Chapter 14, Problem 36SGR

a.

To determine

To find: the length of the diagonal.

Answer to Problem 36SGR

Diagonal $AC\approx 127.28feet.$

Explanation of Solution

Given information: The infield of a baseball diamond is a square with side length

90 feet.

Calculation:

Given the data:

AB=BC= 90 feet.

  

Find the length of the diagonal using the Pythagorean Theorem:

  $\begin{array}{l}A{C}^2=A{B}^2+B{C}^2=A{B}^2+A{B}^2=2A{B}^2\\ AC=\sqrt{2A{B}^2}=\sqrt{2\times {90}^2}=90\sqrt{2}\approx \mathrm{127.28.}\\ AC\approx 127.28feet.\end{array}$

Therefore, diagonal $AC\approx 127.28feet.$

b.

To determine

To write: the ratio for $\sin {45}^{\circ }$ using the lengths of the baseball diamond.

Answer to Problem 36SGR

  $\sin {45}^{\circ }=\frac{\sqrt{2}}{2}.$

Explanation of Solution

Given information: The infield of a baseball diamond is a square with side length

90 feet.

Calculation:

Given the data:

AB=BC= 90 feet.

  

Find the length of the diagonal using the Pythagorean Theorem:

  $\begin{array}{l}A{C}^2=A{B}^2+B{C}^2=A{B}^2+A{B}^2=2A{B}^2\\ AC=\sqrt{2A{B}^2}=\sqrt{2\times {90}^2}=90\sqrt{2}.\\ \angle CAB=\angle ACB={45}^{\circ }.\\ \sin {45}^{\circ }=\frac{AB}{AC}=\frac{90}{90\sqrt{2}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}.\\ \sin {45}^{\circ }=\frac{\sqrt{2}}{2}.\end{array}$

Therefore, $\sin {45}^{\circ }=\frac{\sqrt{2}}{2}.$

c.

To determine

To verify: the ratio wrote in part (b) using the formula $\sin \frac{\theta }{2}=\pm \sqrt{\frac{1-\cos \theta }{2}}$ .

Answer to Problem 36SGR

Explanation of Solution

Given information: The infield of a baseball diamond is a square with side length

90 feet.

Calculation:

Compute $\sin {45}^{\circ }$ , using the half −angle formula $\sin \frac{\theta }{2}=\pm \sqrt{\frac{1-\cos \theta }{2}}$ .

  $\begin{array}{l}\theta ={90}^{\circ }\\ \sin {45}^{\circ }=\sin \frac{90}{2}=\pm \sqrt{\frac{1-\cos {90}^{\circ }}{2}}=\pm \sqrt{\frac{1-0}{2}}=\pm \sqrt{\frac{1}{2}}=\pm \frac{\sqrt{2}}{2}\\ \sin {45}^{\circ }=\frac{\sqrt{2}}{2}.(\text{Because}\text{the}\text{angle}\text{is}\text{in}\text{Quadrant}\text{I,}\text{the}\text{sine}\text{is}\text{positive})\end{array}$

Got the same result as in part (b).

Hence verified.