Chapter 8, Problem 9CR

To determine

To calculate: The dimensions of triangle, where $a=20,c=15$ and angle $C={40}^{\circ }$

Answer to Problem 9CR

Solution:

The dimensions of first triangle are angles $A\approx {58.99}^0,B\approx {81.01}^0$and $C={40}^{\circ }$ with sides $a=20,b=23.049$and $c=15$

The dimensions of another triangle are $A\approx {121.01}^0,B\approx {18.99}^0$and $C={40}^0$ with sides $a=20,b=7.592$and $c=15$

Explanation of Solution

Given information:

  $a=20,c=15$ and angle $C={40}^{\circ }$

Formula used:

The sum of the angles of any triangle equals ${180}^0$, that is $\left(A+B+C={180}^0\right)$

Law of cosine:

For the triangle with sides $a,b,c$ and opposite angles $A,B,C$, respectively

  $c^2=a^2+b^2-2ab\cos C$ … (1)

  $b^2=a^2+c^2-2ac\cos B$ … (2)

  $a^2=b^2+c^2-2bc\cos A$ … (3)

Calculation:

To solve triangle, first find the side $b$ by using equation (1) of law of cosines

Here, $a=20,c=15$, and angle $C={40}^{\circ }$

Substitute these values in $c^2=a^2+b^2-2ab\cos C$, to get value of $b$

  ${\left(15\right)}^2={\left(20\right)}^2+b^2-2\left(20\right)b\cos \left({40}^{\circ }\right)$

Simplify the above equation

  $\Rightarrow 225=400+b^2-40\cdot b\cos \left({40}^{\circ }\right)$

  $\Rightarrow 40\cdot b\cos \left({40}^{\circ }\right)=400+b^2-225$

  $\Rightarrow b^2-40\cdot b\cos \left({40}^{\circ }\right)+175=0$

  $\Rightarrow b^2-30.6417b+175=0$

Solve the above quadratic equation by using quadratic formula.

Compare equation with standard quadratic equation to find $a,b$ and $c$

Therefore, $a=1,b=-30.6417$ and $c=175$

  $\Rightarrow b=\frac{-\left(-30.6417\right)\pm \sqrt{{\left(-30.6417\right)}^2-4\left(1\right)\left(175\right)}}{2\cdot 1}$

  $\Rightarrow b=\frac{30.6417\pm \sqrt{938.9137-700}}{2}$

  $\Rightarrow b=\frac{30.6417\pm \sqrt{238.9137}}{2}$

  $\Rightarrow b=\frac{30.6417\pm \sqrt{238.9137}}{2}$

  $\Rightarrow b=23.049$ and $b=7.592$ .

There are two values of side $b$ ; therefore, two possible triangles exist.

To find angles $A,B$, when $b=23.049$

By using the cosine rule $a^2=b^2+c^2-2bc\cos A$

By rearranging the above equation,

  $\cos A=\frac{b^2+c^2-a^2}{2bc}$

Here, $a=20,b=23.049$ and $c=15$

Therefore,

  $\cos A=\frac{{\left(23.049\right)}^2+{\left(15\right)}^2-{\left(20\right)}^2}{2\left(23.049\right)\left(15\right)}$

  $\cos A=\frac{531.256+225-400}{691.47}$

  $\cos A=\frac{356.256}{691.47}$

Taking ${\cos }^{-1}$ on both sides,

  $\Rightarrow A={\cos }^{-1}\left(\frac{356.256}{691.47}\right)$

  $\Rightarrow A\approx {58.99}^0$

To find the angle $B$ ,

Here, $A\approx {58.99}^0,C={40}^0$

The sum of the angles of any triangle equals ${180}^0$, that is $\left(A+B+C={180}^0\right)$

  $\Rightarrow {58.99}^{\circ }+{40}^{\circ }+B={180}^0$

  $\Rightarrow {98.99}^0+B={180}^0$

  $\Rightarrow B\approx {81.01}^0$

Therefore, dimensions of first triangle are angles $A\approx {58.99}^0,B\approx {81.01}^0$ and $C={40}^{\circ }$ with sides $a=20,b=23.049$ and $c=15$

To find angles $A,B$, when $b=7.592$

By using the cosine rule $a^2=b^2+c^2-2bc\cos A$

By rearranging the above equation,

  $\cos A=\frac{b^2+c^2-a^2}{2bc}$

Here, $a=20,b=7.592$ and $c=15$

Therefore,

  $\cos A=\frac{{\left(7.592\right)}^2+{\left(15\right)}^2-{\left(20\right)}^2}{2\left(7.592\right)\left(15\right)}$

  $\Rightarrow \cos A=\frac{57.6384+225-400}{227.76}$

  $\Rightarrow \cos A=\frac{-117.3615}{227.76}$

Taking ${\cos }^{-1}$ on both sides,

  $\Rightarrow A={\cos }^{-1}\left(\frac{-117.3615}{227.76}\right)$

  $\Rightarrow A\approx {121.01}^0$

Now, to find the angle $B$

Here, $A\approx {121.01}^0$ and $C={40}^0$

The sum of the angles of any triangle equals ${180}^0$, that is $\left(A+B+C={180}^0\right)$

  $\Rightarrow {121.01}^{\circ }+{40}^{\circ }+B={180}^0$

  $\Rightarrow {161.01}^0+B={180}^0$

  $\Rightarrow B\approx {18.99}^0$ .

Thus, the dimensions of triangle are $A\approx {121.01}^0,B\approx {18.99}^0$ and $C={40}^0$ with sides $a=20,b=7.592$ and $c=15$