b = (2)/(k) , and s-c = (3)/(k)` Adding, we get `3s - (a +b +c) = (6)/(k)` or `s= (6)/(k) " " a = (5)/(k)` and `b = (4)/(k) " or " (a)/(b) = (5)/(4)` |
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| 84. |
In triangle ABC, if `cos^(2)A + cos^(2)B - cos^(2) C = 1`, then identify the type of the triangle |
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Answer» Correct Answer - Right angled triangle `cos^(2) A + cos^(2) B - cos^(2) C = 1` or `1 - sin^(2) A + 1 - sin^(2) B - 1 + sin^(2) C = 1` or `sin^(2) A + sin^(2) B = sin^(2) C rArr a^(2) + b^(2) = c^(2)` Thus, the triangle is right angled at C |
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| 85. |
If the angles A,B,C of a triangle are in A.P. and sides a,b,c, are in G.P., then prove that `a^2, b^2,c^2`are in A.P. |
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Answer» Given `2B = A + C` or `3B = pi " or " B = pi//3` (i) Also a, b, c are in G.P. `rArr b^(2) = ac` (ii) Now, `cos B = cos 60^(@) = (1)/(2) = (c^(2) a^(2) -b^(2))/(2ca)` or `ca = c^(2) + a^(2) - b^(2)` or `2b^(2) = c^(2) + a^(2)` [by using Eq. (ii)] Hence, `a^(2), b^(2), c^(2)` are in A.P. |
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| 86. |
If the sides of a triangle are a, b and `sqrt(a^(2) + ab + b^(2))`, then find the greatest angle |
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Answer» Correct Answer - `120^(@)` Sides are `a, b, sqrt(a^(2) + ab + b^(2))` Then the greatest side `c = sqrt(a^(2) + ab + b^(2))` Let the angle opposite to the greatest side be C. `rArr cos C = (a^(2) + b^(2) - (a^(2) + ab + b^(2)))/(2ab) = (-1)/(2)` `rArr C = 120^(@)` |
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| 87. |
In any triangle ABC, find the least value of `(r_(1) + r_(2) + r_(3))/(r)` |
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Answer» Correct Answer - 9 We have already proved that `(1)/(r_(1)) + (1)/(r_(2)) + (1)/(r_(3)) = (1)/(r)` Now using A.M. `ge` H.M. we get `(r_(1) + r_(2) + r_(3))/(3) ge (3)/((1)/(r_(1)) + (1)/(r_(2)) + (1)/(r_(3))) = 3r` `rArr (r_(1) + r_(2) + r_(3))/(r) ge 9` |
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| 88. |
The sides of a triangle are `x^2+x+1, 2x+1 and x^2-1.` Prove that the greatest angle is `120^0` |
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Answer» Let `a = x^(2) + x + 1, b = 2x + 1, and c = x^(2) - 1`. First, we have to decide which side is the greatest. We know that in a triangle, the length of each side is greater than zero. Therefore, we have `b = 2x + 1 gt 0 and c = x^(2) - 1 0`. Thus, `x gt -(1)/(2) and x^(2) gt 1` `rArr x gt - (1)/(2) and x lt -1 " or " x gt 1` `rArr x gt 1` `a = x^(2) + x + 1 = (x + (1)/(2))^(2) + ((3)/(4))` is always positive. Thus, all sides a, b, and c are positive when `x gt 1`. Now `x gt 1 " or" x^(2) gt x` or `x^(2) + x + 1 gt 2x + 1 rArr a gt b` Also, when `x gt 1` `x^(2) + x + 1 gt x^(2) - 1 rArr a gt c` Thus, `a = x^(2) + x + 1` is the greatest side and the angle A opposite to this side is the greatest angle `:. cos A = (b^(2) + c^(2) -a^(2))/(2bc)` `= ((2x + 1)^(2) + (x^(2) - 1)^(2) - (x^(2) + x + 1)^(2))/(2(2x + 1) (x^(2) -1))` `= (-2x^(3) - x^(2) + 2x + 1)/(2(2x^(3) + x^(2) - 2x -1)) = -(1)/(2) = cos 120^(@)` `rArr A = 120^(@)` |
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| 89. |
In `Delta ABC, (sin A (a - b cos C))/(sin C (c -b cos A))=`A. `-2`B. `-1`C. 0D. 1 |
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Answer» Correct Answer - D `(sinA (a -b cos C))/(sinC (c -b cos A))` `= (sin A (b cos C + c cos B - cos C))/(sin C (a cos B + b cos A - b cos A))` `= (sin A (c cos B))/(sin C (a cos B))` `=1 " " ("as " c sin A = a sin C)` |
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| 90. |
If `a = sqrt3, b = (1)/(2) (sqrt6 + sqrt2), and c = sqrt2`, then find `angle A` |
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Answer» Using Cosine Rule `cos A = (b^(2) + c^(2) -a^(2))/(2bc)` `= ((1//4) (8 + 4 sqrt3) + 2 - 3)/(sqrt12 + sqrt4) = (1 + sqrt3)/(2(1 + sqrt3)) = (1)/(2)` So, `A = (pi)/(3)` |
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| 91. |
In `Delta ABC`, prove that `(a - b)^(2) cos^(2).(C)/(2) + (a + b)^(2) sin^(2).(C)/(2) = c^(2)` |
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Answer» `L.H.S. = (a^(2) + b^(2) - 2ab) cos^(2).(C)/(2) + (a^(2) + b^(2) + 2 ab) sin^(2). (C)/(2)` `= a^(2) + b^(2) + 2ab (sin^(2).(C)/(2) - cos^(2).(C)/(2))` `= a^(2) + b^(2) - 2ab cos C` `= c^(2)` (Using consine Rule) |
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| 92. |
In ` A B C ,=`if `(a+b+c)(a-b+c)=3a c ,`then find `/_Bdot` |
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Answer» We have `(a + c)^(2) - b^(2) = 3ac` or `a^(2) + c^(2) - b^(2) = ac` But `cos B = (a^(2) + c^(2) -b^(2))/(2 ac) = (1)/(2)` So, `B = (pi)/(3)` |
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| 93. |
In triangle ABC, line joining the circumcenter and orthocenter is parallel to side AC, then the value of tan A tan C is equal toA. `sqrt3`B. 3C. `3sqrt3`D. none of these |
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Answer» Correct Answer - B The distance of circumcenter from side AC is `R cos B`, and the distance of orthocenter from side AC is `2R cos A cos C`. Therefore, `R cos B = 2R cos A cos C` `:. Cos (A +C) = 2 cos A cos C` `:. sin A sin C = 3 cos A cos C` `:. tan A tan C = 3` |
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| 94. |
In triangle, ABC if `2a^(2) b^(2) + 2b^(2) c^(2) = a^(4) + b^(4) + c^(4)`, then angle B is equal toA. `45^(@)`B. `135^(@)`C. `120^(@)`D. `60^(@)` |
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Answer» Correct Answer - A::B `2a^(2)b^(2) + 2b^(2)c^(2) = a^(4) + b^(4) + c^(4)` Also, `(a^(2) -b^(2) + c^(2)) = a^(4) + b^(4) + c^(4) - 2(a^(2) b^(2) + b^(2) c^(2) - c^(2) a^(2))` `:. (a^(2) -b^(2) + c^(2))^(2) = 2c^(2) a^(2)` `:. (a^(2) -b^(2) + c^(2))/(2ca) = +- (1)/(sqrt2) = cosB` or `B = 45^(@) or 135^(@)` |
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| 95. |
Prove that `(r_(1+r_2))/1=2R` |
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Answer» `r_(1) + r_(2) = 4R (sin.(A)/(2) cos.(B)/(2) cos.(C)/(2) + sin.(B)/(2) cos.(A)/(2) cos.(C)/(2))` `= 4R cos.(C)/(2) (sin.(A)/(2) cos.(B)/(2) + sin.(B)/(2) cos.(A)/(2))` `= 4R cos.(C)/(2) sin ((A + B)/(2))` `= 4R (cos^(2).(C)/(2)) = 2R (1 + cos C)` or `(r_(1) + r_(2))/(1 +cos C) = 2R` |
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| 96. |
Prove that `(r+r_1)tan((B-C)/2)+(r+r_2)tan((C-A)/2)+(r+r_3)tan((A-B)/2)=0` |
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Answer» `(r + r_(1)) tan ((B -C)/(2))` `= [4R sin.(A)/(2) sin.(B)/(2) sin.(C)/(2) + 4R sin.(A)/(2) cos.(B)/(2) cos.(C)/(2)] xx tan ((B- C)/(2))` `=4R sin.(A)/(2) cos ((B -C)/(2)) tan ((B -C)/(2))` `=4R sin.(A)/(2) sin ((B-C)/(2))` `= 2R (sin B - sin C)`...(i) Similarly, `(r + r_(2)) tan ((C -A)/(2)) = 2R (sin C - sin A)`...(ii) `(r + r_(3)) tan ((A - B)/(2)) = 2R (sin A - sin B)`...(iii) On adding Eqs. (i) (ii) and (iii), we get the result `rArr (I_(1) I_(2))/(cos.(C)/(2)) = (4R sin.(A)/(2))/(sin.(A)/(2))` `rArr I_(1) I_(2) = 4 R cos.(C)/(2)` Similarly, `I_(2) I_(3) = 4 R cos.(A)/(2) and I_(1) I_(3) = 4R cos.(B)/(2)` |
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| 97. |
If the distance between incenter and one of the excenter of anequilateral triangle is 4 units, then find the inradius of the triangle. |
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Answer» Distance between excenter `I_(1) and` incenter I is `II_(1) = 4R sin.(A)/(2) = 4` (given) Since triangle, we have `4 = 4R sin.(pi)/(6)` or `R = 2` `:. R = 2r = 2` or `r = 1` |
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