JEE Main 2012 12 May Online
Exam JEE Main
Year 2012
Date 12 May
Mode Online
Subject Mathematics
Chapter Properties of Triangles JEE Main 2012 · 12 May · Online · Mathematics · Properties of Triangles
If in a triangle A B C, (b+c)/(11)=(c+a)/(12)=(a+b)/(13) , then A...
If in a triangle
A B C , b + c 11 = c + a 12 = a + b 13 A B C, \frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13} A BC , 11 b + c = 12 c + a = 13 a + b , then
cos A \cos \mathrm{A} cos A is equal to
Options A 5 / 7 5 / 7 5/7 B 1 / 5 1 / 5 1/5 C 35 / 19 35 / 19 35/19 D 19 / 35 19 / 35 19/35 Correct answer B. 1 / 5 1 / 5 1/5
Step-by-step solution
Let
b + c 11 = c + a 12 = a + b 13 = K \frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=K 11 b + c = 12 c + a = 13 a + b = K in a triangle
A B C A B C A BC .
⇒ b + c = 11 K , c + a = 12 K , a + b = 13 K \Rightarrow b+c=11 K, c+a=12 K, a+b=13 K ⇒ b + c = 11 K , c + a = 12 K , a + b = 13 K On solving these, we get
a = 7 K , b = 6 K , c = 5 K a=7 K, b=6 K, c=5 K a = 7 K , b = 6 K , c = 5 K
Now we know,
cos A = b 2 + c 2 − a 2 2 b c = 36 K 2 + 25 K 2 − 49 K 2 2 ( 6 K ) 5 K = 1 5 \begin{aligned}& \cos A=\frac{b^2+c^2-a^2}{2 b c}\\
& =\frac{36 K^2+25 K^2-49 K^2}{2(6 K) 5 K}=\frac{1}{5}
\end{aligned} cos A = 2 b c b 2 + c 2 − a 2 = 2 ( 6 K ) 5 K 36 K 2 + 25 K 2 − 49 K 2 = 5 1
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