Question

The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60°. Find the angles of the parallelogram.

Answer 1

​Given: In parallelogram ABCD, DP⊥ AB, AQ ⊥ BC and ∠PDQ = 60°

In quadrilateral DPBQ, by angle sum property, we have

$$\begin{gathered} \angle PDQ+\angle DPB+\angle B+\angle BQD=360° \\ \Rightarrow 60°+90°+\angle B+90°=360° \\ \Rightarrow \angle B=360°-240° \\ \Rightarrow \angle B=120° \end{gathered}$$

Therefore, $\angle B=120°$

Now,
$\angle B=\angle D=120°$ (Opposite angles of a parallelogram are equal.)

$\angle A+\angle B=180°$ (Adjacent angles of a parallelogram are supplementary.)

$$\begin{gathered} \Rightarrow \angle A+120°=180° \\ \Rightarrow \angle A=180°-120° \\ \Rightarrow \angle A=60° \end{gathered}$$

Also,
$\angle A=\angle C=60°$ (Opposite angles of a parallleogram are equal.)

So, the angles of a parallelogram are $60°,120°,60°\mathrm{and}120°.$