Question

top is 12 meters across: straight line down is 7.81 meters, slope is 2:1. what is the base

Answer 1

Answer on Question #70730 – Math – Trigonometry

Question

Top is 12 meters across: straight line down is 7.81 meters, slope is 2:1. What is the base?

Solution

$$AB = 12$$ $$AD = 7.81$$

Angles $\angle BAD$ and $\angle ADC$ are equal to $90{}^{\circ}$ (we assume the minimum distance is between AB and DC so the angle $\angle ADC = 90{}^{\circ}$, and ABCD is a trapezium then the angle $\angle DAB = 90{}^{\circ}$, or else we can't find the base). BH is a height of the trapezium; ABHD is a rectangle, so $AB = HD$, $AD = BH$.

The slope is 2:1 so $\tan(\angle BCH) = 1/2$ (or $\tan(\angle BCH) = 2$, that depends on what you mean when you say that the slope is 2:1).

If $\tan(\angle BCD) = 1/2$ then $\tan(\angle BCD) = \tan(\angle BCH) = BH / CH = 1/2$, so $CH = 2 \times BH = 2 \times 7.81 = 15.62$ (meters).

The base is $DC = DH + CH = 12 + 15.62 = 27.62$ (meters).

If $\tan(\angle BCD) = 2$ then $\tan(\angle BCD) = \tan(\angle BCH) = BH / CH = 2$, so $CH = BH / 2 = 7.81 : 2 = 3.905$ (meters).

The base is $DC = DH + CH = 12 + 3.905 = 15.905$ (meters).

Answer: 27.62 meters or 15.905 meters.

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