In triangle ABC, angle A is 30 degrees, AC is 12 cm, AB is 10 cm. A line A is drawn through vertex C parallel to AB. Find the distance from point B to line AC Distance between lines A and AB
The distance from point B to the AC side is the height BD.
Then the triangle ABD is rectangular in which the angle BAD = 30 according to the condition.
The leg BD lies against an angle of 30, then BD = AB / 2 = 10/2 = 5 cm.
The distance from point B to straight line “a” is the perpendicular BK, which is equal to the height CH of the triangle ABC. Then, in a right-angled triangle ABC, the CH leg lies opposite the angle 30, then CH = AC / 2 = 12/2 = 6 cm.
ВK = CH = 6 cm.
Answer: From point B to straight AC 5 cm, between straight lines a and AB 6 cm.
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