Logo
Jeeify
NVAMediumJEE 2023Properties of Triangles

JEE Mathematics 2023 Question with Solution

If the line x=y=zx=y=z intersects the line xsin⁡A+ysin⁡B+zsin⁡C−18=0x\sin A+y\sin B+z\sin C-18=0 and xsin⁡2A+ysin⁡2B+zsin⁡2C−9=0,x\sin 2A+y\sin 2B+z\sin 2C-9=0, where A,B,CA,B,C are the angles of a triangle ABCABC, then 80(sin⁡A2sin⁡B2sin⁡C2)80\left(\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\right) is equal to

Answer

Correct answer:5

Step-by-step solution

Standard Method

Given: The line x=y=zx=y=z intersects both equations

xsin⁡A+ysin⁡B+zsin⁡C−18=0x\sin A+y\sin B+z\sin C-18=0

and

xsin⁡2A+ysin⁡2B+zsin⁡2C−9=0.x\sin 2A+y\sin 2B+z\sin 2C-9=0.

Find: The value of 80(sin⁡A2sin⁡B2sin⁡C2)80\left(\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\right).

Let x=y=z=tx=y=z=t. Then the intersection point must satisfy both equations.

From the first equation,

t(sin⁡A+sin⁡B+sin⁡C)−18=0t(\sin A+\sin B+\sin C)-18=0

so

t=18sin⁡A+sin⁡B+sin⁡C.t=\frac{18}{\sin A+\sin B+\sin C}.

From the second equation,

t(sin⁡2A+sin⁡2B+sin⁡2C)−9=0t(\sin 2A+\sin 2B+\sin 2C)-9=0

so

t=9sin⁡2A+sin⁡2B+sin⁡2C.t=\frac{9}{\sin 2A+\sin 2B+\sin 2C}.

Equating the two values of tt,

18sin⁡A+sin⁡B+sin⁡C=9sin⁡2A+sin⁡2B+sin⁡2C.\frac{18}{\sin A+\sin B+\sin C}=\frac{9}{\sin 2A+\sin 2B+\sin 2C}.

Hence,

2(sin⁡2A+sin⁡2B+sin⁡2C)=sin⁡A+sin⁡B+sin⁡C.2(\sin 2A+\sin 2B+\sin 2C)=\sin A+\sin B+\sin C.

Using sin⁡2A=2sin⁡Acos⁡A\sin 2A=2\sin A\cos A, we get

4(sin⁡Acos⁡A+sin⁡Bcos⁡B+sin⁡Ccos⁡C)=sin⁡A+sin⁡B+sin⁡C.4(\sin A\cos A+\sin B\cos B+\sin C\cos C)=\sin A+\sin B+\sin C.

Now use the standard triangle identities

sin⁡A+sin⁡B+sin⁡C=4cos⁡A2cos⁡B2cos⁡C2\sin A+\sin B+\sin C=4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}

and

sin⁡2A+sin⁡2B+sin⁡2C=4sin⁡Asin⁡Bsin⁡C.\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C.

Then

2⋅4sin⁡Asin⁡Bsin⁡C=4cos⁡A2cos⁡B2cos⁡C2.2\cdot 4\sin A\sin B\sin C=4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}.

So,

2sin⁡Asin⁡Bsin⁡C=cos⁡A2cos⁡B2cos⁡C2.2\sin A\sin B\sin C=\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}.

Using

sin⁡A=2sin⁡A2cos⁡A2,sin⁡B=2sin⁡B2cos⁡B2,sin⁡C=2sin⁡C2cos⁡C2,\sin A=2\sin\frac{A}{2}\cos\frac{A}{2},\quad \sin B=2\sin\frac{B}{2}\cos\frac{B}{2},\quad \sin C=2\sin\frac{C}{2}\cos\frac{C}{2},

we obtain

2⋅8sin⁡A2sin⁡B2sin⁡C2cos⁡A2cos⁡B2cos⁡C2=cos⁡A2cos⁡B2cos⁡C2.2\cdot 8\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}=\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}.

Cancelling the nonzero factor cos⁡A2cos⁡B2cos⁡C2\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2},

16sin⁡A2sin⁡B2sin⁡C2=1.16\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}=1.

Therefore,

sin⁡A2sin⁡B2sin⁡C2=116.\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}=\frac{1}{16}.

Hence,

80(sin⁡A2sin⁡B2sin⁡C2)=80×116=5.80\left(\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\right)=80\times \frac{1}{16}=5.

Therefore, the required numerical value is 55.

The source solution reaches the same final value, although one intermediate identity there is incorrect; the final answer is still 55.

Common mistakes

  • Setting x=y=zx=y=z but forgetting to use a common parameter tt in both equations. This is wrong because the same intersection point must satisfy both equations simultaneously. Write x=y=z=tx=y=z=t first, then derive two expressions for the same tt and equate them.

  • Using the incorrect identity cos⁡A+cos⁡B+cos⁡C=1+4sin⁡A2sin⁡B2sin⁡C2\cos A+\cos B+\cos C=1+4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}. This is wrong because the correct triangle identity is cos⁡A+cos⁡B+cos⁡C=1+4sin⁡A2sin⁡B2cos⁡C2\cos A+\cos B+\cos C=1+4\sin\frac{A}{2}\sin\frac{B}{2}\cos\frac{C}{2} in a specific symmetric form only after using angle-sum constraints carefully. Instead, use the safer identities for sin⁡A+sin⁡B+sin⁡C\sin A+\sin B+\sin C and sin⁡2A+sin⁡2B+sin⁡2C\sin 2A+\sin 2B+\sin 2C.

  • Cancelling factors such as cos⁡A2cos⁡B2cos⁡C2\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2} without checking whether they can be zero. In a triangle, 00

Practice more Properties of Triangles questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Properties of Triangles

Related questions