Question

State wether its true or not, $\angle BOC={90}^0-\frac{\angle BAC}{2}$ ?

• A. True
• B. False

Answer 1

The correct option is B False

(Refer to Image)

$2a=x+z$

$\Rightarrow 2a+y=180$

$\Rightarrow 180-y=x+z$

$\Rightarrow 2b=x+y$

$b=\frac{x+y}{2};a=\frac{x+z}{2}$

$\Rightarrow a+b+k=180$

$\Rightarrow k=180-a-b$

$=180-(x+\frac{y+z}{2})$

$\Rightarrow k=180-x-\left(\frac{y+z}{2}\right)$

$\Rightarrow \angle BOC=180-\angle BAC-\left(\frac{\angle ABC+\angle ACB}{2}\right)$

$\therefore$ Given statement is false.