Chapter 11.8, Problem 4BE

(a)

To determine

Find the distance below Earth surface where vertex of the cone is located.

Answer to Problem 4BE

10,000 ft.

Explanation of Solution

Given:

A good approximation of the detailed landing procedure uses a Heading Alignment Cone with vertex below the surface of the Earth.A typical radius of the cone at a height of 30,000 ft. above the Earth’s surface is 20,000 ft.At a height of 12,000 ft., which is a typical height for Q, the radius of the cone is 14,000 ft.

Calculation:

The given situation is :

  

Consider $\Delta AQI\text{ and }\Delta ADC$ :

Since $\overline{DC}\parallel \overline{QI}$ , the corresponding angles are congruent.

  $\angle ADC\cong \angle AQI$

  $\angle QIA\cong \angle DCA$

So, by AA Similarity Postulate , $\Delta AQI\sim \Delta ADC$

Hence,

  $\begin{array}{l}\frac{QI}{DC}=\frac{AQ}{AD}\\ \Rightarrow \frac{14,000}{20,000}=\frac{12,000+x}{30,000+x}\\ \Rightarrow \frac{7}{10}=\frac{12,000+x}{30,000+x}\\ \Rightarrow 7(30,000+x)=10(12,000+x)\\ \Rightarrow 210,000+x=120,000+10x\\ \Rightarrow 210,000-120,000=10x-x\mathrm{.........}[\text{subtract (120,000+x) from each side]}\\ \Rightarrow 90,000=9x\\ \Rightarrow x=\frac{90,000}{9}\mathrm{.................}[\text{divide each side by 9]}\\ \Rightarrow x=10,000\end{array}$

Hence, the vertex is 10,000 ft. below the Earth’s Surface.

(b)

To determine

Find the radius of cone at a height of 15,000 ft. above Earth’s Surface.

Answer to Problem 4BE

12,500 ft.

Explanation of Solution

Given:

A good approximation of the detailed landing procedure uses a Heading Alignment Cone with vertex below the surface of the Earth.A typical radius of the cone at a height of 30,000 ft. above the Earth’s surface is 20,000 ft.At a height of 12,000 ft., which is a typical height for Q, the radius of the cone is 14,000 ft.

Calculation:

From part (a) , the vertex is 10,000 ft. below Earth’s Surface.

The given situation is :

  

Consider $\Delta ADC\text{ and }\Delta AJK$ :

Since $\overline{DC}\parallel \overline{JK}$ , the corresponding angles are congruent.

  $\angle ADC\cong \angle AJK$

  $\angle DCA\cong \angle JKA$

So, by AA Similarity Postulate , $\Delta ADC\sim \Delta AJK$

Hence,

  $\begin{array}{l}\frac{JK}{DC}=\frac{JA}{DA}\\ \Rightarrow \frac{JK}{20,000}=\frac{15,000+10,000}{30,000+10,000}\\ \Rightarrow \frac{JK}{20,000}=\frac{25,000}{40,000}\\ \Rightarrow JK=\frac{25,000}{40,000}\cdot 20,\mathrm{000................}[\text{multiply each side by 20,000]}\\ \Rightarrow JK=12,500\end{array}$

Hence, the radius of cone at a height of 15,000 ft. above Earth’s Surface is 12,500 ft.

(c)

To determine

Find the height above Earth’s Surface at which the radius of the cone is 12,000 ft.

Answer to Problem 4BE

14,000 ft.

Explanation of Solution

Given:

A good approximation of the detailed landing procedure uses a Heading Alignment Cone with vertex below the surface of the Earth.A typical radius of the cone at a height of 30,000 ft. above the Earth’s surface is 20,000 ft.At a height of 12,000 ft., which is a typical height for Q, the radius of the cone is 14,000 ft.

Calculation:

From part (a) , the vertex is 10,000 ft. below Earth’s Surface.

The given situation is :

  

Consider $\Delta ADC\text{ and }\Delta AJK$ :

Since $\overline{DC}\parallel \overline{JK}$ , the corresponding angles are congruent.

  $\angle ADC\cong \angle AJK$

  $\angle DCA\cong \angle JKA$

So, by AA Similarity Postulate , $\Delta ADC\sim \Delta AJK$

Hence,

  $\begin{array}{l}\frac{JA}{DA}=\frac{JK}{DC}\\ \Rightarrow \frac{JA}{40,000}=\frac{12,000}{20,000}\\ \Rightarrow JA=\frac{12,000}{20,000}\cdot 40,\mathrm{000................}[\text{multiply each side by 40,000]}\\ \Rightarrow JA=24,000\\ h=JA-EA\\ \Rightarrow h=24,000-10,000\\ \Rightarrow h=14,000\end{array}$

Hence, at a height of 14,000 ft. above Earth’s Surface the radius of the cone is 12,000 ft .