Chapter 14.3, Problem 50PPSE

Solution

a.

To calculate: The measure of $\angle C$ if a regular pentagon in inscribed in a circle of radius 10 cm. The given figure is:

  

The resultant answer is $\angle C={72}^{\circ }$ .

Given information:

A regular pentagon in inscribed in a circle of radius 10 cm.

  

Formula used: The sum of interior angles of a regular polygon of n -sides is $\left(n-2\right)\times {180}^{\circ }$ .

Calculation:

The given regular polygon is a pentagon which means it has 5 sides then the value of n is 5.

Use the sum formula of interior angles:

  $\left(n-2\right)\times {180}^{\circ }$

Substitute the value of n ,

  $\begin{array}{c}\left(5-2\right)\times {180}^{\circ }=3\times {180}^{\circ }\\ ={540}^{\circ }\end{array}$

The measure of each angle of the pentagon is:

  $\frac{{540}^{\circ }}{5}={108}^{\circ }$

The measure of $\angle Q\text{ and }\angle S$ is ${\text{ 108}}^{\circ }$ each. Half of it will be ${54}^{\circ }$ each.

Now consider the triangle $QCS$ .

  $\begin{array}{l}\frac{1}{2}\angle Q+\frac{1}{2}\angle S+\angle C={180}^{\circ }\\ {54}^{\circ }+{54}^{\circ }+\angle C={180}^{\circ }\\ \angle C={180}^{\circ }-{108}^{\circ }\\ \angle C={72}^{\circ }\end{array}$

Therefore, the value is $\angle C={72}^{\circ }$ .

b.

To calculate: The length of the diagonal $PS$ .

The resultant answer is $PS=\text{19}\text{.0211 cm}$ .

Given information:

The given pentagon is:

  

Formula used: In a triangle $ABC$ let $a,b,\text{ and }c$ represents the length of the sides opposite $\angle A,\angle B,\text{ and }\angle C$ respectively, the law of cosines are:

  $\begin{array}{l}{a}^2=b^2+c^2-2bc\cos A\\ b^2=c^2+a^2-2ca\cos B\\ c^2=a^2+b^2-2ab\cos C\end{array}$

Calculation:

Consider that $PQ=QS$ .

From cosine theorem,

  $\begin{array}{c}P{Q}^2=r^2+r^2-2r^2\cos \left({72}^{\circ }\right)\\ P{Q}^2=2r^2\left(1-\cos \left({72}^{\circ }\right)\right)\\ P{Q}^2=2\cdot {10}^2\left(1-\cos \left({72}^{\circ }\right)\right)\\ P{Q}^2=138.1966\\ PQ=11.7557\end{array}$

From part (a) $\angle PQS={108}^{\circ }$ .

Length of $PS$ is:

  $\begin{array}{c}P{S}^2=P{Q}^2+Q{S}^2-2\left(PQ\right)\left(QS\right)\cos \left({108}^{\circ }\right)\\ P{S}^2=2P{Q}^2\left(1-\cos \left({108}^{\circ }\right)\right)\\ P{S}^2=2P{Q}^2\left(1-\cos \left({108}^{\circ }\right)\right)\\ P{S}^2=2P{Q}^{2}2{\sin }^2\left(\frac{{180}^{\circ }}{2}\right)\\ PS=2PQ\sin \left({54}^{\circ }\right)\\ PS=19.0211\end{array}$

Therefore, the length of $PS\text{is 19}\text{.0211 cm}$ .