If $a, b$ and $c$ are the longest of the sides of a right angled triangle $A B C$ with $c = 2 a$ and $b^{2} - 3 a^{2} = 0$ then $∠ A B C =$ _____

60^{o}

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30^{o}

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45^{o}

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90^{o}

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Solution

The correct option is A $60^{o}$
$\begin{matrix} C = 2 a, b^{2} = 3 a^{2} \to (i) b^{2} + a^{2} = c^{2} p y t h a g o r o u s t h e o r e m f r o m (i) 3 a^{2} + a^{2} = {(2 a)}^{2} b^{2} = 3 a^{2} {(\frac{b}{a})}^{2} = \frac{3}{1} = {(\frac{\sqrt{3}}{1})}^{2} tan B = \sqrt{3} = tan 60^{0} ∠ A B C = 60^{0} \end{matrix}$

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Similar questions

△ A B C, ∠ A = 90^{o}, ∠ B = 60^{o}, ∠ C = 30^{o}

△ A B C, ∠ B = 90^{o}, ∠ A = 45^{o} = ∠ C

sec 45^{o} =

a^{2} + b^{2} + c^{2} = c a + a b \sqrt{3}

A = 90^{o}, B = 60^{o}, C = 30^{o}

α^{2} + β^{2} = 0, α, β \in R

α = β = 0

△ A B C, ∠ B = 90^{o}, ∠ A = 45^{o} = ∠ C

c o s e c 45^{o} =

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