In a ABC- a-5- c-2-6- - C-600- If b1- b2 are two possible values of third side- then the value of b1-b2b1b2-In ABC- usual notations are given-

Given : a=5, c=2√(6), ∠ C=60^0 using cosine formula : cos C=a^2+b^2 c^22ab ⇒12=25+b^2 2410b ⇒ b^2 5b+1=0 ⇒b_1+b_2b_1b_2=5 ...

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Standard XII

Mathematics

Circum Radius

In a ABC, a=5...

Question

In a $△ A B C, a = 5, c = 2 \sqrt{6}, ∠ C = 60^{0} .$ If $b_{1}, b_{2}$ are two possible values of third side, then the value of $\frac{b_{1} + b_{2}}{b_{1} b_{2}} =$
(In $△ A B C,$ usual notations are given.)

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Solution

The correct option is B $5$
Given : $a = 5, c = 2 \sqrt{6}, ∠ C = 60^{0}$
using cosine formula :
$cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b} \Rightarrow \frac{1}{2} = \frac{25 + b^{2} - 24}{10 b} \Rightarrow b^{2} - 5 b + 1 = 0$
$\Rightarrow \frac{b_{1} + b_{2}}{b_{1} b_{2}} = 5$

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In a △ABC,a=5,c=2√6,∠C=600. If b1,b2 are two possible values of third side, then the value of b1+b2b1b2= (In △ABC, usual notations are given.)

The correct option is B 5Given : a=5,c=2√6,∠C=600 using cosine formula : cosC=a2+b2−c22ab⇒12=25+b2−2410b⇒b2−5b+1=0 ⇒b1+b2b1b2=5

In a $△ A B C, a = 5, c = 2 \sqrt{6}, ∠ C = 60^{0} .$ If $b_{1}, b_{2}$ are two possible values of third side, then the value of $\frac{b_{1} + b_{2}}{b_{1} b_{2}} =$
(In $△ A B C,$ usual notations are given.)

The correct option is B $5$
Given : $a = 5, c = 2 \sqrt{6}, ∠ C = 60^{0}$
using cosine formula :
$cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b} \Rightarrow \frac{1}{2} = \frac{25 + b^{2} - 24}{10 b} \Rightarrow b^{2} - 5 b + 1 = 0$
$\Rightarrow \frac{b_{1} + b_{2}}{b_{1} b_{2}} = 5$