Parallel lines AB and CD intersect with line MK, respectively, at points N and P. The angle KPD is four times

| August 10, 2021 | education

Parallel lines AB and CD intersect with line MK, respectively, at points N and P. The angle KPD is four times the sum of the angles MNB and MPD. Find all formed corners. 4. In triangle ABC, angle C is 60 °. Point D is marked on the AC side so that the BDC angle is 60 °, the ABD angle is 30 °, and CD = 5cm. Find AC and the distance from point D to side AB.

1. Construction in the figure.

Because straight lines are parallel, then the angle KPD = 4 * (MNB + MPD) = 4 * 2 * MNB = 8 * MNB, then MNB = MPD = ANK = CPN = KPD / 8.

Angle KNB = KPD = CPN = ANP = 180 ° – MNB = 180 ° – KPD / 8.

2. Because angle BDC = C = 60 °, and also considering that the sum of the angles in the triangle is 180 °, then the angle DBC = 180 ° – (60 ° + 60 °) = 180 ° – 120 ° = 60 °, therefore, the triangle BDC is equilateral, therefore BD = BC = DC = 5 cm.

Angle B = ABD + DBC = 30 ° + 60 ° = 90 °, therefore, triangle ABC is rectangular, AC is hypotenuse.

AB = BC * tg C = 5 * tan 60 ° = 5 * √3 cm.

AC = √ (AB² + BC²) = √ (75 + 25) = 10 cm.

AD = AC – DC = 10 – 5 = 5 cm.

Angle A = 90 ° – C = 90 ° – 60 ° = 30 °.

KD = AD * sin A = 5 * sin 30 ° = 2.5 cm.

Answer: AC = 10 cm, KD = 2.5 cm.



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