Chapter 8.5, Problem 28WE

To determine

To calculate: to the nearest degree.

Answer to Problem 28WE

So the $\angle TQS=a\sqrt{3}$.

Explanation of Solution

Given information:

Side of cube is a.

Formula used:

Pythagorean theorem:

  , where $h$ , $p$ , and $b$ are hypotenuse, perpendicular, and base in a right − angled triangle, respectively.

Calculation: 

In triangle SPQ,

  $\begin{array}{l}S{P}^2+Q{P}^2=S{Q}^2,\\ \Rightarrow a^2+a^2=S{Q}^2,\\ \Rightarrow S{Q}^2=2a^2,\\ \Rightarrow SQ=a\sqrt{2}.\end{array}$

In triangle SQT,

  $\begin{array}{l}S{T}^2+S{Q}^2=Q{T}^2,\\ $ \Rightarrow a^2+{(a\sqrt{2})}^2=Q{T}^2,$\Rightarrow Q{T}^2=a^2+2a^2,\\ \Rightarrow Q{T}^2=3a^2,\\ \Rightarrow QT=a\sqrt{3}.\end{array}$

Now, again in triangle SQT,

  $\cos \left(\angle TQS\right)=\frac{\text{base}}{\text{hypotenuse}}=\frac{b}{h}=\frac{SQ}{QT}=\frac{a\sqrt{2}}{a\sqrt{3}}=\sqrt{\frac{2}{3}}=0.81649658092\approx 0.8164$ ,

  $\Rightarrow {\text{ cos}}^{-1}\left(\cos \left(\angle TQS\right)\right)=\angle TQS={\cos }^{-1}\left(0.8164\right)=35.2739732\approx 35°$ .

Hence, to the nearest degree, $\angle TQS$ is of $35°$ .