Chapter 6.5, Problem 31E

  $\left(a\right)$

To determine

To show: Two triangles ABC and $A\text{'}B\text{'}C\text{'}$ are possible when $\angle A=40°,a=15\text{ and }b=20$ .

Answer to Problem 31E

The two triangles are provided below.

For triangle ABC , conditions are

  $\angle A=40°,\angle B=59°,\angle C=81°,a=15,b=20\text{ and }c=23.05$ .

  

For triangle $A\text{'}B\text{'}C\text{'}$ , conditions are

  $\angle A\text{'}=40°,\angle B\text{'}=121°,\angle C\text{'}=19°,a=15,b=20\text{ and }c=7.56$

  

Explanation of Solution

Given information:

The $\angle A=40°,a=15\text{ and }b=20$ .

Formula used:

The sine rule for a triangle is $ABC$ with length of its sides as a, b and c, is $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$ .

Calculation:

Consider the conditions, $\angle A=40°,a=15\text{ and }b=20$ .

Recall that the sine rule for a triangle is $ABC$ with length of its sides as a, b and c, is $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$ .

From the above $△ABC$ , with help of sine rule, where $\angle A=40°,a=15\text{ and }b=20$

  $\begin{array}{c}\frac{\sin A}{a}=\frac{\sin B}{b}\\ \frac{\sin 40°}{15}=\frac{\sin B}{20}\\ \sin B=\frac{20\sin 40°}{15}\\ \sin B=0.8571\end{array}$

Simplify it further as,

  $\begin{array}{c}\sin B=0.8571\\ B={\sin }^{-1}\left(0.8571\right)\\ B=59°\end{array}$

Now, there are two possible values for the angle B since the sine function is positive in first and second quadrant. So, value of angle B lies between $0°\text{ and 180}°$ .

So, one value is $B=59°$ , and another is,

  $\begin{array}{c}B=180°-59°\\ =121°\end{array}$

Therefore, $\angle B=59°\text{ and }\angle B\text{'}=121°$ .

Now, similarly there will be two possible values for angle C also. Recall that sum of all angles of a triangle is $180°$

Now, for triangle ABC ,

  $\begin{array}{c}\angle B+\angle A+\angle C=180°\\ \angle C=180°-40°-59°\\ =81°\end{array}$

Recall that the sine rule for a triangle is $ABC$ with length of its sides as a, b and c, is $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$ .

Apply it to compute the length of side c ,

  $\begin{array}{c}\frac{\sin A}{a}=\frac{\sin C}{c}\\ c=\frac{a\sin C}{\sin A}\\ c=\frac{15\sin \left(81°\right)}{\sin \left(40°\right)}\\ c=23.05\end{array}$

Now, for triangle, $A\text{'}B\text{'}C\text{'}$ . Recall that sum of all angles of a triangle is $180°$ ,

  $\begin{array}{c}\angle B\text{'}+\angle A+\angle C\text{'}=180°\\ \angle C\text{'}=180°-40°-121°\\ =19°\end{array}$

Recall that the sine rule for a triangle is $ABC$ with length of its sides as a, b and c, is $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$ .

Apply it to compute the length of side c ,

  $\begin{array}{c}\frac{\sin A}{a}=\frac{\sin C}{c}\\ c=\frac{a\sin C}{\sin A}\\ c=\frac{15\sin \left(19°\right)}{\sin \left(40°\right)}\\ c=7.56\end{array}$

Therefore, for triangle ABC , conditions are

  $\angle A=40°,\angle B=59°,\angle C=81°,a=15,b=20\text{ and }c=23.05$ .

  

Therefore, for triangle $A\text{'}B\text{'}C\text{'}$ , conditions are

  $\angle A\text{'}=40°,\angle B\text{'}=121°,\angle C\text{'}=19°,a=15,b=20\text{ and }c=7.56$

  

  $\left(b\right)$

To determine

To show: The area of triangles ABC and $A\text{'}B\text{'}C\text{'}$ is proportional to the sine of the angles C and C’ $\frac{\text{area of }△ABC}{\text{area of }△A\text{'}B\text{'}C\text{'}}=\frac{\sin C}{\sin C\text{'}}$ .

Explanation of Solution

Given information:

For triangle ABC , conditions are

  $\angle A=40°,\angle B=59°,\angle C=81°,a=15,b=20\text{ and }c=23.05$ .

For triangle $A\text{'}B\text{'}C\text{'}$ , conditions are

  $\angle A\text{'}=40°,\angle B\text{'}=121°,\angle C\text{'}=19°,a=15,b=20\text{ and }c=7.56$

Formula used:

The area of a triangle is denoted by $A=\frac{1}{2}ab\sin \theta$ where a and b are the length of its sides and $\theta$ is the included angle.

Proof:

It is provided that for triangle ABC , conditions are

  $\angle A=40°,\angle B=59°,\angle C=81°,a=15,b=20\text{ and }c=23.05$ .

  

Recall that the area of a triangle is denoted by $A=\frac{1}{2}ab\sin \theta$ where a and b are the length of its sides and $\theta$ is the included angle.

Therefore, for side $\angle A=40°,b=20\text{ and }c=23.05$

  $\begin{array}{c}\text{Area}=\frac{1}{2}bc\sin \theta \\ =\frac{1}{2}\left(20\right)\left(23.05\right)\sin 40°\\ =148.163\end{array}$

For triangle $A\text{'}B\text{'}C\text{'}$ , conditions are

  $\angle A\text{'}=40°,\angle B\text{'}=121°,\angle C\text{'}=19°,a=15,b=20\text{ and }c=7.56$ .

  

Recall that the area of a triangle is denoted by $A=\frac{1}{2}ab\sin \theta$ where a and b are the length of its sides and $\theta$ is the included angle.

Therefore, for side $\angle A\text{'}=40°,b=20\text{ and }c=7.56$

  $\begin{array}{c}\text{Area}=\frac{1}{2}bc\sin \theta \\ =\frac{1}{2}\left(20\right)\left(7.56\right)\sin 40°\\ =48.595\end{array}$

Now,

  $\begin{array}{c}\frac{\text{area of }△ABC}{\text{area of }△A\text{'}B\text{'}C\text{'}}=\frac{148.163}{48.595}\\ =3.05\end{array}$

And,

  $\begin{array}{c}\frac{\sin C}{\sin C\text{'}}=\frac{\sin 81°}{\sin 19°}\\ =3.03\end{array}$

Hence, it is shown that the area of triangles ABC and $A\text{'}B\text{'}C\text{'}$ is proportional to the sine of the angles C and C’ $\frac{\text{area of }△ABC}{\text{area of }△A\text{'}B\text{'}C\text{'}}=\frac{\sin C}{\sin C\text{'}}$ .