In ABC if - C- 60 0- then 1 - a-c - 1 - b-c - k - a-b-c where k-

Since, #160;cos C =cosπ #160;/ 3 #160; = 1 / 2 ∴ a ^ 2 + b ^ 2 + c ^ 2 =ab Adding bc+ac to both sides, we have #160; b ^ 2 +bc+ a ^ ...

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Standard X

Mathematics

Similarity of Right-angled Triangles

In ABC if ...

Question

In $△ A B C$ if $∠ C = 60^{0}$ , then $\frac{1}{a + c} + \frac{1}{b + c} = \frac{k}{a + b + c}$ where $k =$

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Solution

Since, $cos C = cos \frac{π}{3} = \frac{1}{2}$
$∴ a^{2} + b^{2} + c^{2} = a b$
Adding $b c + a c$ to both sides, we have
$b^{2} + b c + a^{2} + a c = a b + b c + c a + c^{2}$
$\Rightarrow b (b + c) + a (a + c) = (a + c) (b + c)$
$\Rightarrow \frac{b}{a + c} + \frac{a}{b + c} = 1 \Rightarrow 1 + \frac{b}{a + c} + 1 + \frac{a}{b + c} = 3$
$\Rightarrow \frac{a + b + c}{a + c} + \frac{a + b + c}{b + c} = 3$
$∴ \frac{1}{a + c} + \frac{1}{b + c} = \frac{3}{a + b + c}$
$∴ k = 3$

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In △ABC if ∠C=600, then 1a+c+1b+c=ka+b+c where k=

Since, cosC=cosπ3=12∴a2+b2+c2=abAdding bc+ac to both sides, we have b2+bc+a2+ac=ab+bc+ca+c2⇒b(b+c)+a(a+c)=(a+c)(b+c)⇒ba+c+ab+c=1⇒1+ba+c+1+ab+c=3⇒a+b+ca+c+a+b+cb+c=3∴1a+c+1b+c=3a+b+c∴k=3

In $△ A B C$ if $∠ C = 60^{0}$ , then $\frac{1}{a + c} + \frac{1}{b + c} = \frac{k}{a + b + c}$ where $k =$