Chapter 6.2, Problem 8PSB

To determine

To find: The given line segment AB is perpendicular to given plane ‘ m’ or not

Answer to Problem 8PSB

The given line segment $\overline{\text{AB}}$ is not perpendicular to given plane m.

Explanation of Solution

Given information:

Two given line segments are perpendicular, i.e., $\overline{\text{AB}}\perp \overline{\text{BD}}$

Magnitude of angle 1 $\left(\angle \text{ABD}\right)=\frac{2}{3}\text{x + 56}$ and

Magnitude of angle 2 $\left(\angle \text{ABC}\right)=2\text{x}-\text{10}$

Formula used:

Congruent angles in a triangle are always equal in magnitude

Calculation:

Since given

  $\overline{\text{AB}}\perp \overline{\text{BD}}$

Further,

In $\Delta s\text{ ABC \& ABD}$ , we have

  $\overline{\text{AB}}\perp \overline{\text{BC}}$

  $\begin{array}{l}\Rightarrow \angle \text{ABD=}\angle \text{ABC}\\ \Rightarrow \frac{2}{3}\text{x + 56=2x }-\text{10}\\ \Rightarrow \frac{2}{3}\text{x}-2\text{x=}-\text{10}-\text{56}\\ \Rightarrow -\frac{4}{3}\text{x=}-\text{66}\\ \Rightarrow \text{x}=49.5\\ \therefore \angle \text{ABD}\\ \text{=}\frac{2}{3}\times 49.5+56=2\times 16.5+56\\ =89°\\ \text{and}\\ \angle \text{ABC}\\ \text{=2}\times \text{49}\text{.5 }-\text{10}\\ \text{=99}-\text{10= 89}°\end{array}$

Since,

  $\angle \text{ABD=}\angle \text{ABC}\ne \text{90}°$

Hence,

The given line segment $\overline{\text{AB}}$ is not the perpendicular to given plane m.