Write the measures of the sides of $\Delta XYZ$ in ascending order,given that $\angle X = 80^{\circ}$ and $\angle Y = 30^{\circ}$.
(A) In $\Delta XYZ$,the sum of angles is $180^{\circ}$.
Therefore,$\angle Z = 180^{\circ} - (80^{\circ} + 30^{\circ}) = 180^{\circ} - 110^{\circ} = 70^{\circ}$.
The angles in ascending order are $\angle Y (30^{\circ}) < \angle Z (70^{\circ}) < \angle X (80^{\circ})$.
Since the side opposite to the smaller angle is smaller,the sides in ascending order are $XZ < XY < YZ$.
Therefore,$\angle Z = 180^{\circ} - (80^{\circ} + 30^{\circ}) = 180^{\circ} - 110^{\circ} = 70^{\circ}$.
The angles in ascending order are $\angle Y (30^{\circ}) < \angle Z (70^{\circ}) < \angle X (80^{\circ})$.
Since the side opposite to the smaller angle is smaller,the sides in ascending order are $XZ < XY < YZ$.
Explore More
Similar Questions
In $\Delta PQR$,$\angle Q = \angle R$ and $PQ = 6.5 \, cm$,then find $PR$. (in $, cm$)
Easy
View SolutionAn exterior angle of a triangle is $105^{\circ}$ and its two interior opposite angles are equal. Each of these equal angles is
Easy
View SolutionIn $\Delta PQR$,$PQ = PR$ and $\angle R = 40^{\circ}$,then $\angle P = \dots$ (in $^{\circ}$)
Easy
View SolutionIn $\Delta PQR$,$\angle P = 42^{\circ}$ and $\angle Q = 75^{\circ}$,then find $\angle R$. (in $^{\circ}$)
Easy
View SolutionPoint $P$ lies in the interior of $\Delta ABC$. Prove that $PB + PC < AB + AC$.
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo