In the parallelogram from the vertex of an obtuse angle equal to 150 degrees, two unequal heights are drawn

In the parallelogram from the vertex of an obtuse angle equal to 150 degrees, two unequal heights are drawn, the sum of the lengths of which is 10. Find the perimeter of the parallelogram.

1. A, B, C, D – the tops of the parallelogram. Angle D = 150 °. ВK – the height drawn to AD. BH is the height drawn to the CD. BK + BH = 10 units of measurement.

2. Angle АBК = 150 – angle СBК = 150 ° – 90 ° = 60 °.

3. Angle СBН = 150 – angle АBН = 150 ° – 90 ° = 60 °.

4. BK: AB = cosine of angle ABK = cosine 60 ° = 1/2.

AB = BK: 1/2 = 2BK.

5. BH: BC = cosine of angle СBH = cosine 60 ° = 1/2.

BC = BH: 1/2 = 2BH.

6. The perimeter of the parallelogram is = 2 (AB + BC) = 2 (2BK + 2BH) = 4 (BK + BH) 4 x 10 = 40 units.



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