In a triangle ABC- if - A -600- a -5- b -4- then c is a root of the equationA- c2-5 c-9-0B- c 2-4 c -9-0C- c2-10 c-25-0D- c2-5 c-41-0

The correct option is B c2 4c 9 = 0 Given, Δ A=60^0,a=5,b=4 ∴ cos A=b^2+c^2 a^2/2bc ⇒cos 60^0=1/2=16+c^2 25/8c ⇒ 4c=c^2 9 ⇒ c^2 4c 9=0 ...

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Standard XII

Mathematics

Cosine Rule

In a triangle...

Question

$In a triangle ABC, if ∠ A = 60^{0}, a = 5, b = 4, then c is a root of the equation$

c² - 5c - 9 = 0

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c² - 4c - 9 = 0

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c² - 10c + 25 = 0

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c² - 5c - 41 = 0

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Solution

The correct option is B
c² - 4c - 9 = 0

$G i v e n, Δ A = 60^{0}, a = 5, b = 4 ∴ c o s A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} \Rightarrow c o s 60^{0} = \frac{1}{2} = \frac{16 + c^{2} - 25}{8 c} \Rightarrow 4 c = c^{2} - 9 \Rightarrow c^{2} - 4 c - 9 = 0$

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In a triangle ABC, if∠A=600,a=5,b=4,then c is a root of the equation

The correct option is B c2 - 4c - 9 = 0 Given,ΔA=600,a=5,b=4∴cosA=b2+c2−a22bc⇒cos 600=12=16+c2−258c⇒4c=c2−9⇒c2−4c−9=0

$In a triangle ABC, if ∠ A = 60^{0}, a = 5, b = 4, then c is a root of the equation$

The correct option is B
c² - 4c - 9 = 0

$G i v e n, Δ A = 60^{0}, a = 5, b = 4 ∴ c o s A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} \Rightarrow c o s 60^{0} = \frac{1}{2} = \frac{16 + c^{2} - 25}{8 c} \Rightarrow 4 c = c^{2} - 9 \Rightarrow c^{2} - 4 c - 9 = 0$