Write the measures of its sides in ascending order in each of the following triangles:
$(1) \text{In } \Delta ABC, \angle A = 50^{\circ} \text{ and } \angle B = 60^{\circ}$

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(A) In $\Delta ABC$,the sum of angles is $180^{\circ}$.
Therefore,$\angle C = 180^{\circ} - (50^{\circ} + 60^{\circ}) = 180^{\circ} - 110^{\circ} = 70^{\circ}$.
We know that the side opposite to the smaller angle is smaller and the side opposite to the larger angle is larger.
The angles in ascending order are $\angle A < \angle B < \angle C$ $(50^{\circ} < 60^{\circ} < 70^{\circ})$.
Therefore,the sides in ascending order are $BC < AC < AB$.

Explore More

Similar Questions

Write the measures of the sides of $\Delta XYZ$ in ascending order,given that $\angle X = 80^{\circ}$ and $\angle Y = 30^{\circ}$.

In the given figure,$AM$ and $BN$ are both perpendicular to $AB$. $MN$ intersects $AB$ at $P$. Also,$P$ is the midpoint of $AB$. Prove that $AM = BN$ and $P$ is the midpoint of $MN$.

In $\Delta XYZ$,$\angle Y = 90^{\circ}$ and $XY = YZ$,then $\angle X = \dots$ (in $^{\circ}$)

Prove that the sum of any two sides of a triangle is greater than twice the median with respect to the third side.

Bisectors of the angles $B$ and $C$ of an isosceles triangle with $AB = AC$ intersect each other at $O$. $BO$ is produced to a point $M$. Prove that $\angle MOC = \angle ABC$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo