Chapter 55, Problem 27A

To determine

(a)

The value of angle E and angle F in the given figure.

Answer to Problem 27A

The angle E and angle F are equal to $\angle E=48°39\text{'}$ and $\angle F=24°03\text{'}$

Explanation of Solution

Given information:

The given value angle 1 is $\angle 1=107°18\text{'}$

The given value angle 2 is $\angle 2=41°21\text{'}$

Given figure is

Calculation:

The point P on the circle is the tangent point. Thus, the straight-line EF becomes tangent to the circle. The line joining center and the tangent point is perpendicular to the tangent. In other words, the angle EPO or the angle OPF is 90^o.

In triangle OPE,

  $\begin{array}{l}\angle OEP+\angle EPO+\angle POE=180°\\ \angle EPO=90°0\text{'}\\ \angle POE=\angle 2=41°21\text{'}\\ 90°+41°21\text{'}+\angle OEP=180°\\ \angle OEP=180°-\left(90°+41°21\text{'}\right)\\ \angle OEP=48°39\text{'}\end{array}$

The value of angle E is $\angle E=48°39\text{'}$

Now, let us calculate the value of angle F. The value of angle F can be calculated by considering triangle OPF.

Now, the angle $\angle 1=107°18\text{'}$ and $\angle 2=41°21\text{'}$, so the angle POF is

  $\begin{array}{l}\angle POF=\angle 1-\angle 2\\ \angle POF=107°18\text{'}-41°21\text{'}\\ \angle POF=65°57\text{'}\end{array}$

  $\begin{array}{l}\angle OPF+\angle POF+\angle PFO=180°\\ \angle POF=65°57\text{'}\\ \angle OPF=90°0\text{'}\\ 90°+65°57\text{'}+\angle PFO=180°\\ \angle PFO=180°-\left(90°+65°57\text{'}\right)\\ \angle PFO=24°03\text{'}\end{array}$

The value of angle F is $\angle F=24°03\text{'}$

Now, the angle E and angle F are equal to $\angle E=48°39\text{'}$ and $\angle F=24°03\text{'}$

Conclusion:

Thus, the angle E and angle F are equal to $\angle E=48°39\text{'}$ and $\angle F=24°03\text{'}$

To determine

(b)

The value of angle E and angle F in the given figure.

Answer to Problem 27A

The angle E and angle F are equal to $\angle E=41°40\text{'}$ and $\angle F=31°02\text{'}$

Explanation of Solution

Given information:

The given value angle 1 is $\angle 1=107°18\text{'}$

The given value angle 2 is $\angle 2=48°20\text{'}$

Given figure is

Calculation:

The point P on the circle is the tangent point. Thus, the straight-line EF becomes tangent to the circle. The line joining center and the tangent point is perpendicular to the tangent. In other words, the angle EPO or the angle OPF is 90^o.

In triangle OPE,

  $\begin{array}{l}\angle OEP+\angle EPO+\angle POE=180°\\ \angle EPO=90°0\text{'}\\ \angle POE=\angle 2=48°20\text{'}\\ 90°+48°20\text{'}+\angle OEP=180°\\ \angle OEP=180°-\left(90°+48°20\text{'}\right)\\ \angle OEP=41°40\text{'}\end{array}$

The value of angle E is $\angle E=41°40\text{'}$

Now, let us calculate the value of angle F. The value of angle F can be calculated by considering triangle OPF.

Now, the angle $\angle 1=107°18\text{'}$ and $\angle 2=48°20\text{'}$, so the angle POF is

  $\begin{array}{l}\angle POF=\angle 1-\angle 2\\ \angle POF=107°18\text{'}-48°20\text{'}\\ \angle POF=58°58\text{'}\end{array}$

  $\begin{array}{l}\angle OPF+\angle POF+\angle PFO=180°\\ \angle POF=58°58\text{'}\\ \angle OPF=90°0\text{'}\\ 90°+58°58\text{'}+\angle PFO=180°\\ \angle PFO=180°-\left(90°+58°58\text{'}\right)\\ \angle PFO=31°02\text{'}\end{array}$

The value of angle F is $\angle F=31°02\text{'}$

Now, the angle E and angle F are equal to $\angle E=41°40\text{'}$ and $\angle F=31°02\text{'}$

Conclusion:

Thus, the angle E and angle F are equal to $\angle E=41°40\text{'}$ and $\angle F=31°02\text{'}$