Chapter 8, Problem 8CT

To determine

The solution of a triangle, for which two sides and one angle are givenas $a=8,b=4,C={70}^0$ .

Answer to Problem 8CT

Solution:

The third side of the triangle is $c=7.6230$, and other angles are $A={80.46}^0,B={29.54}^0$ .

Explanation of Solution

Given information:

The triangle with two sides $a=8,b=4,$ and an angle $C={70}^0$ .

To solve the triangle, for which two sides and one angle are given,use the law of cosine rule.

Here, $a=8,b=4$, and $C={70}^0$ .

Law of cosine:

  $c^2=a^2+b^2-2ab\cdot \cos C$

By substituting values of $a,b$, and $C$ ,

  $\Rightarrow c^2=8^2+4^2-2\cdot 8\cdot 4\cdot \cos \left(70\right)$

  $\Rightarrow c^2=64+16-64\cdot \left(0.3420\right)$

  $\Rightarrow c^2=58.1107$

  $\Rightarrow c=7.6230$

To find angle $B$ :

  $\cos B=\frac{a^2+c^2-b^2}{2ac}$

  $\cos B=\frac{{\left(8\right)}^2+{\left(7.6230\right)}^2-{\left(4\right)}^2}{2\left(8\right)\left(7.6230\right)}$

  $\cos B=\frac{\left(64\right)+\left(58.1107\right)-\left(16\right)}{121.9686}=\frac{106.1107}{121.9686}$

  $B={\cos }^{-1}\left(\frac{106.1107}{121.9686}\right)\approx {29.54}^0$ .

To find angle $C$ :

  $\cos A=\frac{b^2+c^2-a^2}{2bc}$

  $\cos A=\frac{{\left(4\right)}^2+{\left(7.6230\right)}^2-{\left(8\right)}^2}{2\left(4\right)\left(7.6230\right)}$

  $\cos A=\frac{\left(16\right)+\left(58.1101\right)-\left(64\right)}{60.984}=\frac{10.1101}{60.984}$

  $A={\cos }^{-1}\left(\frac{10.1107}{60.984}\right)\approx {80.46}^0$ .

Thus, the third side of triangle is $c=7.6230$, and other angles are $A={80.46}^0,B={29.54}^0$ .