Chapter 52, Problem 30A

Use the angle values given to find ∠2.
a. If ∠1 = 37°, find ∠2.
b. If ∠1 = 29°, find ∠1.

To determine

(a)

Value of $\angle 2$.

Answer to Problem 30A

Value of $\angle 2$ =56⁰.

Explanation of Solution

Given information:

  $\angle 1$ =37^O

  

Calculation:

$\angle FBC$ is equal to the $\angle 1$ because of opposite angle.

So, $\angle FBC$ =37^o

Now,

  $\begin{array}{l}\angle B=\angle FBC+52\\ \angle B=37+52\\ \angle B={89}^0\end{array}$

And

  $\begin{array}{l}\angle IAB={90}^0\\ and\angle IAB+\angle BAF={180}^0\\ So\angle BAF={180}^0-\angle IAB\\ \angle BAF={180}^0-{90}^0\\ \angle BAF={90}^0\end{array}$

Also,

  $\begin{array}{l}\angle HEF={84}^0\\ and\angle HEF+\angle FED={180}^0\\ So\angle FED={180}^0-\angle HEF\\ \angle FED={180}^0-{84}^0\\ \angle FED={96}^0\end{array}$

Now consider a rectangular ABDE.

Sum of the angles of a rectangle is 360^o.

  $\begin{array}{l}\angle A={90}^0\\ \angle B={89}^0\\ \angle E={96}^0\\ So\angle D={360}^0-\angle A-\angle B-\angle E\\ \angle D=360-90-89-96\\ \angle D={85}^0\end{array}$

Now consider a line HDGE.

  $\begin{array}{l}\angle CDE={85}^0\\ and\angle CDE+\angle CDG={180}^0\\ So\angle CDG={180}^0-\angle CDE\\ \angle CDG={180}^0-{85}^0\\ \angle CDG={95}^0\end{array}$

Now consider a line BCD.

  $\begin{array}{l}\angle BCG={151}^0\\ and\angle BCG+\angle GCD={180}^0\\ So\angle GCD={180}^0-\angle BCG\\ \angle GCD={180}^0-{151}^0\\ \angle GCD={29}^0\end{array}$

Now consider a triangle CDG. The sum of angles is 180⁰.

  $\begin{array}{l}\angle GCD={29}^0\\ \angle CDG={95}^0\\ \angle CGD+\angle GCD+\angle CDG={180}^0\\ \angle CGD=180-\angle GCD-\angle CDG\\ \angle CGD=180-29-95\\ \angle 2=\angle CGD={56}^0\end{array}$

To determine

(b)

Value of $\angle 2$.

Answer to Problem 30A

Value of $\angle 2$ =64⁰.

Explanation of Solution

Given information:

  

  $\angle 1$ =29^O

Calculation:

$\angle FBC$ is equal to the $\angle 1$ because of opposite angle.

So, $\angle FBC$ =29^o

Now,

  $\begin{array}{l}\angle B=\angle FBC+52\\ \angle B=29+52\\ \angle B={81}^0\end{array}$

And,

  $\begin{array}{l}\angle IAB={90}^0\\ and\angle IAB+\angle BAF={180}^0\\ So\angle BAF={180}^0-\angle IAB\\ \angle BAF={180}^0-{90}^0\\ \angle BAF={90}^0\end{array}$

Also,

  $\begin{array}{l}\angle HEF={84}^0\\ and\angle HEF+\angle FED={180}^0\\ So\angle FED={180}^0-\angle HEF\\ \angle FED={180}^0-{84}^0\\ \angle FED={96}^0\end{array}$

Now consider a rectangular ABDE.

Sum of the angles of a rectangle is 360^o.

  $\begin{array}{l}\angle A={90}^0\\ \angle B={81}^0\\ \angle E={96}^0\\ So\angle D={360}^0-\angle A-\angle B-\angle E\\ \angle D=360-90-81-96\\ \angle D={93}^0\end{array}$

Now consider a line HDGE.

  $\begin{array}{l}\angle CDE={93}^0\\ and\angle CDE+\angle CDG={180}^0\\ So\angle CDG={180}^0-\angle CDE\\ \angle CDG={180}^0-{93}^0\\ \angle CDG={87}^0\end{array}$

Now consider a line BCD.

  $\begin{array}{l}\angle BCG={151}^0\\ and\angle BCG+\angle GCD={180}^0\\ So\angle GCD={180}^0-\angle BCG\\ \angle GCD={180}^0-{151}^0\\ \angle GCD={29}^0\end{array}$

Now consider a triangle CDG, sum of angles is 180⁰_.

  $\begin{array}{l}\angle GCD={29}^0\\ \angle CDG={87}^0\\ \angle CGD+\angle GCD+\angle CDG={180}^0\\ \angle CGD=180-\angle GCD-\angle CDG\\ \angle CGD=180-29-87\\ \angle 2=\angle CGD={64}^0\end{array}$