Chapter 8, Problem 7CT

To determine

The solution of the triangle whose two sides and one angle are $a=3,b=7$, and $A={40}^0$ .

Answer to Problem 7CT

Solution:

There is no trianglewith$a=3,b=7$, and $A={40}^0$ .

Explanation of Solution

Given information:

The two sides are $a=3,b=7$, andone angle of triangle is $A={40}^0$ .

To solve the triangle withtwo sides and one angle given, use the law of sines.

Here, $a=3,b=7$, and $A={40}^0$ .

Law of sines:

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

To find the angle $B$ using the law of sine rule:

  $\frac{\sin A}{a}=\frac{\sin B}{b}$

  $\Rightarrow \frac{\sin {40}^0}{3}=\frac{\sin B}{7}$

  $\Rightarrow \sin B=\frac{7\cdot \sin {40}^0}{3}$

  $\Rightarrow \sin B=\frac{4.4995}{3}$

  $\Rightarrow \sin B=1.4998$

  $\Rightarrow B={\sin }^{-1}\left(1.4998\right)$

Since, $1.4998$ is not in the domain of inverse sine function, value of ${\sin }^{-1}\left(1.4998\right)$ is not defined.

That is, there is no such angle $B$ for which $\sin B>1$ .

Therefore, there is no triangle with $a=3,b=7$, and $A={40}^0$ .