# 10.8 Vectors  (Page 6/22)

 Page 6 / 22

The dot product of two vectors involves multiplying two vectors together, and the result is a scalar.

## Dot product

The dot product    of two vectors $\text{\hspace{0.17em}}v=⟨a,b⟩\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}u=⟨c,d⟩\text{\hspace{0.17em}}$ is the sum of the product of the horizontal components and the product of the vertical components.

$v\cdot u=ac+bd$

To find the angle between the two vectors, use the formula below.

$\mathrm{cos}\text{\hspace{0.17em}}\theta =\frac{v}{|v|}\cdot \frac{u}{|u|}$

## Finding the dot product of two vectors

Find the dot product of $\text{\hspace{0.17em}}v=⟨5,12⟩\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}u=⟨-3,4⟩.$

Using the formula, we have

$\begin{array}{l}v\cdot u=⟨5,12⟩\cdot ⟨-3,4⟩\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=5\cdot \left(-3\right)+12\cdot 4\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-15+48\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=33\hfill \end{array}$

## Finding the dot product of two vectors and the angle between them

Find the dot product of v 1 = 5 i + 2 j and v 2 = 3 i + 7 j . Then, find the angle between the two vectors.

Finding the dot product, we multiply corresponding components.

$\begin{array}{l}{v}_{1}\cdot {v}_{2}=⟨5,2⟩\cdot ⟨3,7⟩\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=5\cdot 3+2\cdot 7\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=15+14\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=29\hfill \end{array}$

To find the angle between them, we use the formula $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta =\frac{v}{|v|}\cdot \frac{u}{|u|}.$

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{v}{|v|}\cdot \frac{u}{|u|}=⟨\frac{5}{\sqrt{29}}+\frac{2}{\sqrt{29}}⟩\cdot ⟨\frac{3}{\sqrt{58}}+\frac{7}{\sqrt{58}}⟩\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{5}{\sqrt{29}}\cdot \frac{3}{\sqrt{58}}+\frac{2}{\sqrt{29}}\cdot \frac{7}{\sqrt{58}}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{15}{\sqrt{1682}}+\frac{14}{\sqrt{1682}}=\frac{29}{\sqrt{1682}}\hfill \\ \begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.707107\hfill \\ {\mathrm{cos}}^{-1}\left(0.707107\right)=45°\hfill \end{array}\hfill \end{array}$

## Finding the angle between two vectors

Find the angle between $\text{\hspace{0.17em}}u=⟨-3,4⟩\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}v=⟨5,12⟩.$

Using the formula, we have

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\theta ={\mathrm{cos}}^{-1}\left(\frac{u}{|u|}\cdot \frac{v}{|v|}\right)\hfill \\ \left(\frac{u}{|u|}\cdot \frac{v}{|v|}\right)=\frac{-3i+4j}{5}\cdot \frac{5i+12j}{13}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(-\frac{3}{5}\cdot \frac{5}{13}\right)+\left(\frac{4}{5}\cdot \frac{12}{13}\right)\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\frac{15}{65}+\frac{48}{65}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{33}{65}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\theta ={\mathrm{cos}}^{-1}\left(\frac{33}{65}\right)\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}={59.5}^{\circ }\hfill \end{array}$

## Finding ground speed and bearing using vectors

We now have the tools to solve the problem we introduced in the opening of the section.

An airplane is flying at an airspeed of 200 miles per hour headed on a SE bearing of 140°. A north wind (from north to south) is blowing at 16.2 miles per hour. What are the ground speed and actual bearing of the plane? See [link] .

The ground speed is represented by $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ in the diagram, and we need to find the angle $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ in order to calculate the adjusted bearing, which will be $\text{\hspace{0.17em}}\text{\hspace{0.17em}}140°+\alpha \text{\hspace{0.17em}}.$

Notice in [link] , that angle $\text{\hspace{0.17em}}BCO\text{\hspace{0.17em}}$ must be equal to angle $\text{\hspace{0.17em}}AOC\text{\hspace{0.17em}}$ by the rule of alternating interior angles, so angle $\text{\hspace{0.17em}}BCO\text{\hspace{0.17em}}$ is 140°. We can find $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ by the Law of Cosines:

$\begin{array}{l}{x}^{2}={\left(16.2\right)}^{2}+{\left(200\right)}^{2}-2\left(16.2\right)\left(200\right)\mathrm{cos}\left(140°\right)\hfill \\ {x}^{2}=45,226.41\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=\sqrt{45,226.41}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=212.7\hfill \end{array}$

The ground speed is approximately 213 miles per hour. Now we can calculate the bearing using the Law of Sines.

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\mathrm{sin}\text{\hspace{0.17em}}\alpha }{16.2}=\frac{\mathrm{sin}\left(140°\right)}{212.7}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha =\frac{16.2\mathrm{sin}\left(140°\right)}{212.7}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=0.04896\hfill \\ {\mathrm{sin}}^{-1}\left(0.04896\right)=2.8°\hfill \end{array}$

Therefore, the plane has a SE bearing of 140°+2.8°=142.8°. The ground speed is 212.7 miles per hour.

Access these online resources for additional instruction and practice with vectors.

## Key concepts

• The position vector has its initial point at the origin. See [link] .
• If the position vector is the same for two vectors, they are equal. See [link] .
• Vectors are defined by their magnitude and direction. See [link] .
• If two vectors have the same magnitude and direction, they are equal. See [link] .
• Vector addition and subtraction result in a new vector found by adding or subtracting corresponding elements. See [link] .
• Scalar multiplication is multiplying a vector by a constant. Only the magnitude changes; the direction stays the same. See [link] and [link] .
• Vectors are comprised of two components: the horizontal component along the positive x -axis, and the vertical component along the positive y -axis. See [link] .
• The unit vector in the same direction of any nonzero vector is found by dividing the vector by its magnitude.
• The magnitude of a vector in the rectangular coordinate system is $\text{\hspace{0.17em}}|v|=\sqrt{{a}^{2}+{b}^{2}}.\text{\hspace{0.17em}}$ See [link] .
• In the rectangular coordinate system, unit vectors may be represented in terms of $i$ and $j$ where $\text{\hspace{0.17em}}i\text{\hspace{0.17em}}$ represents the horizontal component and $\text{\hspace{0.17em}}j\text{\hspace{0.17em}}$ represents the vertical component. Then, v = a i + b j   is a scalar multiple of $\text{\hspace{0.17em}}v\text{\hspace{0.17em}}$ by real numbers $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}b.\text{\hspace{0.17em}}$ See [link] and [link] .
• Adding and subtracting vectors in terms of i and j consists of adding or subtracting corresponding coefficients of i and corresponding coefficients of j . See [link] .
• A vector v = a i + b j is written in terms of magnitude and direction as $\text{\hspace{0.17em}}v=|v|\mathrm{cos}\text{\hspace{0.17em}}\theta i+|v|\mathrm{sin}\text{\hspace{0.17em}}\theta j.\text{\hspace{0.17em}}$ See [link] .
• The dot product of two vectors is the product of the $\text{\hspace{0.17em}}i\text{\hspace{0.17em}}$ terms plus the product of the $\text{\hspace{0.17em}}j\text{\hspace{0.17em}}$ terms. See [link] .
• We can use the dot product to find the angle between two vectors. [link] and [link] .
• Dot products are useful for many types of physics applications. See [link] .

A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5)  and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
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