Question

Construct $∆$DEF, such that, DE = 6.5 cm, $\angle$E = 50$°$, $\angle$F = 30$°$; and draw EM $\perp$ DF; measure the length EM.

Answer 1

$$\begin{gathered} \angle \mathrm{DEF}+\angle \mathrm{EDF}+\angle \mathrm{EFD}=180° \\ \Rightarrow \angle \mathrm{EDF}=180°-\angle \mathrm{EFD}-\angle \mathrm{DEF} \\ \Rightarrow \angle \mathrm{EDF}=180°-30°-50° \\ \Rightarrow \angle \mathrm{EDF}=100° \end{gathered}$$

Steps of construction:
1. Draw a line segment DE=6.5 cm.
2. Draw $\angle \mathrm{DEY}=50°$.
3. Draw another angle $\angle \mathrm{EDX}=100°$.
4. The rays EY and DX intersects at point F.
5. Draw a perpendicular EM on DF.

Thus, $∆$DEF is the required triangle. When measured, EM = 6.4 cm.