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Standard XII

Mathematics

Combination

An isosceles ...

Question

An isosceles $△ A B C$ is inscribed in a circle $x^{2} + y^{2} = a^{2}$ with the vertex $A$ at $(a, 0)$ and the base angles $B$ and $C$ each equal to $75^{0}$ then length of the base $B C$ is

\frac{a}{2}

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a

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\frac{2 a}{\sqrt{3}}

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\frac{\sqrt{3} a}{2}

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Solution

The correct option is A $a$
$∠ B = ∠ C = 75^{0}$
$\Rightarrow ∠ B A C = 30^{0} \Rightarrow ∠ B O C = 60^{0}$
$\Rightarrow B O C$ is an equilateral triangle
$\Rightarrow B C = O B =$ the radius of the circle
$\Rightarrow B C = a$

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An isosceles △ABC is inscribed in a circle x2+y2=a2 with the vertex A at (a,0) and the base angles B and C each equal to 750 then length of the base BC is

The correct option is A a∠B=∠C=750⇒∠BAC=300⇒∠BOC=600⇒BOC is an equilateral triangle⇒BC=OB= the radius of the circle⇒BC=a

An isosceles $△ A B C$ is inscribed in a circle $x^{2} + y^{2} = a^{2}$ with the vertex $A$ at $(a, 0)$ and the base angles $B$ and $C$ each equal to $75^{0}$ then length of the base $B C$ is

The correct option is A $a$
$∠ B = ∠ C = 75^{0}$
$\Rightarrow ∠ B A C = 30^{0} \Rightarrow ∠ B O C = 60^{0}$
$\Rightarrow B O C$ is an equilateral triangle
$\Rightarrow B C = O B =$ the radius of the circle
$\Rightarrow B C = a$