Chapter 8.4, Problem 43AYU

a.

To determine

Show that: $Areaof\Delta OAC=\frac{1}{2}\sin \alpha \cos \alpha$

Explanation of Solution

Given information:

Refer to the figure. If $\left|OA\right|=1$ :

  

Calculation:

Here, we know that according to law of sines:

  $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

Now, as a triangle whose angles are $A,B,C$ are given as, $A+B+C=180°$

Now, for $\Delta OAC,\angle AOC+\angle OCA+\angle OAC=180°$ . Thus, from the figure,

  $\angle AOC=\alpha$ And $\angle OCA=90°$

  $\alpha +90+\angle OAC=180°$

Now, we will subtract $\alpha$ and $90°$ from both sides of equation:

  $\begin{array}{l}\angle OAC=180°-90°-\alpha \\ \angle OAC=90°-\alpha \\ \therefore \angle OAC=90°-\alpha \end{array}$

Here, we will find side $OC$ using the equality $\frac{\sin A}{a}=\frac{\sin B}{b}$

Now, substitute $\left|OC\right|$ for $b$ , $\angle OCA=90°$ for $A,\left|OA\right|=1$ for $a$ , and $\angle OAC=90°-\alpha$ for $B$ .

  $\frac{\sin 90°}{1}=\frac{\sin \left(90°-\alpha \right)}{\left|OC\right|}$

As, we know that $\sin 90°=1and\sin \left(90°-\alpha \right)=\cos \alpha$

  $\therefore 1=\frac{\cos \alpha }{\left|OC\right|}$

Now, we will multiply both sides of the equation, $1=\frac{\cos \alpha }{\left|OC\right|}by\left|OC\right|$

  $\left|OC\right|=\cos \alpha$

Here we will find the area of the triangle $OAC$ by using the formula $K=\frac{1}{2}ab\sin C$

Put $\left|OC\right|=1$ for $a$ , $\left|OC\right|=\cos \alpha$ for $b$ , and $\angle AOC=\alpha$ for $C$ in the equation:

  $\begin{array}{l}K=\frac{1}{2}\times 1\times \cos \infty \times \sin \infty \\ K=\frac{1}{2}\sin \alpha \cos \alpha \end{array}$

Hence, the area of $\Delta OAC=\frac{1}{2}\sin \alpha \cos \alpha$

b.

To determine

Show that: $Areaof\Delta OCB=\frac{1}{2}{\left|OB\right|}^2\sin \beta \cos \beta$

Explanation of Solution

Given information:

Refer to the figure. If $\left|OA\right|=1$ :

   Calculation:

Here, we will find the area of the triangle $OCB$ by using the formula $K=\frac{1}{2}ab\sin C$

Put $\left|OC\right|=\cos \alpha$ for $a$ , $\left|OB\right|$ for $b$ , and $\angle COB=\alpha$ for $C$ in the equation:

  $K=\frac{1}{2}\left|OB\right|\sin \beta \cos \infty$

Hence, the area of $\Delta OAC=\frac{1}{2}\sin \infty \cos \alpha$

Now, for $\Delta OCB,\angle COB+\angle OBC+\angle OCB=180°$ . Thus, from the figure,

  $\angle COB=\beta$ And $\angle OCB=90°$

  $\beta +90+\angle OBC=180°$

Now, we will subtract $\beta$ and $90°$ from both sides of equation:

  $\begin{array}{l}\angle OBC=180°-90°-\beta \\ \angle OBC=90°-\beta \\ \therefore \angle OAC=90°-\beta \end{array}$

Here, we will find the value of $\cos \infty$ using the equality $\frac{\sin A}{a}=\frac{\sin B}{b}$

Now, substitute $\left|OB\right|$ for $b$ , $\angle OBC$ for $A$ , and $\angle OBC=90°-\beta$ for $B$ .

  $\frac{\sin \left(90°-\beta \right)}{\cos \alpha }=\frac{\sin 90°}{\left|OB\right|}$

As, we know that $\sin 90°=1and\sin \left(90°-\beta \right)=\cos \beta$

  $\therefore \frac{\cos \beta }{\cos \alpha }=\frac{1}{\left|OB\right|}$

Now, we will multiply both sides of the equation, $by\left|OC\right|$ and $\cos \alpha$

  $\left|OB\right|\cos \beta =cos\alpha$

Now, we will substitute $\left|OB\right|\cos \beta$ for $\cos \alpha$ in the following formula:

  $\begin{array}{l}K=\frac{1}{2}\left|OB\right|\cos \infty \sin \beta \\ K=\frac{1}{2}\left|OB\right|\left|OB\right|\cos \beta \sin \beta \\ K=\frac{1}{2}{\left|OB\right|}^2\sin \beta \cos \beta \end{array}$

Hence, the area of $\Delta OCB=\frac{1}{2}{\left|OB\right|}^2\sin \beta \cos \beta$ .

c.

To determine

Show that: $Areaof\Delta OAB=\frac{1}{2}\left|OB\right|\sin \left(\alpha +\beta \right)$

Explanation of Solution

Given information:

Refer to the figure. If $\left|OA\right|=1$ :

   Calculation:

Here, we will find the area of the triangle $OAB$ by using the formula $K=\frac{1}{2}ab\sin C$

Now, from the figure $\angle AOB=\alpha +\beta$ substitute $\left|OA\right|=1$ for $a$ , $\left|OB\right|$ for $b$ and $\angle AOB=\alpha +\beta$ for $C$ in the equation:

  $\begin{array}{l}K=\frac{1}{2}\times 1\times \left|OB\right|\times \sin \left(\alpha +\beta \right)\\ K=\frac{1}{2}\times \left|OB\right|\sin \left(\alpha +\beta \right)\end{array}$

Hence, the area of $\Delta OAB=\frac{1}{2}\left|OB\right|\sin \left(\alpha +\beta \right)$ .

d.

To determine

Show that: $\left|OB\right|=\frac{\cos \alpha }{\cos \beta }$

Explanation of Solution

Given information:

Refer to the figure. If $\left|OA\right|=1$ :

   Calculation:

Here, we will find side $\left|OB\right|$ by using the equality $\frac{\sin A}{a}=\frac{\sin B}{b}$ .

Now, substitute $\left|OB\right|\text{for b, }\left|OC\right|=\cos \alpha \text{for a,}\angle OBC=90-\beta \text{for A,}\angle OCB\text{for B}$

  $\left|OB\right|$ for $b$ ,for $a$ , $\frac{\sin \left(90°-\beta \right)}{\cos \alpha }=\frac{\sin 90°}{\left|OB\right|}$

As, we know that $\sin 90°=1and\sin \left(90°-\beta \right)=\cos \beta$

  $\therefore \frac{\cos \beta }{\cos \alpha }=\frac{1}{\left|OB\right|}$

Now, we will multiply both sides of the equation, $by\left|OB\right|$ and $\cos \alpha$

  $\left|OB\right|\cos \beta =cos\alpha$

Now, we will divide both sides of the equation by $\cos \beta$ .

  $\left|OB\right|=\frac{\cos \alpha }{\cos \beta }$

Hence, $\left|OB\right|=\frac{\cos \alpha }{\cos \beta }$.

e.

To determine

Show that: $\sin \left(\alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta$ .

Explanation of Solution

Given information:

Refer to the figure. If $\left|OA\right|=1$ :

   Calculation:

Now, from the figure $Areaof\Delta OAB=Areaof\Delta OAC+Areaof\Delta OCB$ . Thus,

  $\frac{1}{2}\left|OB\right|\sin \left(\alpha +\beta \right)=\frac{1}{2}\sin \infty \cos \alpha +\frac{1}{2}{\left|OB\right|}^2\sin \beta \cos \beta$

Now, substitute $\frac{\cos \alpha }{\left|OB\right|}$ for $\cos \beta$ and $\left|OB\right|\cos \beta$ for $\cos \alpha$ in the equation:

  $\begin{array}{l}\frac{1}{2}\left|OB\right|\sin \left(\alpha +\beta \right)=\frac{1}{2}\sin \infty \cos \alpha +\frac{1}{2}{\left|OB\right|}^2\sin \beta \cos \beta \\ \frac{1}{2}\left|OB\right|\sin \left(\alpha +\beta \right)=\frac{1}{2}\sin \infty \left|OB\right|\cos \beta +\frac{1}{2}{\left|OB\right|}^2\sin \beta \frac{\cos \alpha }{\left|OB\right|}\\ \frac{1}{2}\left|OB\right|\sin \left(\alpha +\beta \right)=\frac{1}{2}\left|OB\right|\sin \infty \cos \beta +\frac{1}{2}\left|OB\right|\sin \beta \cos \alpha \end{array}$

Now, we will divide both sides of the equation by $\frac{1}{2}\left|OB\right|$ .

  $\sin \left(\alpha +\beta \right)=\sin \infty \cos \beta +\cos \alpha \sin \beta$

Hence, $\sin \left(\alpha +\beta \right)=\sin \infty \cos \beta +\cos \alpha \sin \beta$.