Chapter 6.4, Problem 45E

To determine

The interior angles of thetriangle with the given verticesusing vectors?

Answer to Problem 45E

The interior angles of thetriangle are $\text{41}{\text{.63}}^0,\text{43}{\text{.61}}^0\text{ and 94}{\text{.76}}^0$

Explanation of Solution

Given info:

  $(-3,0),(2,2),(0,6)$

Formula used:

If $\theta$ is the angle between two nonzero vectors u and v then

  $\cos \theta =\frac{u.v}{\Vert u\Vert \Vert v\Vert }$

Calculation:

Let the triangle be PQR, with $P=(-3,0),Q=(2,2)\text{ and }R=(0,6)$

We have

  $\begin{array}{l}\overrightarrow{PQ}=\langle 5,2\rangle \\ \overrightarrow{PR}=\langle 3,6\rangle \\ \overrightarrow{QR}=\langle -2,4\rangle \end{array}$

Finding the angle,

  $\begin{array}{l}\cos P=\frac{15+12}{\sqrt{5^2+2^2}\times \sqrt{3^2+6^2}}=\frac{27}{\sqrt{29}\times \sqrt{45}}\\ P=\text{41}{\text{.63}}^0\\ \cos Q=\frac{-10+8}{\sqrt{5^2+2^2}\times \sqrt{{\left(-2\right)}^2+4^2}}=\frac{-2}{\sqrt{29}\times \sqrt{20}}\\ Q=\text{94}{\text{.76}}^0\\ R={180}^0-\text{41}{\text{.63}}^0-\text{94}{\text{.76}}^0=\text{43}{\text{.61}}^0\end{array}$

The interior angles of thetriangle are $\text{41}{\text{.63}}^0,\text{43}{\text{.61}}^0\text{ and 94}{\text{.76}}^0$

Conclusion:

Thus,theinterior angles of thetriangle are $\text{41}{\text{.63}}^0,\text{43}{\text{.61}}^0\text{ and 94}{\text{.76}}^0$