Question

The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is ${60}^o$. Find the angles of the parallelogram.

• A. $120,60,120,60$
• B. $110,70,110,70$
• C. $125,55,125,55$
• D. $115,65,115,65$

Answer 1

The correct option is A $120,60,120,60$

Given-

$\square ABCD$ is a parallelogram.

$\angle B$ & $\angle D$ are obtuse.

$BM$ & $DN$ are altitudes from B & D on AD & AB, respectively, intersecting at O.

$\angle DOM=BON$ ...(vertically opposite angles$={60}^o$)

To find out-

The angles of the parallelogram ABCD.

Solution-

$\angle DOM+\angle NOM={180}^o$ ...(linear pair)

$⟹{60}^o+\angle NOM={180}^o$

$⟹\angle NOM={120}^o$ .........(i)

Now, $\angle AMO=\angle DNO={90}^O$ ...(both are perpendiculares on AD & AB, respectively.)$

$\therefore$ In the quadrilateral $AMON$,

$\angle AMO+\angle DNO={180}^O$

$\therefore \angle NOM+\angle MAN={{360}^o-180}^O={180}^o$

(sum of the angles of a quadrilateral $={360}^o)$

$⟹\angle MAN={180}^O-{120}^o$ ...(from i )

i.e $\angle A={60}^o$

$\therefore \angle A=\angle C={60}^o$ ....(opposite angles of a paralleogram)

$\therefore \angle D+\angle B={360}^o-{120}^o={240}^o$

$\therefore \angle D=\angle B=\frac{{240}^o}{2}={120}^o$ ...(opposite angles of a paralleogram)

Ans- $\angle A=\angle C={60}^o,\angle D=\angle B={120}^o$