Chapter 8.2, Problem 63AYU

To determine

Make up three problems involving oblique triangles.

Answer to Problem 63AYU

  

Explanation of Solution

Given information

Make up three problems involving oblique triangles. One should result in one triangle, the second in two triangles, and the third in no triangle.

Calculation:

Let us consider an oblique triangle $a=3,b=2andA=40°,$ figured as:

  

By using low of sine in the given oblique triangle, we get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\\ \frac{\sin 40°}{3}=\frac{\sin B}{2}=\frac{\sin C}{c}\mathrm{.....}\left(1\right)\end{array}$

Taking first two in $\left(1\right)$ , we get

  $\begin{array}{l}\sin B=\frac{2}{3}\times \sin 40°\\ \sin B\approx \frac{2}{3}\times 0.64279\\ \sin B\approx 0.428525\\ B\approx {\sin }^{-1}\left(0.428525\right)or180°-{\sin }^{-1}\left(0.428525\right)\\ B\approx 25.37399°or180°-25.37399°\\ B\approx 25.37399°or154.62601°\\ ifB\approx 154.62601°,thenA+B40°+154.62601°\approx 194.62601°>180°\end{array}$

It means this choice of angle $B$ is wrong. So that

  $B\approx 25.37399°\mathrm{.......}\left(2\right)$

By the property of triangle, we get

  $\begin{array}{l}A+B+C=180°\\ C=180°-\left(A+B\right)\\ C\approx 180°-\left(40°+25.37399°\right)\end{array}$ Using $\left(2\right)$ and given condition

  $C\approx 114.626°\mathrm{.....}\left(3\right)$

From $\left(1\right)and\left(3,\right)weget$

  $\begin{array}{l}\frac{\sin 40°}{3}=\frac{\sin 114.626°}{c}\\ c=3\times \frac{\sin 114.626°}{\sin 40°}\\ c\approx 3\times \frac{0.90905}{0.64279}\\ c\approx 4.24268\end{array}$

Hence, it forms only one triangle figured as:

  

Let us consider an oblique triangle $a=6,b=8andA=35°,$ figured as:

  

By using low of sine for the given oblique triangle, we get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\\ \frac{\sin 35°}{6}=\frac{\sin B}{8}=\frac{\sin C}{c}\mathrm{......}\left(1\right)\end{array}$

Taking first two in $\left(1\right)$ , we get

  $\begin{array}{l}\sin B=\frac{8}{6}\times \sin 35°\\ \sin B\approx \frac{4}{3}\times 0.57358\\ \sin B\approx 0.76477\\ B\approx {\sin }^{-1}\left(0.76477\right)or180°{\sin }^{-1}\left(0.76477\right)\\ B\approx 49.88641°or180°-49.88641°\\ B\approx 49.88641°or130.11359°\\ ifB\approx 49.88641°,thenA+B\approx 35°+49.88641°\approx 84.88641°<180°\end{array}$

And if $B\approx 130.11359°,tthenA+B\approx 35°+130.11359°<180°$

It means both choices of the angle $B$ can be taken. Let

  $B_1\approx 49.88641°and{B}_2\approx 130.11359°\mathrm{.....}\left(2\right)$

For $A+B_1\approx 49.88641°,$ by the property of triangle, we get

  $\begin{array}{l}A+B_1+C=180°\\ C=180°-\left(A+B_1\right)\end{array}$

  $C\approx 180°-\left(35°+49.88641°\right)$ Using $\left(2\right)$

  $C\approx 95.11359°=C_1\left(say\right)\mathrm{.....}\left(3\right)$

For $B_2\approx 130.11359°,$ by the property of triangle, we get

  $\begin{array}{l}A+B_2+C=180°\\ C=180°-\left(A+B_2\right)\\ C\approx 180°-\left(35°+130.11359°\right)\\ C\approx 14.88641°=C_2\left(say\right)\mathrm{.......}\left(4\right)\end{array}$

From $\left(1\right)and\left(3,\right)$ we get

  $\begin{array}{l}\frac{\sin 35°}{6}=\frac{\sin 95.11359°}{c}\\ c=6\times \frac{\sin 95.11359°}{\sin 35°}\\ c\approx 6\times \frac{0.99602}{0.57358}\\ c\approx \mathrm{10.419.....}\left(5\right)\end{array}$

From $\left(1\right)and\left(4,\right)$ we get

  $\begin{array}{l}\frac{\sin 35°}{6}=\frac{\sin 14.88641°}{c}\\ c=6\times \frac{\sin 14.88641°}{\sin 35°}\\ c\approx 6\times \frac{0.2569}{0.57358}\\ c\approx \mathrm{2.687.....}\left(6\right)\end{array}$

Hence, there can be two triangle form, given as

  

Let us consider the triangle $a=2,c=1andC=50°,$ figured as by using law of sine for the given triangle, we get

  $\begin{array}{l}\frac{\sin A}{2}=\frac{\sin B}{b}=\frac{\sin C}{1}\\ \frac{\sin A}{2}=\frac{\sin B}{b}=\frac{\sin 50°}{1}\end{array}$

Taking first and last, we get

  $\begin{array}{l}\sin A=2\times \sin 50°\\ \sin A\approx 2\times 0.76604\\ \sin A\approx 1.532>1\end{array}$

It is never be possible for a triangle. There can be no any triangle form from such given information; the given information can be figure as: