Tick the correct answer and justify : In $∆ABC, AB = 6\sqrt3\ cm, AC = 12\ cm$ and $BC = 6\ cm$. The angle B is:
(a) $120^o$
(b) $60^o$
(c) $90^o$
(d) $45^o$
Given:
In $∆ABC, AB = 6\sqrt3\ cm, AC = 12\ cm$ and $BC = 6\ cm$.
To do:
We have to find angle B.
Solution:
$AC^{2}=(12)^2$
$=144$
$A B^{2}+B C^{2}=(6 \sqrt{3})^{2}+(6)^{2}$
$=108+36$
$=144$
$AC^2=A B^{2}+B C^{2}$
This implies,
Triangle ABC is a right angle triangle, right angled at $B$.
$\angle B =90^{\circ}$
Related Articles
- Construct a $∆ABC$ in which $AB + AC = 5.6\ cm, BC = 4.5\ cm$ and $\angle B = 45^o$.
- Construct a $∆ABC$ in which $BC = 3.6\ cm, AB + AC = 4.8\ cm$ and $\angle B = 60^o$.
- Construct a $\triangle ABC$ in which $BC = 3.4\ cm, AB - AC = 1.5\ cm$ and $\angle B = 45^o$.
- Construct $\vartriangle ABC$ in which $BC=7\ cm,\ \angle B=75^{o}$ and $AB+AC=12\ cm$.
- Using ruler and compasses only, construct a $∆ABC$, given base $BC = 7\ cm, \angle ABC = 60^o$ and $AB + AC = 12\ cm$.
- Using ruler and compasses only, construct a $\triangle ABC$, from the following data:$AB + BC + CA = 12\ cm, \angle B = 45^o$ and $\angle C = 60^o$.
- Construct a $\vartriangle ABC$ in which $AB\ =\ 6\ cm$, $\angle A\ =\ 30^{o}$ and $\angle B\ =\ 60^{o}$, Construct another $\vartriangle AB’C’$ similar to $\vartriangle ABC$ with base $ AB’\ =\ 8\ cm$.
- Construct $∆ABC$ such that $AB=2.5\ cm$, $BC=6\ cm$ and $AC=6.5\ cm$. Measure $\angle B$.
- Construction a quadrilateral $LION$ with $LI=6.5\ cm$, $IO=7.2\ cm$, $\angle I=90^o$, $\angle O = 60^o$ and $\angle N=105^o$.
- ABCD is a trapezium in which \( A B \| C D \). The diagonals \( A C \) and \( B D \) intersect at \( O . \) If \( O A=6 \mathrm{~cm}, O C=8 \mathrm{~cm} \) find \( \frac{\text { Area }(\Delta A O B)}{\text { Area }(\Delta C O D)} \).
- ABCD is a trapezium in which \( A B \| C D \). The diagonals \( A C \) and \( B D \) intersect at \( O . \) If \( O A=6 \mathrm{~cm}, O C=8 \mathrm{~cm} \) find \( \frac{\text { Area }(\Delta A O D)}{\text { Area }(\Delta C O D)} \).
- Let ABC be a right triangle in which $AB = 6\ cm, BC = 8\ cm$ and $\angle B = 90^o$. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
- In a $\triangle ABC$, if $\angle A = 120^o$ and $AB = AC$. Find $\angle B$ and $\angle C$.
- In case of negative work, the angle between the force and displacement is:(a) $0^{o}$(b) $45^{o}$(c) $90^{o}$(d) $180^{o}$
- In $\triangle ABC$, side $AB$ is produced to $D$ so that $BD = BC$. If $\angle B = 60^o$ and $\angle A = 70^o$, prove that $AD > AC$.
Kickstart Your Career
Get certified by completing the course
Get Started