MCQ
In $\triangle\text{ABC, }\text{AB}=6\sqrt{3 }\text{cm},\text{AC}=12\text{cm}$ and $\text{BC}=6\text{cm}.$ Then, $\angle\text{B}$ is:
- A$45^\circ$
- B$60^\circ$
- ✓$90^\circ$
- D$120^\circ$
✓
Answer
Correct option: C.
$90^\circ$
In $\triangle\text{ABC},$
$\text{AB}=6\sqrt{3 }\text{cm},\text{AC}=12\text{cm}$ and $\text{BC}=6\text{cm}$
$\text{AB}^2+\text{BC}^2$
$=3\sqrt{3}^2+6^2$
$=108+36$
$=144$
$\text{AC}^2=12^2=144$
$\Rightarrow\text{AB}^2+\text{BC}^2=\text{AC}^2$
So, by the Converse of Pythagoras theorem,
$\triangle\text{ABC}$ is a right angled triangle and since $AC$ is the hypotenuse,
$\angle\text{B}$ which is opposite $\text{AC} = 90^\circ.$
$\text{AB}=6\sqrt{3 }\text{cm},\text{AC}=12\text{cm}$ and $\text{BC}=6\text{cm}$
$\text{AB}^2+\text{BC}^2$
$=3\sqrt{3}^2+6^2$
$=108+36$
$=144$
$\text{AC}^2=12^2=144$
$\Rightarrow\text{AB}^2+\text{BC}^2=\text{AC}^2$
So, by the Converse of Pythagoras theorem,
$\triangle\text{ABC}$ is a right angled triangle and since $AC$ is the hypotenuse,
$\angle\text{B}$ which is opposite $\text{AC} = 90^\circ.$
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