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

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(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 $XZ < XY < YZ$.

Explore More

Similar Questions

It is given that $\Delta PQR \cong \Delta EDF$. Is it true to say that $PR = EF$? Give a reason for your answer.

State whether each of the following statements is true or false:
$(1)$ In $\Delta XYZ$,if $XY > XZ$,then $\angle Z > \angle Y$.
$(2)$ In $\Delta ABC$ and $\Delta PQR$,if $\frac{AB}{PR} = \frac{BC}{QP} = \frac{CA}{RQ} = 1$,then $\Delta ABC \cong \Delta RPQ$.

In $\triangle PQR$,$\angle R = \angle P$,$QR = 4 \, cm$,and $PR = 5 \, cm$. Then the length of $PQ$ is (in $cm$):

$ABC$ is an isosceles triangle with $AB = AC$. $BD$ and $CE$ are its two medians. Show that $BD = CE$.

In the figure,$AD$ is the bisector of $\angle BAC$. Prove that $AB > BD$.

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