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Trignometric Ratios, Functions and Equations Test - 4

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Trignometric Ratios, Functions and Equations Test - 4
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  • Question 1
    2 / -0.83

    If in a triangle  ABC, (s −a) (s −b) = s (s −c), then  angle C is equal to

    Solution

    Thus, C/2 = 45 º ⇒C = 90 º

  • Question 2
    2 / -0.83

    if A = 45,B = 75, then a+c √2 is equal to,

    Solution

    ∠C = 180 −45 −75 = 60

    sin 75 = sin (45 + 30) = sin45.cos30 + sin30.cos45
    sin 75  = 1/√2 x  √3/2 + 1/2  1/√2


  • Question 3
    2 / -0.83

    The angles of a triangle are as 1 : 2 : 7, then ratio of greatest side to least side is

    Solution

    Let the angles be x, 2x, and 7x respectively.
    ⇒x + 2x + 7x = 180 °[angle sum property of triangle]
    ⇒10x = 180 °
    ⇒x = 18 °
    Hence, the angles are : 18 °,36 °,126 °


    ⇒Greatest Side = K sin 126 °
    ⇒Smallest side = K sin 18 °
    So, required ratio =K sin 126 °/ K sin 18 °
    = sin(90 °+36 °)/sin18 °= cos36 °/sin18 °  [sin(90 °+x)=cosx]
    = (√5+1)/(√5 −1)

  • Question 4
    2 / -0.83

    If the sides of a triangle are 13, 7, 8 the greatest angle of the triangle is

    Solution

    a = 13, b = 7, c = 8

     By cosine formula,



    ⇒A = 120 º= 2 π/3

  • Question 5
    2 / -0.83

    There exists a triangle ABC satisfying the conditions

    Solution

    The sine formula is  
    a/sinA = b/sinB  
    ⇒a sin B = b sin A
    (a) b sin A = a  
    ⇒a sin B = a  
    ⇒B = π/2  
    Since, ∠A  <π/2 therefore, the triangle is possible.
    ∴The triangle is possible.

  • Question 6
    2 / -0.83

    In a  ΔABC, (b +c) cos A + (c + a) cos B + (a + b) cos C is equal to

    Solution

    (b+c) cos A  + (c + a) cos B + (a + b) cos C
    ⇒ b cos A + c cos A + c cos B + a cos B + a cos C + b cos C
    ⇒(b cos C + c cos B) + (c cos A + a cos C) + (a cos B + b cos A) ...(1)
    Using projection formula,
    a = (b cos C + c cos B) ...(2)
    b = (c cos A + a cos C) ...(3)
    c = (a cos B + b cos A) ...(4)

    On adding the projection formula, we get the initial expression
    i.e.  (2) + (3) + (4) = (1)
    ∴(b + c) cos A + (c + a) cos B + (a + b) cos C = a + b + c

  • Question 7
    2 / -0.83

    If the angles of a triangle ABC are in A.P., then

    Solution

    Let the angles be a,a+d,a+2d.
    Then,
    a+a+d+a+2d=180 º
    a+d=60 º
    So, ∠B=60 º,
    By cosine formula,

    ⇒b2   = a2   + c2   −ac

  • Question 8
    2 / -0.83

    The perimeter of a triangle ABC is 6 times the arithmetic mean of the sines of its angles. If the side b is 2, then the angle B is

  • Question 9
    2 / -0.83

    Solution



  • Question 10
    2 / -0.83

    The area of a triangle is  80cm2 and its perimeter is 8 cm. The radius of its inscribed circle is

    Solution

    Area of a triangle = 80 cm2 , a + b + c = 8 cm
    s = perimeter/2=(a + b + c)/2 = 8/2 = 4 cm

    Radius of inscribed circle (r) = (area of the triange)/s = 80/4 = 20 cm

  • Question 11
    2 / -0.83

    If the angles of a triangle be in the ratio 1 : 4 : 5, then the ratio of the greatest side to the smallest side is

  • Question 12
    2 / -0.83

    If a = 4, b = 3 and A = 60∘ , then c is a root of the equation

    Solution


    c2   −7 = 3c  ⇒c2   −3c −7 = 0

    Replace c by x : x2 - 3x - 7 = 0

  • Question 13
    2 / -0.83

    If  b2 +c2   = 3a2 , then cot B + cot C - cot A =

  • Question 14
    2 / -0.83

  • Question 15
    2 / -0.83

    If the sides of a triangle be 2x + 1, x2 - 1 and  x2  + x  + 1, then the greatest angle is

  • Question 16
    2 / -0.83

    If in a triangle ABC, acos2 (c/2) + ccos2 (A/2) = 3b/2, then its sides a, b, c are in

  • Question 17
    2 / -0.83

    In a  ΔABC, if a sin A = b sin B then the triangle is

  • Question 18
    2 / -0.83

    If in a  ΔABC, a cos A = b cos B, then the triangle is

  • Question 19
    2 / -0.83

    In a  ΔABC, is equal to  

  • Question 20
    2 / -0.83

    In a triangle ABC, a = 2b and  ∠A = 3 ∠B then angle A is

  • Question 21
    2 / -0.83

    In a triangle ABC, AD is the median A to BC, then its length is equal to

  • Question 22
    2 / -0.83

    If cos A + cos B = , then the sides of the triangle ABC are in

    Solution

    cos A + cos B = 4 sin2 (C/2 ​)
    ⇒2 cos (A+B)/2 ​cos (A −B)/2 ​= 4 sin2 (C/2 ​)

    ∵A + B + C = π⇒A + B = π−C

    ⇒cos (π−C)/2 ​cos (A −B)/2 ​= 2 sin2 (C/2 ​)
    ⇒sin C/2 ​cos (A −B)/2 ​= 2 sin2 (C/2 ​)
    ⇒cos (A −B)/2 = 2 sin (C/2 ​)
    ⇒cos C/2 ​cos (A −B)/2 = 2 sin (C/2 ​) cos (C/2)​
    ⇒cos (π−(A+B)​)/2 cos (A −B)/2 = sin C
    ⇒2 sin (A+B)/2 ​cos (A −B)/2 ​= sin C
    ⇒sin A + sin B = 2 sin C

    ∵a/sinA ​= b/sinB ​= c/sinC ​= k
    ⇒sinA = ak, sin B = bk , sin C = ck

    ⇒ak + bk = 2(ck)
    ⇒a+b=2c

    Therefore the sides of triangle a,b,c are in A.P.

  • Question 23
    2 / -0.83

    ABC is an equilateral triangle of each side a (> 0). The inradius of the triangle is

  • Question 24
    2 / -0.83

    In any  ΔABC, the expression (a + b + c) (a + b –c) (b + c –a) (c + a –b) is equal to

  • Question 25
    2 / -0.83

    In any  ΔABC, if C =90∘ , then tan B/2  is equal to

    Solution


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