Chapter 8.2, Problem 63AYU

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

To determine

To find: Three problems involving oblique triangles. One should result in one triangle, the second in two triangles, and the third in no triangle.

Answer to Problem 63AYU

Solution:

Case1. The given measurements represent only one triangle.

Case 2. The given measurement represents two triangles.

Case 3. The given measurement represents no such triangle.

Explanation of Solution

Given:

Consider three problem so that it results in one triangle, two triangles or no triangle.

Formula used:

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

Calculation:

Case 1: Consider an oblique triangle $\text{ABC}$ such that $a = 20, b = 18 and A = 54°$

$\frac{\text{sin}A}{a} = \frac{\text{sin}B}{b}$

$\frac{\text{sin}54°}{20} = \frac{\text{sin}B}{18}$

$\text{sin}B = \frac{6\left(\text{sin}32° \right)}{5} \approx 0.7281$

$B = 46.73$°

$b$ is not the longest side hence $\text{b}$ cannot be greater than $90°$

Hence the given measurements represent only one triangle.

Case 2: Consider an oblique triangle $\text{ABC}$ such that $a = 5, b = 6 and A = 32°$

$\frac{\text{sin}A}{a} = \frac{\text{sin}B}{b}$

$\frac{\text{sin}32° }{5} = \frac{\text{sin}B}{6}$

$\text{sin}B = \frac{6\left(\text{sin}32° \right)}{5} \approx 0.6359$

$B = 39.5$° or $B = 140.5°$

Hence there are two triangles with $\text{B} = 39.5°$ and $\text{B} = 140.5°$

Case 3: Consider an oblique triangle $\text{ABC}$ such that $a = 5, b = 9 and A = 42°$

$\frac{\text{sin}A}{a} = \frac{\text{sin}B}{b}$

$\frac{\text{sin}42° }{5} = \frac{\text{sin}B}{9}$

$\text{sin}B = \frac{9\left(\text{sin}42° \right)}{5} \approx 1.2044$

There is no value of $\text{B}$ such that $\text{sin}B = 1.2044$

Hence no such triangles with given measurement exist.