Chapter 9.10, Problem 17PSC

a.

To determine

To calculate: The side PT with altitude PF of 35.

Answer to Problem 17PSC

The side PT is $36.4$ .

Explanation of Solution

Given information:

In a regular square pyramid,

Side AB=BC=CD=AD = 20,

Side PF =35.

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Calculation:

  

Side FT = 10.

Side PT can be calculated by applying Pythagoras Theorem.

In right angled triangle PFT , we get

  $\begin{array}{l}{\left(PT\right)}^2={\left(PF\right)}^2+{\left(FT\right)}^2\\ {\left(PT\right)}^2={\left(35\right)}^2+{\left(10\right)}^2\\ {\left(PT\right)}^2=1225+100\\ {\left(PT\right)}^2=1325\\ PT=\sqrt{1325}\approx 36.4\end{array}$

b.

To determine

To find: The side BP if side AD is 20.

Answer to Problem 17PSC

The side BP is $37.75$ .

Explanation of Solution

Given information:

In a regular square pyramid,

Side AB=BC=CD=AD = 20,

Side PF =35.

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Calculation:

  

Side FT = 10.

Side PT can be calculated by applying Pythagoras Theorem.

In right angled triangle PFT , we get

  $\begin{array}{l}{\left(PT\right)}^2={\left(PF\right)}^2+{\left(FT\right)}^2\\ {\left(PT\right)}^2={\left(35\right)}^2+{\left(10\right)}^2\\ {\left(PT\right)}^2=1225+100\\ {\left(PT\right)}^2=1325\\ PT=\sqrt{1325}\approx 36.4\end{array}$

Side BT = 10.

Side BP can be calculated by applying Pythagoras Theorem.

In right angled triangle PTB , we get

  $\begin{array}{l}{\left(PB\right)}^2={\left(PT\right)}^2+{\left(BT\right)}^2\\ {\left(PB\right)}^2={\left(36.4\right)}^2+{\left(10\right)}^2\\ {\left(PB\right)}^2=1325+100\\ {\left(PB\right)}^2=1425\\ PB=\sqrt{1425}\approx 37.75\end{array}$

c.

To determine

To calculate: The measure of angle PTF if side AD is 20.

Answer to Problem 17PSC

The measure of angle PTF is ${74}^{\circ }$ .

Explanation of Solution

Given information:

In a regular square pyramid,

Side AB=BC=CD=AD = 20,

Side PF =35.

Formula used:

  $\tan \theta =\frac{oppositeside}{adjacentside}$

Calculation:

  

In ΔPFT,

  $\begin{array}{l}tan\angle PTF=\frac{PF}{FT}\\ tan\angle PTF=\frac{35}{10}\\ tan\angle PTF=3.5\\ \angle PTF={\tan }^{-1}\left(3.5\right)\\ \angle PTF={74}^{\circ }\end{array}$

d.

To determine

To calculate: The measure of angle PBF if side BC is 20.

Answer to Problem 17PSC

The measure of angle PBF is ${68}^{\circ }$ .

Explanation of Solution

Given information:

In a regular square pyramid,

Side AB=BC=CD=AD = 20,

Side PF =35.

Formula used:

  $\sin \theta =\frac{oppositeside}{hypotenuseside}$

Calculation:

  

In ΔPFB,

  $\begin{array}{l}\sin \angle PBF=\frac{PF}{PB}\\ \sin \angle PBF=\frac{35}{37.75}\\ \sin \angle PBF=0.9271\\ \angle PBF={\sin }^{-1}\left(0.9271\right)\\ \angle PBF={68}^{\circ }\end{array}$