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

Let the parallelogram be ABCD, in which `angleADC and angleABC ` are obture angles. Now, DE and DF are two altitudes of parallelogram and angles between them is `60^(@)`.

Now, BEDF is a quadrilateral, in which `angleBED=angleBFD=90^(@)`
`therefore" "angleFBE=360^(@)-(angleFDE+angleBED+angleBFD)`
`therefore" "angleFBE=360^(@)-(angleFDE+angleBED+angleBFD)`
`=360^(@)-240^(@)=120^(@)`
Since, ABCD is parallelogram. ltBrgt `therefore" "angleADC= 120^(@)`
Now, `" "angleA+angleB=180^(@)" "` [sum of two cointerior angles is `180^(@)`] ltBrgt `therefore" "angleA=180^(@)-angleB`
`5=180^(@)-120^(@)" "[becauseangleFBE=angleB]` ltBrgt `rArr" "angleA=60^(@)`
Also, `" "angleC=angleA=60^(@)`
Hence, angles of the parallelogram are `60^(@), 120^(@), 60^(@) and 120^(@),` respectively.