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Properties of Triangle question

2010 · Shift 1 · Q39
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  5. /2010 · Shift 1 · Q39

Properties of Triangle question

2010 · Shift 1 · Q39

JEE AdvancedMathematicsProperties of TriangleMCQ+4 / −1
If the angles A,BA, BA,B and CCC of a triangle are in an arithmetic progression and if a,ba, ba,b and ccc denote the lengths of the sides opposite to A,BA, BA,B and CCC respectively, then the value of the expression acsin⁡2C+casin⁡2A{a \over c}\sin 2C + {c \over a}\sin 2Aca​sin2C+ac​sin2A is
  1. A
    12{1 \over 2}21​
  2. B
    32{{\sqrt 3 } \over 2}23​​
  3. C
    111
  4. D
    3{\sqrt 3 }3​
View written solutionFree

Correct answer: D

  1. Since the angles of the triangle are in arithmetic progression, we have 2B=A+C.2B=A+C.2B=A+C. But in any triangle, A+B+C=π.A+B+C=\pi.A+B+C=π. Combining these: A+(A+C)+C=π⇒2A+2C=π⇒A+C=π2.A+(A+C)+C=\pi \Rightarrow 2A+2C=\pi \Rightarrow A+C=\frac{\pi}{2}.A+(A+C)+C=π⇒2A+2C=π⇒A+C=2π​. Hence, B=A+C2=π3.B=\frac{A+C}{2}=\frac{\pi}{3}.B=2A+C​=3π​.

    So the three angles can be written as A=π3−d,B=π3,C=π3+dA=\frac{\pi}{3}-d,\quad B=\frac{\pi}{3},\quad C=\frac{\pi}{3}+dA=3π​−d,B=3π​,C=3π​+d for some ddd.

  2. By the Sine Rule, asin⁡A=csin⁡C\frac{a}{\sin A}=\frac{c}{\sin C}sinAa​=sinCc​ so ac=sin⁡Asin⁡C,ca=sin⁡Csin⁡A.\frac{a}{c}=\frac{\sin A}{\sin C},\qquad \frac{c}{a}=\frac{\sin C}{\sin A}.ca​=sinCsinA​,ac​=sinAsinC​.

  3. Therefore,

    =\frac{\sin A}{\sin C}(2\sin C\cos C)+\frac{\sin C}{\sin A}(2\sin A\cos A).$$ Simplifying, $$=2\sin A\cos C+2\sin C\cos A.$$

  4. Use the identity sin⁡xcos⁡y+sin⁡ycos⁡x=sin⁡(x+y).\sin x\cos y+\sin y\cos x=\sin(x+y).sinxcosy+sinycosx=sin(x+y). Hence, 2sin⁡Acos⁡C+2sin⁡Ccos⁡A=2sin⁡(A+C).2\sin A\cos C+2\sin C\cos A=2\sin(A+C).2sinAcosC+2sinCcosA=2sin(A+C).

  5. Since A+C=π−B=π−π3=2π3A+C=\pi-B=\pi-\frac{\pi}{3}=\frac{2\pi}{3}A+C=π−B=π−3π​=32π​, 2sin⁡(A+C)=2sin⁡2π3=2⋅32=3.2\sin(A+C)=2\sin\frac{2\pi}{3}=2\cdot \frac{\sqrt{3}}{2}=\sqrt{3}.2sin(A+C)=2sin32π​=2⋅23​​=3​.

  6. Thus the value of the expression is 3.\boxed{\sqrt{3}}.3​​.

So the correct option is D.

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