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  • ABC is right angled at B such that <img alt="\angle BCA = 2\angle BAC" src="https://learn.careers360.com/latex-image/?%5

\DeltaABC is right angled at B such that \angle BCA = 2\angle BAC. Then

Option: 1

AC = BC


Option: 2

AC = 2BC


Option: 3

AC = 3BC


Option: 4

None of the above


Answers (1)

best_answer

CONSTRUCTION: Produce BC to D such that BD = BC. Join AD.

\\\text{Let } \angle B A C=x^{\circ}. \text{Then, } \angle B C A=2 x^{\circ} \\\\\text{In } \triangle A B C\text{ and } \triangle A B D,\text{ we have} \\\\B C=B D \quad\text{ (by construction) } \\\\A B=A B \quad\text{( common )} \\\\\quad \angle A B C=\angle A B D=90^{\circ} \\ \\\therefore \quad \triangle A B C \cong \triangle A B D \quad \text { (SAS-criteria) }

\\ \therefore \quad \angle C A B=\angle D A B=x^{\circ} \quad \ldots (i) (c.p.c.t.) \\\\\text{and }A C=A D \quad \ldots (ii) (c.p.c.t.). \\\\\text{In } \triangle C A D,\text{ we have} \\\\\angle C A D=\angle C A B+\angle D A B=\left(x^{\circ}+x^{\circ}\right)=2 x^{\circ} \\ \\\angle A C D=\angle A C B=2 x^{\circ}

But, we know that the sides opposite to equal angles are equal.

\therefore \quad \angle A C D=\angle C A D \Rightarrow A D=C D

From (ii) and (iii), we get

A C=C D \Rightarrow A C=2 B C \quad[\because C D=B C+B D=2 B C]

Hence, AC = 2BC

Posted by

Devendra Khairwa

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