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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.
| 1. |
If the angles of a triangle are in the ratio `4:1:1,`then the ratio of the longest side to the perimeter isA. `sqrt3 : (2 + sqrt3)`B. `1 : 6`C. `1 : 2 + sqrt3`D. `2 : 3` |
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Answer» Correct Answer - A Given that `4A + A + A = 180^(@)` or `A = 30^(@)` Angle are `120^(@), 30^(@), 30^(@)` `rArr (sin 120^(@))/(a) - (sin 30^(@))/(b) = (sin 30^(@))/(c) = 2R` (say) `rArr (a)/(a + b + c) = (sin 120^(@))/(sin 120^(@) + sin 30^(@) + sin 30^(@)) = (sqrt3)/(2 + sqrt3)` |
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| 2. |
A triangle with integral sides has perimeter 8 cm. Then find the area of the triangle |
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Answer» Correct Answer - `2 sqrt2 cm^(2)` The only possible set of sides can be 3, 3, 2 (as sum of any two side must be greater than the third one) So, `Delta = sqrt(s(s -a) (s-b) (s-c))` `= sqrt(4 xx 1 xx 1 xx 2) = 2sqrt2 cm^(2)` |
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| 3. |
The area of the circle and the area of a regular polygon of `n`sides and of perimeter equal to that of the circle are in the ratio of`tan(pi/n):pi/n`(b) `cos(pi/n):pi/n``sinpi/n :pi/n`(d)`cot(pi/n):pi/n`A. `tan((pi)/(n)): (pi)/(n)`B. `cos ((pi)/(n)) : (pi)/(n)`C. `sin.(pi)/(n): (pi)/(n)`D. `cot((pi)/(n)): (pi)/(n)` |
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Answer» Correct Answer - A Let r be the radius of the circle and `A_(1)` be its area. Then `A_(1) = pir^(2)`. Since the perimeter of the circle is same as the perimeter of a regular polygon of n sides, we have `2pi r = na`, Where a is the length of one side of the regular polygon. Thus, `a = (2pi r)/(n)` Let `A_(2)` be the area of the polygon. Then, `A_(2) = (1)/(4) na^(2) cot ((pi)/(n)) = (pi^(2) r^(2))/(n) cot ((pi)/(n))` `rArr A_(1) : A_(2) = pi r^(2) : (pi^(2) r^(2))/(n) cot ((pi)/(n)) = tn ((pi)/(n)) : (pi)/(n)` |
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| 4. |
The sides of a triangle are in A.P. and its area is `(3)/(5)` th of an equilateral triangle of the same perimeter. Find the greatest angle of the triangle |
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Answer» Correct Answer - `120^(@)` Let the sides be `x - d, x, x + d`. Then `s = (3x)/(2), (s -a) = (x)/(2) + d`, `(s-b) = (x)/(2), (s-c) = (x)/(2) -d`. Area of triangle `= sqrt((3x)/(2).((x)/(2) + d).(x)/(2).((x)/(2) -d))` `= (x)/(2) sqrt(3((x^(2))/(4) -d^(2))) = (x)/(4) sqrt(3(x^(2) -4d^(2)))` The area of equilateral triangle whose perimeter is `3x " is " (sqrt3x^(2))/(4)` Given, `(3)/(5) xx (sqrt3x^(2))/(4) = (x)/(4) sqrt(3(x^(2) -4d^(2)))` or `(9)/(25) xx (3x^(4))/(16) = (x^(2))/(16) xx 3 (x^(2) -4d^(2))` or `x^(2) - (9x^(2))/(25) = 4d^(2)` or `16 x^(2) = 100d^(2)` or `x = (5)/(2) d` Therefore, the sides of triangle measure `((5d)/(2) -d), (5d)/(2), ((5d)/(2) + d)` or `(3d)/(2), (5d)/(2), (7d)/(2)` Hence, the ratio of sides is `3 : 5 : 7` For the greatest angle, `cos theta = (3^(2) + 5^(2) -7^(2))/(2 xx 3 xx 5) = (9 + 25 - 49)/(30) = (-1)/(2)` `:. theta = 120^(2)` |
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| 5. |
In equilateral triangle ABC with interior point D, if the perpendiculardistances from D to the sides of 4,5, and 6, respectively, are given, thenfind the area of ` A B Cdot` |
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Answer» Let the side equilateral triangleABC be a Area of triangle, `Delta = (a xx 4 + a xx 5 + a xx 6)/(2)` or `(a (4 + 5 + 6))/(2) = (sqrt3)/(4) a^(2)` or `(15)/(2) = (sqrt3a)/(4)` or `a = (30)/(sqrt3) = 10 sqrt3` or `Delta = (sqrt3)/(4) xx 100 xx 3` `= 75 sqrt3` |
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| 6. |
In `Delta ABC` if `AB = x, BC = x + 1, angleC = (pi)/(3)`, then the less integer value of x isA. 6B. 7C. 8D. none of these |
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Answer» Correct Answer - B Using cosine rule, we get `x^(2) = (x + 1)^(2) + b^(2) -2 (x + 1) b cos.(pi)/(3)` `rArr 0 = 2x + 1 + b^(2) - (x +1) b` `rArr b^(2) -(x +1) b + 2x + 1 = 0` Since b is real, we have `rArr (x + 1)^(2) - 4 (2x +1) ge 0` `rarr x^(2) -6x - 3 ge 0` `rArr x ge 3 + sqrt12` The least integral value of x is 7 |
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| 7. |
In triangle ABC, `b^(2) sin 2C + c^(2) sin 2B = 2bc` where `b = 20, c = 21`, then inradius =A. 4B. 6C. 8D. 9 |
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Answer» Correct Answer - B `b^(2) sin 2C + c^(2) sin 2B = 2bc` `rArr b^(2) (2 sin C cos C) + c^(2) (2 sin B cos B) = 2bc` `rArr b^(2) (c cos C) + c^(2) (b cos B) = 2Rbc` `rArr b cos C + c cos B = 2R` `rArr a = 2R` `rArr A = 90^(@)` `:. a^(2) = b^(2) + c^(2) = 400 + 441 = 841` `rarr a = 29` Now, `r = (2Delta)/(2s) = (20 xx 21)/(20 + 21 + 29) = (20 xx 21)/(70) = 6` |
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| 8. |
In `DeltaABC, R, r, r_(1), r_(2), r_(3)` denote the circumradius, inradius, the exradii opposite to the vertices A,B, C respectively. Given that `r_(1) :r_(2): r_(3) = 1: 2 : 3` The sides of the triangle are in the ratioA. `1 : 2 : 3`B. `3 : 5 : 7`C. `1 : 5 : 9`D. `5 : 8 : 9` |
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Answer» Correct Answer - D `(Delta)/(s-a) : (Delta)/(s-b): (Delta)/(s-c) = 1: 2:3` Let `(Delta)/((s-a)/(1)) = (Delta)/((s-b)/(2)) = (Delta)/((s-c)/(3)) = (Delta)/(6k)` `rArr (1)/(s-a) = (1)/(6k), (1)/(s-b)= (1)/(3k), (1)/(s-c) = (1)/(2k)` `rArr s-a = 6k, s-b = 3k, s-c = 2k` `rArr s= 11k` `:. a = 5k, b= 8k, c = 9k` Hence, ratio of sides is `5: 8: 9` |
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| 9. |
In `DeltaABC, R, r, r_(1), r_(2), r_(3)` denote the circumradius, inradius, the exradii opposite to the vertices A,B, C respectively. Given that `r_(1) :r_(2): r_(3) = 1: 2 : 3` The greatest angle of the triangle is given byA. `cos^(-1) ((1)/(30))`B. `cos^(-1)((1)/(3))`C. `cos^(-1)((1)/(10))`D. `cos^(-1) ((1)/(5))` |
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Answer» Correct Answer - C Since c is the greatest side, C is the greatest angle `cos C = (a^(2) + b^(2) -c^(2))/(2ab) = (25 + 64 - 81)/(2 xx 5 xx 8) = (1)/(10)` |
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| 10. |
In `DeltaABC, R, r, r_(1), r_(2), r_(3)` denote the circumradius, inradius, the exradii opposite to the vertices A,B, C respectively. Given that `r_(1) :r_(2): r_(3) = 1: 2 : 3` The value of `R : r` isA. `5 : 2`B. `5 : 4`C. `5 : 3`D. `3 : 2` |
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Answer» Correct Answer - A Area of the triangle, `Delta k^(2) sqrt(11 xx 6 xx 3 xx 2) = 6 sqrt11 k^(2)` So, `r = (Delta)/(s) = (6)/(sqrt11) k` And, `R = (abc)/(4Delta) = (5 xx 8 xx 9)/(4 xx 6 xx sqrt11) k = (15)/(sqrt11) k` `:. R : r = 5 : 2` |
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| 11. |
In triangle ABC, if `A - B = 120^(2) and R = 8r`, where R and r have their usual meaning, then cos C equalsA. `3//4`B. `2//3`C. `5//6`D. `7//8` |
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Answer» Correct Answer - D `R = 8r = 8 (4R sin.(A)/(2) sin.(B)/(2) sin.(C)/(2))` `:. 2 sin.(A)/(2) sin.(B)/(2) sin.(C)/(2) = (1)/(16)` or `(cos.(A -B)/(2) -cos.(A +B)/(2)) sin.(C)/(2) = (1)/(16)` or `sin.(C)/(2) ((1)/(2) - sin.(C)/(2)) = (1)/(16) rArr sin.(C)/(2) = (1)/(4)` or `cos C = 1 -2 sin^(2).(C)/(2) = 1 - (1)/(8) = (7)/(8)` |
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| 12. |
If two sides of a triangle are roots of the equation `x^(2) -7x + 8 = 0` and the angle between these sides is `60^(@)` then the product of inradius and circumradius of the triangle isA. `(8)/(7)`B. `(5)/(3)`C. `(5sqrt2)/(3)`D. 8 |
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Answer» Correct Answer - B Let a and b the roots of `x^(2) -7 x + 8 = 0` Then `a + b = 7, ab = 8` Also, `C = 60^(@)` `rArr c^(2) = a^(2) + b^(2) - ab` `rArr c^(2) = (a +b)^(2) -3ab = 49 -24 = 25` `rArr c = 5` `:. r.R = (abc)/(2(a+b+c)) = (8xx5)/(2(7+5)) = (5)/(3)` |
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| 13. |
Given `b = 2, c = sqrt3, angle A = 30^(@)`, then inradius of `DeltaABC` isA. `(sqrt3 -1)/(2)`B. `(sqrt3 + 1)/(2)`C. `(sqrt3-1)/(4)`D. none of these |
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Answer» Correct Answer - A `b = 2, c = sqrt3, anlge A = 30^(@)` `a = sqrt(b^(2) + c^(2) - 2bc cos A)` `= sqrt(4+3 - 2 xx 2 sqrt3 (sqrt3)/(2)) = 1` `rArr r = (s-a) tan.(A)/(2)` `= (b + c-a)/(2) tan.(A)/(2)` `= (sqrt3 +1)/(2) tan 15^(@)` `= (sqrt3 +1)/(2) (sqrt3 -1)/(sqrt3+1) = (sqrt3 -1)/(2)` |
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| 14. |
The sides of a triangle are in the ratio `1: sqrt3:2.` Then the angles are in the ratioA. `1 : 3 : 5`B. `2: 3 : 4`C. `3 : 2: 1`D. `1 : 2: 3` |
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Answer» Correct Answer - D Sides are in the ratio `1 : sqrt3 : 2` Let `a = k, b = sqrt3 k, and c = 2k` `cosA = (b^(2) + c^(2) - a^(2))/(2bc) = (sqrt3)/(2) rArr A = (pi)/(6)` `cos B = (c^(2) + a^(2) -b^(2))/(2ac) = (1)/(2) rArr B = (pi)/(3)` `rArr C = pi - bar(A + B) = (pi)/(2)` `rArr A: B: C = 1 : 2: 3` |
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| 15. |
In triangle ABC, `sinA sin B + sin B sin C + sin C sin A = 9//4 and a = 2`, then the value of `sqrt3 Delta`, where `Delta` is the area of triangle, is _______ |
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Answer» Correct Answer - 3 The given equation is `4 sin A sin B + 4 sin B sin C + 4 sin C sin A = 9` `rArr 2 cos (A -B) -2 cos (A +B) + 2 cos (B -C)` `- 2 cos (B +C) + 2 cos (C -A) -2 cos (C + A) = 9` or `2 [cos (A -B) + cos (B -C) + cos (C -A)]` `= 9 -2 (cos A + cos B + cos C)` `ge 9 -2 xx (3)/(2) = 6` or `cos (A -B) + cos (B -C) + cos (C -A) ge 3` But `cos (A -B) le1, cos (B -C) le 1, cos (C -A) le 1` or `cos (A -B) = 1, cos (B -C) =1, cos (C -A) = 1` Thus, `A = B = C`, i.e., triangle ABC is an equilateral triangle Hence, `Delta = sqrt3` |
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| 16. |
Suppose `alpha,beta,gammaa n ddelta`are the interior angles of regular pentagon, hexagon, decagon, anddodecagon, respectively, then the value of `|cosalphasecbetacosgammacos e cdelta|`is _________ |
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Answer» Correct Answer - 1 Interior angle of regular polygon of side n is `(180^(@) - (360^(@))/(n))` Hence, `alpha = 180^(@), beta = 120^(@), gamma = 144^(@), delta = 150^(@)` `:. cos alpha = cos 108^(@) = - sin 180^(@) = - ((sqrt5 -1)/(4))` `sec beta = sec 120^(@) = -2` `cos gamma = cos 144^(@) = - cos 36^(@) = - ((sqrt5 + 1)/(4))` `"cosec" delta = "cosec" 150^(@) = +2` `:. |((sqrt5 -1)/(4)) (2) ((sqrt5 +1)/(4)) (-2)| = 1` |
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| 17. |
The sides of triangle ABC satisfy the relations `a + b - c= 2 and 2ab -c^(2) =4`, then the square of the area of triangle is ______ |
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Answer» Correct Answer - 3 `a + b - c = 2` and `2ab - c^(2) = 4` `rArr a^(2) + b^(2) + c^(2) + 2ab - 2bc - 2ca = 4 = 2ab -c^(2)` `rArr (b-c)^(2) + (a-c)^(2) = 0` `rArr a = b = c` Triangle is equilateral, hence `a = 2` `rArr Delta = sqrt3` |
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| 18. |
In an acute angled triangle ABC, `r + r_(1) = r_(2) + r_(3) and angleB gt (pi)/(3)`, thenA. `b + 2c lt 2a lt 2b + 2c`B. `b + 4cc lt 4a lt 2b + 4c`C. `b + 4c lt 4a lt 4b + 4c`D. `b + 3c lt 3a lt 3b + 3c` |
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Answer» Correct Answer - D `r - r_(2) = r_(3) -r_(1)` `rArr (Delta)/(s) - (Delta)/(s-b) = (Delta)/(s-c) -(Delta)/(s-a)` or `(-b)/(s(-b)) = (c-a)/((s-a) (s-c))` or `((s-a) (s-c))/(s(s-b)) = (a-c)/(b)` `rArr tan^(2).(B)/(2) = (a-c)/(b)` But `(B)/(2) in ((pi)/(6), (pi)/(4))`. Therefore, `tan^(2).(B)/(2) in ((1)/(3),1)` `rArr (1)/(3) lt (a-c)/(b) lt 1` or `b lt 3a -3c lt 3b` or `b + 3c lt 3a lt 3b + 3c` |
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| 19. |
If the cotangents of half the angles of a triangle are in A.P., thenprove that the sides are in A.P. |
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Answer» `cot.(A)/(2), cot.(B)/(2), cot.(C)/(2)` are in A.P. `rArr 2 cot.(B)/(2) = cot.(A)/(2) + cot.(C)/(2)` `rArr 2 sqrt((s (s -b))/((s-a) (s-c))) = sqrt((s (s -a))/((s -b) (s-c))) + sqrt((s (s-c))/((s - a) (s -b)))` `rArr 2(s-b) = s-a + s -c` `rArr 2b = a + c` `rArr` a, b, c are in A.P. |
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| 20. |
If `b=3,c=4,a n dB=pi/3,`then find the number of triangles that can be constructed. |
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Answer» We have, `(sin B)/(b) = (sin C)/(c) " or " (sin (pi//3))/(3) = (sin C)/(4)` or `sinC = (2)/(sqrt3) gt 1`, which is not possible Hence, no triangle is possible |
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| 21. |
If `A=30^0, a=7,a n db=8`in ` A B C ,`then find the number of triangles that can be constructed. |
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Answer» We have `(a)/(sin A) = (b)/(sin B)` or `sinB = (b sin A)/(a) = (8 sin 30^(@))/(7) = (4)/(7)` Thus, we have, `b gt a gt b sin A` Hence, angle B has two value given by `sin B = 4//7` |
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| 22. |
If in triangle ABC, `(a=(1+sqrt(3))c m ,b=2c m ,a n d/_C=60^0`, then find the other two angles and the third side. |
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Answer» From `cos C = (a^(2) + b^(2) - c^(2))/(2ab)`, we have `(1)/(2) = ((1 + sqrt3)^(2) + 4 -c^(2))/(2(1 + sqrt3) 2)` or `2 + 2sqrt3 = 1+ 3 + 2sqrt3 + 4 - c^(2)` or `c^(2) = 6 " or " c = sqrt6 cm` Also, `(sin A)/(a) = (sin B)/(b) = (sin C)/(c)` or `(sin A)/(1 + sqrt3) = (sin B)/(2) = (sqrt3//2)/(sqrt6)` `rArr sin B = (1)/(sqrt2) " or " B = 45^(@)` and `A = 180^(@) - (60^(@) + 45^(@)) = 75^(@)` |
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| 23. |
In ` A B C ,s i d e sb , c`and angle `B`are given such that `a`has two valus `a_1a n da_2dot`Then prove that `|a_1-a_2|=2sqrt(b^2-c^2sin^2B)` |
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Answer» `cos B = (c^(2) + a^(2) - b^(2))/(2ca)` or `a^(2) - (2c cos B) a + c^(2) - b^(2) = 0` This equation has roots `a_(1) and a_(2)` `rArr a_(1) + a_(2) = 2c cos B, a_(1) a_(2) = c^(2) - b^(2)` `rArr (a_(1) -a_(2))^(2) = (a_(1) + a_(2))^(2) - 4a_(1) a_(2) = 4c^(2) cos^(2) B - 4(c^(2) - b^(2))` `= 4b^(2) - 4c^(2) sin^(2) B = 4(b^(2) -c^(2) sin^(2) B)` or `|a_(1) -a_(2)| = 2 sqrt(b^(2) - c^(2) sin^(2) B)` |
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| 24. |
Find the value of tan A, if area of `Delta ABC is a^(2) -(b-c)^(2).` |
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Answer» `Delta = (a + b - c) (a - b + c)` `rArr Delta^(2) = [2 (s -b) 2(s -c)]^(2)` or `s(s - a) (s -b) (s-c) = 16 (s-b)^(2) (s-c)^(2)` or `((s-b) (s-c))/(s(s-a)) = (1)/(16)` or `tan.(A)/(2) = (1)/(4)` `rArr tan^(2)A = (2 tan (A//2))/(1 - tan^(2) (A//2)) = (2(1//4))/(1-(1//16)) = (8)/(15)` |
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| 25. |
If in triangle ABC, a, c and angle A are given and `c sin A lt a lt c`, then (`b_(1) and b_(2)` are values of b)A. `b_(1) + b_(2) = 2c cos A`B. `b_(1) + b_(2) = c cos A`C. `b_(1) b_(2) = c^(2) -a^(2)`D. `b_(1) b_(2) = c^(2) + a^(2)` |
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Answer» Correct Answer - A::C From the cosine formula, `cos A = (b^(2) + c^(2) -a^(2))/(2bc)` or `b^(2) - (2c cos A) b + (c^(2) -a^(2)) = 0`, Which is a quadratic equation in b. Therefore, `c sin A lt a lt` Therefore, two triangle will be obtained, But this is possible when two values of the thired side are also obtained. Clearly, two values of sides b will be `b_(1) and b_(2)`. Let these be the roots of the above equation. then, Sum of roots `= b_(1) + b_(2) = 2c cos A and b_(1) b_(2) = c^(2) - a^(2)` |
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| 26. |
There exists triangle ABC satisfyingA. `tan A + tan B + tan C = 0`B. `(sin A)/(2) = (sin B)/(3) = (sin C)/(7)`C. `(a + b)^(2) = c^(2) + ab and sqrt2 (sinA + cos A) = sqrt3`D. `sin A + sin B = (sqrt3 + 1)/(2), cos A cos B = (sqrt3)/(4) = sin A sin B` |
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Answer» Correct Answer - C::D (1) `tan A + tan B + tan C = tan A tan B tan C = 0` Therefore, either A or B or C = 0. This is not possible in a triangle (2) By sine rule, `(a)/(2) = (b)/(3) = (c)/(7) = lamda`(say) `a + b = 5 lamda, c = 7 lamda` i.e., `a + b lt c` This is not possible in a triangle, as the sum of two sides is greater than the third. (3) Given that `(a + b)^(2) = c^(2) + ab` or `a^(2) + b^(2) - c^(2) + ab = 0` or `2ab cos C + ab = 0` or `cos C = -(1)/(2) or angleC = 120^(@)` Also `sqrt2 (sin A + cos A) = sqrt3`(given) or `sin (A + 45^(@)) = (sqrt3)/(2)` or `A = 45^(@) = 60^(@) " " (A + 45^(@) = 120^(@) " out as " angle C )` or `A = 15^(@) and "hence " B = 45^(@)` `:. Delta` is possible (4) Given that `cos a cos B = (sqrt3)/(4) = sin A sin B` `rArr cos (A -B) = (sqrt3)/(2)` or `A - B = (pi)/(6)` and `cos (A +B) = 0` or `A +B = (pi)/(2)` `rArr A = 60^(@), B = 30^(@)` Hence, `sin A + sin B = (sqrt3+1)/(2)`. This is possible in a triangle |
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| 27. |
In a triangle ABC, if a, b, c are in A.P. and `(b)/(c) sin 2C + (c)/(b) sin 2B + (b)/(a) sin 2A + (a)/(b) sin 2B = 2`, then find the value of sin B |
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Answer» Correct Answer - `(1)/(2)` `(b)/(c) sin 2C + (c)/(b) sin 2B + (b)/(a) sin 2A + (a)/(b) sin 2B = 2` `rArr 2 sin B cos C + 2 sin C cos B + 2 sin B cos A + 2 sin B cos B = 2` `rArr sin (B + C) + sin (A + B) = 1` `rArr sin A + sin C = 1` `rArr sin B = (1)/(2) " " ("As " 2 sin B = sin A + sin C)` |
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| 28. |
Prove that `a^(2) sin 2B + b^(2) sin 2A = 4 Delta` |
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Answer» `a^(2) sin 2B + b^(2) sin 2A` `= 4R^(2) [sin^(2) A (2 sin B cos B) + sin^(2) B (2 sin A cos A)]` [using sine rule] `= 8R^(2) sin A sin B (sin A cos B + sin B cos A)` `=8R^(2) sin A sin B sin (A + B)` `= 8R^(2) sin A sin B C = 4Delta` |
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| 29. |
Prove that`(cotA/2+cotB/2)(asin^2B/2+bsin^2A/2)=ccotC/2` |
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Answer» `(cot.(A)/(2) + cot.(B)/(2)) (a sin^(2).(B)/(2) + b sin^(2).(A)/(2))` `= [sqrt((s(s-a))/((s-b)(s-c))) + sqrt((s(s -b))/((s-a) (s-c)))] xx [(a (s -c) (s-a))/(ca) + (b(s-b) (s-c))/(bc)]` `=sqrts. [((s-a) + (s-b))/(sqrt((s-a)(s-b)(s-c)))] ((s-c))/(c) [s -a + s - b]` `= sqrts. [(c(s-c))/(sqrt((s-a) (s-b) (s-c)))]` `= csqrt((s(s-c))/((s-a) (s-b))) = c cot.(C)/(2)` |
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| 30. |
In ` A B C ,`let `L ,M ,N`be the feet of the altitudes. The prove that `sin(/_M L N)+sin(/_L M N)+sin(/_M N L)=4sinAsinBsinC` |
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Answer» Using properties of pedal triangle, we have `angle MLN = 180^(@) - 2A` `angle LMN = 180^(@) - 2B` `anlge MNL = 180^(@) - 2C` `rArr sin (angle MLN) + sin (angle LMN) + sin (angle MNL)` `= sin 2A + sin 2B + sin 2C` `= 4 sin A sin B sin C` |
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| 31. |
Prove that `r_1+r_2+r_3-r=4R` |
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Answer» `r_(1) + r_(2) + r_(3) - r = (Delta)/(s -a) + (Delta)/(s -b) + (Delta)/(s-c) -(Delta)/(s)` `= Delta [(s -b + s-a)/((s -a) (s-b)) + ((sis + c))/(s(s -c))]` `= Delta [(c)/((s-a)(s-b)) + (c)/(s(s -c))]` `= Deltac [(s(s-c) + (s -a) (s-b))/(s(s -a) (s-b) (s-c))]` `= (Deltac)/(Delta^(2)) [2s^(2) -s (a + b + c) + ab]` `= ((c)/(Delta)) [2s^(2) -s (2s) + ab]` `= 4 ((abc)/(4Delta)) = 4R` |
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| 32. |
If in a triangle `r_1=r_2+r_3+r ,`prove that the triangle is right angled. |
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Answer» We have `r_(1) = r_(2) + r_(3) + r` or `r_(1) - r = r_(2) + r_(3)` `rArr (Delta)/(s -a) -(Delta)/(s) = (Delta)/(s -b) + (Delta)/(s -c)` or `(Delta a)/(s(s -a)) = (Delta (2s -b -c))/((s-b) (s-c)) = (Delta a)/((s-b) (s-c))` or `s(s -a) = (s -b) (s -c)` or `s^(2) - sa = s^(2) - (b + c) s + bc` or `2s (b + c-a) = 2bc` or `(a + b + c) (b + c-a) = 2bc` or `(b + c)^(2) - a^(2) = 2bc` or `b^(2) + c^(2) = a^(2)` Hence, the triangle is right angled |
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| 33. |
If in a triangle ABC, `(1 + cos A)/(a) + (1 + cos B)/(b) + (1+ cos C)/(c) = (k^(2) (1 + cos A) (1 + cos B) (1 + cos C))/(abc)`, then k is equal toA. `(1)/(2 sqrt2R)`B. 2RC. `(1)/(R)`D. none of these |
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Answer» Correct Answer - B `L.H.S. = (1 + cos A)/(a) + (1 + cos B)/(b) + (1 + cos C)/(c)` `= (2 cos^(2).(A)/(2))/(2R sin A) + (2 cos^(2).(B)/(2))/(2R sin B) + (2 cos^(2).(C)/(2))/(2R sin C)` `= (1)/(2R) (cot.(A)/(2) + cot.(B)/(2) + cot.(C)/(2))` Now, `(1 + cos A)/(a) .(1 + cos B)/(b) .(1 + cos C)/(c)` `= (cos^(2).(A)/(2))/(R sin A).(cos^(2).(B)/(2))/(R sin B) .(cos^(2)/(C)/(2))/(R sin C)` `= (1)/(8R^(3)) cot.(A)/(2) cot.(B)/(2) cot.(C)/(2)` |
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| 34. |
Given that `Delta = 6, r_(1) = 3, r_(3) = 6` Inradius is equal toA. 2B. 1C. 1.5D. 2.5 |
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Answer» Correct Answer - B 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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| 35. |
Given that `Delta = 6, r_(1) = 3, r_(3) = 6` Circumradius R is equal toA. 2.5B. 3.5C. 1.5D. none of these |
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Answer» Correct Answer - A 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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| 36. |
If `Delta` is the area of a triangle with side lengths ` a, b, c,` then show that as `Delta leq 1/4 sqrt((a + b + c) abc)` Also, show that the equality occurs in the above inequality if and only if `a = b = c`. |
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Answer» We have to prove that `Delta le (1)/(4) sqrt((a + b + c)abc)` or `Delta le (1)/(4) sqrt(2 s abc)` or `Delta^(2) le (1)/(16) 2s abc` or `Delta^(2) le (1)/(16) 2s Delta 4R` or `rs le (1)/(2) sR` Hence, `R ge 2R` [which is always true in `Delta_(2)`] Alternative Method: In triangle, sum of two sides is greater than the third side So `a + b gt c, b + c gt a and c + a gt b` Now consider quantities `a + b -c, b + c -a, c + a -b` Using A.M. `ge` G.M. we get `((a + b -c) + (b + c -a))/(2) ge sqrt((a + b -c) (b + c -a))` or `b ge sqrt((a + b -c) (b + c -a))` Similarly we get `c ge sqrt((c +a -b) (b + c -a))` and `a ge sqrt((a + b -c) (c + a -b))` Multiplying we get `abc ge (a + b -c) (b + c -a) (c + a -b)` `rArr abc ge (2s -2a) (2s -2b) (2s -2c)` `rArr sabc ge 8s (s -a) (s-b) (s -c)` `rArr (a + b + c) abc ge 16 Delta^(2)` `rArr Delta le (1)/(4) sqrt((a + b + c) abc)` |
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| 37. |
If area of `DeltaABC (Delta)` and angle C are given and if c opposite to given angle is minimum, thenA. `a = sqrt((2Delta)/(sinC))`B. `b = sqrt((2Delta)/(sinC))`C. `a = (4Delta)/(sinC)`D. `b = (4Delta)/(sin^(2)C)` |
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Answer» Correct Answer - A::B `c^(2) = a^(2) + b^(2) - 2ab cos C` `=(a - b)^(2) + 2ab (1-cos C)` `=(a -b)^(2) + (4Delta)/(sinC) (1-cosC)` For c to be minium, `a = b` Also, `Delta = (1)/(2) ab sin C` `rArr a^(2) = (2Delta)/(sin C) = b^(2)` |
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| 38. |
The base BC of `DeltaABC` is fixed and the vertex A moves, satisfying the condition `cot.(B)/(2) + cot.(C)/(2) = 2 cot.(A)/(2)`, thenA. `b + c = a`B. `b + c = 2a`C. vertex A moves along a straight lineD. vertex A moves along an ellipse |
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Answer» Correct Answer - B::D Given `(s(s-b))/(Delta) + ((s-c))/(Delta) = (2s(s-a))/(Delta)` `rArr s-b + s - c = 2 (s-a)` `rArr b+c = 2a` So, locus of vertex A is an ellipse |
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| 39. |
If in a triangle `(r)/(r_(1)) = (r_(2))/(r_(3))`, thenA. `A = 90^(@)`B. `B = 90^(@)`C. `C = 90^(@)`D. none of these |
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Answer» Correct Answer - C `(r)/(r_(1)) = (r_(2))/(r_(3))` or `r r_(3) = r_(1) r_(2)` `rArr (Delta)/(s) (Delta)/(s-c) = (Delta)/(s-a) (Delta)/(s-b)` or `((s-a)(s-b))/(s(s-c)) = 1` `rArr tan^(2).(C)/(2) = 1 " or " tan.(C)/(2) =1` or `tan^(2).(C)/(2) = 45^(@) " or " C = 90^(@)` |
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| 40. |
In a triangle ABC, if the sides a,b,c, are roots of `x^3-11 x^2+38 x-40=0,`then find the value of `(cosA)/a+(cosB)/b+(cosC)/c` |
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Answer» Here a, b, c are roots of equation `x^(3) - 11 x^(2) + 38 x - 40 = 0` Therefore, `a + b + c = 11, ab + bc + ac = 38, and abc = 40` `rArr (cos A)/(a) + (cos B)/(b) + (cos C)/(c) = (a^(2) + b^(2) + c^(2))/(2 abc)` `= ((a + b + c)^(2) - 2 (ab + bc + ac))/(2abc)` `= (11^(2) - 76)/(80) = (45)/(80) = (9)/(16)` |
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| 41. |
If in a triangle, `(1-(r_(1))/(r_(2))) (1 - (r_(1))/(r_(3))) = 2`, then the triangle isA. right angledB. isoscelesC. equilateralD. none of these |
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Answer» Correct Answer - A We have `(1-(s-b)/(s-a)) (1-(s-c)/(s-a)) = 2` or `2(b-a) (c-a) = 4 (s-a)^(2)` or or `2(bc-ac -ab +a^(2)) = (2s-2a)^(2)` or `a(bc-ac-ab+a^(2)) = (b+c -a)^(2)` or `a^(2) = b^(2) + c^(2)` Hence, triangle is right angled |
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| 42. |
In `Delta ABC, (cot. (A)/(2) + cot. (B)/(2)) (a sin.^(2) (B)/(2) + b sin.^(2) (A)/(2))=`A. `cot C`B. `c cot C`C. `cot.(C)/(2)`D. `c cot.(C)/(2)` |
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Answer» Correct Answer - D `{cot.(A)/(2) + cot.(B)/(2)} {a sin^(2).(B)/(2) + b sin^(2).(A)/(2)}` `= {(cos.(C)/(2))/(sin.(A)/(2) sin.(B)/(2))} {a sin^(2).(B)/(2) + b sin^(2).(A)/(2)}` `= {cos.(C)/(2)}{a (sin.(B)/(2))/(sin.(A)/(2)) + b (sin.(A)/(2))/(sin.(B)/(2))}` `= sqrt((s(s-c))/(ab)) {a(sqrt((s-a)(s-c))/(ac))/(sqrt(((s-b)(s-c))/(bc))) + b(sqrt((s-b)(s-c))/(bc))/(sqrt(((s-a)(s-c))/(ac)))}` `= sqrt((s(s-c))/(ab)) {sqrt(((s-a)/(s-b)) ab) + sqrt(((s-b)/(s-a)) ab)}` `= sqrt(2(s-c)) {(s-a+ s-b)/(sqrt((s-a) (s-b)))}` `= sqrt(2-(s-c)) {(2s-a-b)/sqrt((s-a) (s-b))}` `= c sqrt((s(-c))/((s-a)(s-b))) = c cot.(C)/(2)` |
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| 43. |
In `Delta ABC, "cot"(A)/(2) + "cot" (B)/(2) + "cot" (C)/(2)` is equal toA. `(Delta)/(r^(2))`B. `((a + b + c)^(2))/(abc) 2R`C. `(Delta )/(r)`D. `(Delta)/(Rr)` |
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Answer» Correct Answer - A `cot.(A)/(2) + cot.(B)/(2) + cot.(C)/(2)` `= (s(s-a))/(Delta) + (s(s-b))/(Delta) + (s-(s -c))/(Delta)` `= (s)/(Delta) [3s - (a + b + c)] = (s^(2))/(Delta) = ((Delta//r)^(2))/(Delta) = (Delta)/(r^(2))` |
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| 44. |
If in ` A B C ,A=pi/7,B=(2pi)/7,C=(4pi)/7`then `a^2+b^2+c^2`must be`R^2`(b) `3R^2`(c) `4R^2`(d)`7R^2`A. `R^(2)`B. `3R^(2)`C. `4R^(2)`D. `7R^(2)` |
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Answer» Correct Answer - D `a^(2) + b^(2) + c^(2) = 4R^(2) (sin^(2) A + sin^(2) B + sin^(2)C)` `= 2R^(2) (3 - (cos 2A + cos 2B + cos 2C))` Now, `cos 2A + cos 2B + cos 2C` `= cos.(2pi)/(7) + cos.(4pi)/(7) + cos.(8pi)/(7)` `= cos.(2pi)/(7) + cos.(4pi)/(7) + cos.(6pi)/(7)` `= (sin.(3pi)/(7))/(sin.(pi)/(7)) cos.(4pi)/(7)` `= (-2 sin.(3pi)/(7) cos.(3pi)/(7))/(2 sin.(pi)/(7))` `= (-sin.(6pi)/(7))/(2sin.(pi)/(7)) = (-1)/(2)` `:. a^(2) +b^(2) + c^(2) = 2R^(2) (3-((-1)/(2))) = 7R^(2)` |
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| 45. |
For any triangle ABC, prove that`a(bcosC-ccosB)=b^2-c^2` |
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Answer» `a (b cos C - c cos B)` `=(b cos C + c cos B) (b cos C - c cos B)` `= b^(2) cos^(2) C - c^(2) cos^(2) B` `= b^(2)cos^(2) C -c^(2) B` `= b^(2) - c^(2) - (b^(2) sin^(2) C - c^(2) sin^(2) B)` `= b^(2) - c^(2)` [as by the sine rule, `b sin C = c sin B`] Alternativey, using cos B and cos C formulae, we can prove the result |
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| 46. |
If `A=75^0,b=45^0,`then prove that `b+csqrt(2)=2a` |
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Answer» `A = 75^(@) , B = 45^(@) rArr C = 60^(@)` Now, `(a)/(sin A) = (b)/(sin B) = (c)/(sin C) = 2R` `rArr b + c sqrt2 = (sin 45^(@))/(sin 75^(@)) a + sqrt2 (sin 60^(@))/(sin 75^(@)) a` `((1)/(sqrt2))/((sqrt3 + 1)/(2 sqrt2)) a + sqrt2 ((sqrt3)/(2))/((sqrt3 + 1)/(2 sqrt2)) a` `= (2)/(sqrt3 + 1) a + (2 sqrt3 a)/(sqrt3 + 1) = 2a` |
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| 47. |
If in a triangle `A B C ,/_C=60^0,`then prove that `1/(a+c)+1/(b+c)=3/(a+b+c)`. |
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Answer» By the cosine formula, we have `c^(2) = a^(2) + b^(2) - 2ab cos C` or `c^(2) = a^(2) + b^(2) - 2ab cos 60^(@) = a^(2) + b^(2) - ab` (i) Now `(1)/(a + c) + (1)/(b + c) - (3)/(a + b + c)` `=[((b + c) (a + b +c) + (a + c) (a + b+ c) -3 (a + c) (b + c))/((a + b) (b + c) (a + b + c))]` `= ((a^(2) + b^(2) - ab) - c^(2))/((a + b) (b + c) (a + b+ c)) = 0` [from Eq. (i)] or `(1)/(a + c) + (1)/(b + c) = (3)/(a + b + c)` |
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| 48. |
In `DeltaABC`, if AB = c is fixed, and `cos A + cosB + 2 cos C = 2` then the locus of vertex C isA. ellipseB. hyperbolaC. circleD. parabola |
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Answer» Correct Answer - A `cos A + cos B = 2(1 - cos C) = 4 sin^(2).(C)/(2)` `rArr 2cos.(A +B)/(2) cos.(A -B)/(2) = 4 sin^(2).(C)/(2)` `rArr cos.(A -B)/(2) = 2 sin.(C)/(2)` `rArr 2 cos.(C)/(2) cos.(A -B)/(2) = 4 sin.(C)/(2) cos.(C)/(2)` `rArr 2 sin.(A + B)/(2) cos.(A -B)/(2) = 2 sin C` `rArr sin A+ sin B = 2 sin C` `rArr a + b = 2c`(constant) So, locus of vertex C is an ellipse with vertices A and B as foci |
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| 49. |
Prove that `("a c o s"A+b cosB+ccosC)/(a+b+c)=r/Rdot` |
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Answer» We have `a cos A + b cos B + c cos C` `= R(2 sin A cos A + 2 sin B cos B + 2 sin A sin C)` `= R (sin 2A + sin 2B + sin 2C)` `= 4R sin A sin B sin C` and `a + b + c = 2R (sin A + sin B + sin C)` `= 8 R cos (A//2) cos (B//2) cos (C//2)` `rArr (a cos A + b cos B + c cos C)/(a + b + c)` `= (4R sin A sin B sin C)/(8R cos A //2 cos B//2 cos C//2)` `= ((2 sin.(A)/(2) cos.(A)/(2))(2sin .(B)/(2) cos.(B)/(2)) (2sin.(C)/(2) cos.(C)/(2)))/(2cos.(A)/(2) cos.(B)/(2) cos.(C)/(2))` `= 4 sin.(A)/(2)sin.(B)/(2) sin.(C)/(2) = (r)/(R)` |
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| 50. |
If in a triangle a `cos^2C/2+cos^2A/2=(3b)/2,`then find the relation between the sides of the triangle. |
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Answer» `a cos^(2).(C)/(2) + cos^(2).(A)/(2) = (3b)/(2)` or `a (1 + cos C) + c (1 + cos A) = 3b` or `a+c + (a cos C + c cos A) = 3b` or `a + c + b = 3b` [bt the projection formula] or `a + c = 2b` Hence, a, b, c and in A.P. |
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