Question

The sides $a,b,c$ of a triangle satisfy the relations $c^2=2ab$ and $a^2+c^2=3b^2$. Then the measure of $\angle BAC$, in degrees, is

• A. $30$
• B. $45$
• C. $60$
• D. $90$

Answer 1

The correct option is B $45$

In $△ABC,$

Sides $a,b,c$

$\Rightarrow$ $c^2=2ab$ [ given ] ----- ( 1 )

$\Rightarrow$ $a+c^2=3b^2$ [ Given ] ----- ( 2 )

$\Rightarrow$ $a^2+2ab=3b^2$ [ From ( 1 ) ]

$\Rightarrow$ $a^2+2ab+b^2=4b^2$ [ Adding $b^2$ on both sides ]

$\Rightarrow$ $(a+b{)}^2=4b^2$

$\Rightarrow$ $a+b=\pm 2b$

Since, $a,b,c$ are sides then they are always positive.

$\therefore$ $a+b=2b$

$\therefore$ $a=b$

Substitute $a=b$ in equation ( 1 ) we get,

$\Rightarrow$ $c^2=2a^2$

$\Rightarrow$ $c=\sqrt{2}a$

We can see,

$c^2=a^2+b^2$

$c^2=a^2+a^2$

$c=\sqrt{2}a$

$\therefore$ We can see $ABC$ is right angled triangle and $c$ is the hypotenuse.

$\Rightarrow$ $\angle C={90}^o$

$\Rightarrow$ $\angle A+\angle B={90}^o$

$\Rightarrow$ $\angle A+\angle A={90}^o$ [ Base angles of an equal sides also an equal ]

$\Rightarrow$ $2\angle A={90}^o$

$\therefore$ $\angle A={45}^o$

$\therefore$ $\angle BAC={45}^o$