Mark the correct alternative in each of the following :

$In a triangle A B C, a = 4, b = 3, ∠ A = 60^{\circ} then c is a root of the equation$

$c^{2} - 3 c - 7 = 0$

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$c^{2} + 3 c + 7 = 0$

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$c^{2} - 3 c + 7 = 0$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$c^{2} + 3 c - 7 = 0$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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Solution

The correct option is A
$c^{2} - 3 c - 7 = 0$

$G i v e n : I n Δ A B C a = 4, b = 3, ∠ A = 60^{\circ} ∴ c o s A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} \Rightarrow c o s 60^{\circ} = \frac{3^{2} + c^{2} - 4^{2}}{2 \times 3 \times c} \Rightarrow \frac{1}{2} = \frac{9 + c^{2} - 16}{6 c} \Rightarrow c^{2} - 3 c - 7 = 0$

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Similar questions

∠ A = 60 °

c^{2} - 3 c - 7 = 0

c^{2} + 3 c + 7 = 0

c^{2} - 3 c + 7 = 0

c^{2} + 3 c - 7 = 0

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

c^{2} - 3 c - 10 = 0

y = m x + c

(x - 3)^{2} + y^{2} = 1

L_{1}

L_{1}

x^{2} + y^{2} = 1

(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})

In the circuit shown in the figure

C_{1} = C, C_{2} = 2 C, C_{3} = 3 C, C_{4} = 4 C .

Then select correct alternate

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