Chapter 13, Problem 12CR

To determine

To calculate: The dimensions of a triangle and find its area where the triangle is

Answer to Problem 12CR

Solution:

The sides of a triangle are $a\approx 6.09,b=5,c=9$, and angles are $A={40}^{\circ },B\approx {31.9}^{\circ },C\approx {108.1}^{\circ }$, and area is approximately $14.46$ square units.

Explanation of Solution

Given information:

The triangle:

Formula used:

1. Law of cosine:

2. $a^2=b^2+c^2-2bc\cos A$, where $a,b,c$ are sides of a triangle, and the angles opposite to the sides $a,b,c$ are $A,B,C$ respectively.

3. Area of triangle $=\frac{1}{2}\times \text{base}\times \text{height}$

Calculation:

Here, from the triangle, $b=5,c=9,$ and $A={40}^{\circ }$.

By using law of cosine,

$\Rightarrow a^2=5^2+9^2-2\left(5\right)\left(9\right)\cos {40}^{\circ }$

$\Rightarrow a^2=106-90\cdot \left(0.7660444431\right)$

$\Rightarrow a^2=106-68.94399988$

$\Rightarrow a^2=37.05600012$

$\Rightarrow a=\pm 6.087363971\approx \pm 6.09$

As the length of the side is never negative, so $a\approx 6.09$.

Now, to find angle $B$,

By using law of cosine,

$\Rightarrow b^2=a^2+c^2-2ac\cos B^{\circ }$

$\Rightarrow 5^2={\left(6.09\right)}^2+9^2-2\left(6.09\right)\left(9\right)\cos B^{\circ }$

$\Rightarrow 25-81-37.0881=-2\left(6.09\right)\left(9\right)\cos B^{\circ }$

$\Rightarrow -93.0881=-109.62\cos B^{\circ }$

$\Rightarrow \cos B^{\circ }=\frac{93.0881}{109.62}$

$\Rightarrow \cos B^{\circ }=\frac{93.0881}{109.62}$

$\Rightarrow \cos B^{\circ }=0.8491890166$

$\Rightarrow B^{\circ }={\cos }^{-1}\left(0.8491890166\right)$

$\Rightarrow B^{\circ }=31.87642838\approx {31.9}^0$

$\Rightarrow B^{\circ }\approx {31.9}^0$

Now, to find angle $C$,

The sum of the angles of the triangle must be equal to ${180}^{\circ }$.

$A+B+C=180$

Substitute $A={40}^{\circ }$ and $B\approx {31.9}^0$, it gives

$\Rightarrow 40+31.9+C=180$

$\Rightarrow C=180-71.9$

$\Rightarrow C={108.1}^{\circ }$

Therefore, the sides of a triangle are $a\approx 6.09,b=5,c=9$, and angles are $A={40}^{\circ },B\approx {31.9}^{\circ },C\approx {108.1}^{\circ }$.

Now, to find area of a triangle, The base of the triangle $=9$

Need to find height of the triangle, say $h$,

Using sine ratio to find $h$ gives $\sin {40}^{\circ }=\frac{h}{5}$

$\Rightarrow h=5\sin {40}^{\circ }$

By using the formula for area of a triangle, Area $=\frac{1}{2}\times \text{9}\times {\text{5 sin40}}^{\circ }$

$=14.46272122\approx 14.46$ square units.

Thus, area of a triangle is approximately $14.46$ square units.

Therefore, the sides of a triangle are $a\approx 6.09,b=5,c=9$, and angles are $A={40}^{\circ },B\approx {31.9}^{\circ },C\approx {108.1}^{\circ }$, and area is approximately $14.46$ square units.