Basic Trigonometric Identities

$\begin{array}{|l|} \hline \sin A=\displaystyle \frac{1}{\csc A}\ \text{and}\\ \csc A=\displaystyle \frac{1}{\sin A}\\ \cos A=\displaystyle \frac{1}{\sec A}\ \text{and}\\ \sec A=\displaystyle \frac{1}{\cos A}\\ \tan A=\displaystyle \frac{\sin A}{\cos A}\\ \tan A=\displaystyle \frac{1}{\cot A}\ \text{and}\\ \cot A=\displaystyle \frac{1}{\tan A}\\ \cot A=\displaystyle \frac{\cos A}{\sin A}\\ \hline\end{array}$

Pythagorean Identities

$\begin{array}{|l|} \hline \sin^2 A+ \cos^2 A=1\\ \tan^2 A+ 1=\sec^2 A\\ 1+ \cot^2 A=\csc^2 A\\ \hline\end{array}$

Trigonometric Ratios of Complementary Angles

$\begin{array}{|l|}\hline \sin \left(90^{\circ}-\alpha\right)=\cos \alpha\\ \cos \left(90^{\circ}-\alpha\right)=\sin \alpha\\ \tan \left(90^{\circ}-\alpha\right)=\cot \alpha\\ \cot \left(90^{\circ}-\alpha\right)=\tan \alpha\\ \sec \left(90^{\circ}-\alpha\right)=\csc \alpha\\ \csc \left(90^{\circ}-\alpha\right)=\sec \alpha\\ \hline \end{array}$



Prove the following identities.

1.           $\cot \theta \sqrt{1-\cos ^{2} \theta}=\cos \theta$

Show/Hide Solution

$$\begin{aligned} &\ \ \ \ \ \cot \theta \sqrt{1-\cos ^{2} \theta}\\\\ &=\cot \theta \sqrt{\sin ^{2} \theta}\\\\ &=\frac{\cos \theta}{\sin \theta} \times \sin \theta\\\\ &=\cos \theta\end{aligned}$$

2.           $\displaystyle \frac{\tan ^{2} \theta+1}{\tan \theta \csc ^{2} \theta}=\tan \theta$

Show/Hide Solution

$$\begin{aligned} &\ \ \ \ \ \frac{\tan ^{2} \theta+1}{\tan \theta \csc ^{2} \theta}\\\\ &=\frac{\sec ^{2} \theta}{\cos \theta} \times \frac{1}{\sin ^{2} \theta}\\\\ &=\frac{1}{\cos ^{2} \theta} \times \sin \theta \cos \theta\\\\ &=\frac{\sin \theta}{\cos \theta}=\tan \theta \end{aligned}$$

3.           $\left(1-\sin ^{2} \theta\right)\left(1+\cot ^{2} \theta\right)=\cot ^{2} \theta$

Show/Hide Solution

$\begin{aligned} &\ \ \ \ \ \left(1-\sin ^{2} \theta\right)\left(1+\cot ^{2} \theta\right)\\\\ &=\cos ^{2} \theta \csc ^{2} \theta\\\\ &=\frac{\cos ^{2} \theta}{\sin ^{2} \theta}\\\\ &=\cot ^{2} \theta\end{aligned}$

4.           $\tan ^{2} \theta-\cot ^{2} \theta=\sec ^{2} \theta-\csc ^{2} \theta$

Show/Hide Solution

$\begin{aligned} &\ \ \ \ \ \tan ^{2} \theta-\cot ^{2} \theta\\\\ & =\left(\sec ^{2} \theta-1\right)-\left(\csc ^{2} \theta-1\right)\\\\ &=\sec ^{2} \theta-1-\csc ^{2} \theta+1\\\\ &=\sec ^{2} \theta-\csc ^{2} \theta\end{aligned}$

5.           $\sin \theta \sec \theta \sqrt{\csc ^{2} \theta-1}=1$

Show/Hide Solution

$\begin{aligned} &\ \ \ \ \ \sin \theta \sec \theta \sqrt{\csc ^{2} \theta-1}\\\\ &=\frac{\sin \theta}{\cos \theta} \times \sqrt{\cot ^{2} \theta}\\\\ &=\tan \theta \cot \theta=1\end{aligned}$

6.           $16 \sec ^{2} \theta+\csc ^{2} \theta=\sec ^{2} \theta \csc ^{2} \theta$

Show/Hide Solution

$\begin{aligned} &\ \ \ \ \ \sec ^{2} \theta+\csc ^{2} \theta \\\\ &=\frac{1}{\cos ^{2} \theta}+\frac{1}{\sin ^{2} \theta}\\\\ &=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\cos ^{2} \theta \sin ^{2} \theta}\\\\ &=\frac{1}{\cos ^{2} \theta \sin ^{2} \theta}\\\\ &=\sec ^{2} \theta \csc ^{2} \theta\end{aligned}$

7.           $\left(1+\tan ^{2} \theta\right)\left(1-\sin ^{2} \theta\right)=1$

Show/Hide Solution

$\begin{aligned} &\ \ \ \ \ \left(1+\tan ^{2} \theta\right)\left(1-\sin ^{2} \theta\right)\\\\ &=\sec ^{2} \theta \cos ^{2} \theta\\\\ &=\frac{1}{\cos ^{2} \theta} \times \cos ^{2} \theta\\\\ &=1\end{aligned}$

8.           $(1+\tan \theta)^{2}+(1-\tan \theta)^{2}=2 \sec ^{2} \theta$

Show/Hide Solution

$\begin{aligned} &\ \ \ \ \ (1+\tan \theta)^{2}+(1-\tan \theta)^{2}\\\\ &=\left(1+2 \tan \theta+\tan ^{2} \theta\right)+\left(1-2 \tan \theta+\tan ^{2} \theta\right)\\\\ &=2+2 \tan ^{2} \theta\\\\ &=2\left(1+\tan ^{2} \theta\right)\\\\ &=2 \sec ^{2} \theta\end{aligned}$

9.           $\sec ^{2} \theta \cot ^{2} \theta-1=\cot ^{2} \theta$

Show/Hide Solution

$\begin{aligned} &\ \ \ \ \ \sec ^{2} \theta \cot ^{2} \theta-1 \\\\ &=\frac{1}{\cos ^{2} \theta} \times \frac{\cos ^{2} \theta}{\sin ^{2} \theta}-1\\\\ &=\frac{1}{\sin ^{2} \theta}-1\\\\ &=\csc ^{2} \theta-1\\\\ &=\cot ^{2} \theta\end{aligned}$

10.           $\displaystyle \frac{1}{1-\sin \theta}+\displaystyle \frac{1}{1+\sin \theta}=2 \sec ^{2} \theta$

Show/Hide Solution

$\begin{aligned} &\ \ \ \ \ \frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}\\\\ &=\frac{1+\sin \theta+1-\sin \theta}{(1-\sin \theta)(1+\sin \theta)}\\\\ &=\frac{2}{1-\sin ^{2} \theta}\\\\ &=\frac{2}{\cos ^{2} \theta}\\\\ &=2 \sec ^{2} \theta\end{aligned}$

11.           $\displaystyle \frac{1}{\sin ^{2} \theta}-\displaystyle \frac{1}{\tan ^{2} \theta}=1$

Show/Hide Solution

$\begin{aligned} &\ \ \ \ \ \frac{1}{\sin ^{2} \theta}-\frac{1}{\tan ^{2} \theta}\\\\ &=\csc ^{2} \theta-\cot ^{2} \theta=1\end{aligned}$

12.           $(\tan \theta+\sec \theta)^{2}=\displaystyle \frac{1+\sin \theta}{1-\sin \theta}$

Show/Hide Solution

$\begin{aligned} &\ \ \ \ \ (\tan \theta+\sec \theta)^{2}\\\\ &=\left(\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}\right)^{2}\\\\ &=\left(\frac{1+\sin \theta}{\cos \theta}\right)^{2}\\\\ &=\frac{(1+\sin \theta)^{2}}{\cos ^{2} \theta}\\\\ &=\frac{(1+\sin \theta)^{2}}{1-\sin ^{2} \theta}\\\\ &=\frac{(1+\sin \theta)^{2}}{(1-\sin \theta)(1+\sin \theta)}\\\\ &=\frac{1+\sin \theta}{1-\sin \theta}\end{aligned}$

13.           $\sin ^{4} \theta-\cos ^{4} \theta=1-2 \cos ^{2} \theta$

Show/Hide Solution

$\begin{aligned} &\ \ \ \ \ \sin ^{4} \theta-\cos ^{4} \theta \\\\ &=\left(\sin ^{2} \theta\right)^{2}-\left(\cos ^{2} \theta\right)^{2}\\\\ &=\left(\sin ^{2} \theta+\cos ^{2} \theta\right)\left(\sin ^{2} \theta-\cos ^{2} \theta\right) \\\\ &=1\left(1-\cos ^{2} \theta-\cos ^{2} \theta\right)\\\\ &=1-2 \cos ^{2} \theta\end{aligned}$

14.           $\displaystyle \frac{\tan ^{2} \theta+1}{\tan ^{2} \theta}=\csc ^{2} \theta$

Show/Hide Solution

$\begin{aligned} &\ \ \ \ \ \frac{\tan ^{2} \theta+1}{\tan ^{2} \theta}\\\\ &=\frac{\tan ^{2} \theta}{\tan ^{2} \theta}+\frac{1}{\tan ^{2} \theta}\\\\ &=1+\cot ^{2} \theta\\\\ &=\csc ^{2} \theta\end{aligned}$

15.           $\sin ^{2} \theta \tan \theta+\cos ^{2} \theta \cot \theta+2 \sin \theta \cos \theta=\tan \theta+\cot \theta$

Show/Hide Solution

$\begin{aligned} &\ \ \ \ \ {{\sin }^{2}}\theta \tan \theta +{{\cos }^{2}}\theta \cot \theta +2\sin \theta \cos \theta \\ &={{\sin }^{2}}\theta \cdot \frac{{\sin \theta }}{{\cos \theta }}+{{\cos }^{2}}\theta \cdot \frac{{\cos \theta }}{{\sin \theta }}+2\sin \theta \cos \theta \\ &=\frac{{{{{\sin }}^{3}}\theta }}{{\cos \theta }}+\frac{{{{{\cos }}^{3}}\theta }}{{\sin \theta }}+2\sin \theta \cos \theta \\ &=\frac{{{{{\sin }}^{4}}\theta +{{{\cos }}^{4}}\theta +2{{{\sin }}^{2}}\theta {{{\cos }}^{2}}\theta }}{{\sin \theta \cos \theta }}\\ &=\frac{{{{{\left( {{{{\sin }}^{2}}\theta +{{{\cos }}^{2}}\theta } \right)}}^{2}}}}{{\sin \theta \cos \theta }}\\ &=\frac{{\left( {{{{\sin }}^{2}}\theta +{{{\cos }}^{2}}\theta } \right)\left( {{{{\sin }}^{2}}\theta +{{{\cos }}^{2}}\theta } \right)}}{{\sin \theta \cos \theta }}\\ &=\frac{{{{{\sin }}^{2}}\theta +{{{\cos }}^{2}}\theta }}{{\sin \theta \cos \theta }}\\ &=\frac{{{{{\sin }}^{2}}\theta }}{{\sin \theta \cos \theta }}+\frac{{{{{\cos }}^{2}}\theta }}{{\sin \theta \cos \theta }}\\ &=\tan \theta +\cot \theta \end{aligned}$

16.           Find the value of acute angle $\alpha$ in each of the following equations:

               (a)    $\cos 2 \alpha=\sin 7 \alpha$

               (b)    $\tan 3 \alpha=\cot 2 \alpha$

               (c)    $\sec \alpha=\csc 5 \alpha$

Show/Hide Solution

$\begin{array}{l} \text {(a)} \quad \cos 2 \alpha=\sin 7 \alpha\\\\ \therefore \quad 2\alpha + 7 \alpha = 90^{\circ}\\\\ \quad\quad 9\alpha= 90^{\circ}\\\\ \therefore \quad \alpha= 90^{\circ}\\\\ \text {(b)} \quad \tan 3 \alpha=\cot 2 \alpha\\\\ \therefore \quad 3\alpha + 2 \alpha = 90^{\circ}\\\\ \quad\quad 5\alpha= 90^{\circ}\\\\ \therefore \quad \alpha= 18^{\circ}\\\\ \text {(c)} \quad \sec \alpha=\csc 5 \alpha\\\\ \therefore \quad \alpha + 5 \alpha = 90^{\circ}\\\\ \quad\quad 6\alpha= 90^{\circ}\\\\ \therefore \quad \alpha= 15^{\circ} \end{array}$

17.           Prove the identity $\cos \left(90^{\circ}-\alpha\right) \tan \left(90^{\circ}-\alpha\right)=\cos \alpha$

Show/Hide Solution

$\begin{aligned} &\ \ \ \ \ \cos (90-\alpha) \tan (90-\alpha)\\\\ &=\sin \alpha \cot \alpha\\\\ &=\sin \alpha \times \frac{\cos \alpha}{\sin \alpha}\\\\ &=\cos \alpha \end{aligned}$

18.           Prove the identity $\sin \left(90^{\circ}-\alpha\right) \sec \left(90^{\circ}-\alpha\right)=\cot \alpha$

Show/Hide Solution

$\begin{aligned} &\ \ \ \ \ \sin (90-\alpha) \sec \left(90^{\circ}-\alpha\right)\\\\ &=\cos \alpha \csc \alpha\\\\ &=\cos \alpha \times \frac{1}{\sin \alpha}\\\\ &=\frac{\cos \alpha}{\sin \alpha}\\\\ &=\cot \alpha\end{aligned}$

1.           State why the two polygons are, are not, similar.


(a)

The two rectangles are not similar because corresponding sides are not proportional.


(b)

The two squares are similar because corresponding sides are proportional and corresponding angles are equal.



(c)

The two triangles are similar because corresponding sides are proportional and corresponding angles are equal.


(d)

The two polygons are not similar because they have different number of sides.

2.           Complete the proportions.

             (a)    If $\triangle A B C \sim \triangle D E F$ then $\displaystyle\frac{A B}{?}=\frac{B C}{?}=\frac{?}{D F}$.

             (b)     If $\triangle G H I \sim \triangle K L M$ then $\displaystyle\frac{?}{H I}=\frac{?}{G H}=\frac{?}{G I}$.


Show/Hide Solution

(a) $\displaystyle\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}$,

(b) $\displaystyle\frac{LM}{HI}=\frac{KL}{GH}=\frac{KM}{GI}$

3.           State whether the proportions are correct for the indicated similar triangles.

           $\begin{array}{l} \text{(a)}\ \triangle A B C \sim \triangle X Y Z.\\\\ \quad\ \ \displaystyle\frac{A B}{X Y}=\displaystyle\frac{B C}{Y Z}\\\\ \text{(b)}\ \triangle D E F \sim \triangle H I J.\\\\ \quad\ \ \displaystyle\frac{D E}{H I}=\displaystyle\frac{E F}{IJ}\\\\ \text{(c)}\ \triangle R S T \sim \triangle L M K.\\\\ \quad\ \ \displaystyle\frac{R T}{L M}=\displaystyle\frac{S T}{M K}\\\\ \text{(d)}\ \triangle X Y Z \sim \triangle U VW.\\\\ \quad\ \ \displaystyle\frac{X Y}{U V}=\displaystyle\frac{X Z}{V W} \end{array}$

Show/Hide Solution

(a) correct

(b) correct

(c) incorrect

(d) correct

4.           Given    :     $\triangle P Q R \sim \triangle U V W$ and lengths of sides are as marked.

Find      :     The values of $x$ and $y$.


Show/Hide Solution

$\begin{array}{l} \triangle P Q R\sim\triangle U V W\ \ \text { (Given) } \\\\ \displaystyle \frac{P Q}{U V}=\displaystyle \frac{Q R}{V W}=\displaystyle \frac{P R}{U W} \\\\ \displaystyle \frac{7}{x}=\displaystyle \frac{10}{y}=\displaystyle \frac{12}{9} \\\\ \displaystyle \frac{7}{x}=\displaystyle \frac{4}{3} \\\\ 4 x=21 \\\\ x=5.25 \\\\ \displaystyle \frac{10}{y}=\displaystyle \frac{4}{3} \\\\ 4 y=30 \\\\ y=7.5 \end{array}$

5.           The measures of two angles of $\triangle X Y Z$ are $82^{\circ}$ and $16^{\circ} .$ Find the measures of the angles of a triangle similar to $\triangle X Y Z$.

Show/Hide Solution

The measures of the first two angles of $\triangle X Y Z$ are $82^{\circ}$ and $16^{\circ}$.

$\therefore$ The measure of the third angle of $\triangle X Y Z$ \[ \begin{array}{l} =180^{\circ}-\left(82^{\circ}+16^{\circ}\right)\\\\ =180^{\circ}-98^{\circ} \\\\ =82^{\circ} \end{array} \] $\therefore$ The measures of the angles of a triangle similar to $\triangle X Y Z$ are $82^{\circ}, 82^{\circ}$ and $16^{\circ}$ .


Graph of the Function $y = |x − h| + k$


The graph of the absolute value function$y = |x − h| + k$ can be seen as the translation of $h$-units horizontally and $k$-units vertically of the graph $y = |x|$.

Graph of the Function $y = -|x − h| + k$


The graph of the absolute value function$y = -|x − h| + k$ can be seen as the translation of $h$-units horizontally and $k$-units vertically of the graph $y = -|x|$.


1.           Compare the graphs of the following functions to the graph of $y=|x|$.

             (a)    $y=|x-3|-2$

             (b)    $y=|x+1|+3$

             (c)    $y=|x-2|+3$

Show/Hide Solution



(a) The graph of $y=|x-3|-2$ is the translation of positive 3 units horizontally and negative 2 units vertically of the graph $y=|x| .$


(b) The graph of $y=|x+1|+3$ is the translation of negative 1 unit horizontally and positive 3 units vertically of the graph $y=|x| .$


(c) The graph of $y=|x-2|+3$ is the translation of positive 2 units horizontally and positive 3 units vertically of the graph $y=|x| .$


2.           Compare the graphs of the following functions to the graph of $y=-|x|$.

             (a)    $y=-|x+3|+2$

             (b)    $y=-|x-4|+1$

             (c)    $y=-|x+4|-1$

Show/Hide Solution



(a) The graph of $y=-|x+3|+2$ is the translation of negative 3 units horizontally and positive 2 units vertically of the graph $y=-|x| .$



(b) The graph of $y=-|x-4|+1$ is the translation of positive 4 units horizontally and positive 1 unit vertically of the graph $y=-|x| .$



(c) The graph of $y=y=-|x+4|-1$ is the translation of negative 4 units horizontally and negative 1 unit vertically of the graph $y=-|x| .$


1.           Draw a set of coordinate axes. Locate the points, $A(2,3)$, $B(2, -4)$ and $C(-4, 3)$. Label each point with its coordinates. Determine whether each of the line segments $AB, BC$ and $CA$ is horizontal or vertical.

Show/Hide Solution


  • $AB$ is a vertical line.
  • $BC$ is neither horizontal nor vertical line.
  • $AC$ is a horizontal line.

2.           Find the missing coordinates in the following table if $M$ is the midpoint of points $P$ and $Q$. $$\begin{array}{|c|c|c|} \hline P & Q & M \\ \hline (2,6) & & (3,3) \\ \hline (3,2) & (-3,-1) & \\ \hline & (0,-1) & (-3,2) \\ \hline (1,5) & & (2.5,3.5) \\ \hline \end{array}$$

Show/Hide Solution

(i) Let the coordinates of the point $Q$ be $(a, b)$.

Since $M$ is the midpoint of $P$ and $Q$, using midpoint formula,

$(3,3) = \left(\displaystyle\frac{a+2}{2},\displaystyle \frac{b+6}{2}\right)$

$\therefore\ \displaystyle \frac{a+2}{2} = 3\ \text{and}\ \displaystyle \frac{b+6}{2} = 3$

$\therefore\ a=4\ \text{and}\ b = 0$

$\therefore\ Q=(4,0)$

(ii) Let the coordinates of the point $M$ be $(x, y)$.

Since $M$ is the midpoint of $P$ and $Q$, using midpoint formula,

$\begin{aligned} (x,y) &= \left(\frac{3-3}{2},\displaystyle \frac{2-1}{2}\right)\\\\ \therefore\ (x,y) &=\left(0,\frac{1}{2}\right) \end{aligned}$

(iii) Let the coordinates of the point $P$ be $(x, y)$.

Since $M$ is the midpoint of $P$ and $Q$, using midpoint formula,

$(-3,2) = \left(\displaystyle\frac{x+0}{2},\displaystyle \frac{y-1}{2}\right)$

$\therefore\ \displaystyle \frac{x}{2} = -3\ \text{and}\ \displaystyle \frac{y-1}{2} = 2$

$\therefore\ x=-6\ \text{and}\ y = 5$

$\therefore\ P=(-6,5)$

(iv) Let the coordinates of the point $Q$ be $(x, y)$.

Since $M$ is the midpoint of $P$ and $Q$, using midpoint formula,

$(2.5,3.5) = \left(\displaystyle\frac{x+1}{2},\displaystyle \frac{y+5}{2}\right)$

$\therefore\ \displaystyle \frac{x+1}{2} = 2.5\ \text{and}\ \displaystyle \frac{y+5}{2} = 3.5$

$\therefore\ x=4\ \text{and}\ y = 2$

$\therefore\ P=(4,2)$


3.           Find the coordinates of the midpoint and the length of the line segment joining these pairs of points. $$\begin{array}{l} \text{(a)}\ (0,0)\ \text{and}\ (4, -4) \\ \text{(b)}\ (1, 5)\ \text{and}\ (3,1) \\ \text{(c)}\ (-3, -3)\ \text{and}\ (0,0) \\ \text{(d)}\ (-1,3)\ \text{and}\ (5, 1) \\ \text{(e)}\ (-1,6)\ \text{and}\ (2, -2) \\ \text{(f)}\ (-3, -4)\ \text{and}\ (3, -1) \end{array}$$

Show/Hide Solution

$\begin{aligned} \text{(a)}\ & \text{The midpoint between } (0,0)\ \text{and }(4,-4)\\\\ &=\left( {\displaystyle \frac{{0+4}}{2},\displaystyle \frac{{0-4}}{2}} \right)\\\\ &=\left( {2,-2} \right)\\\\ & \text{length of segment}\\\\ &=\sqrt{(4-0)^2+(-4-0)^2}\\\\ &= 4\sqrt{2} \end{aligned}$

$\begin{aligned} \text{(b)}\ & \text{The midpoint between }(1,5)\ \text{and }(3,1)\\\\ &=\left( {\displaystyle \frac{{1+3}}{2},\displaystyle \frac{{5+1}}{2}} \right)\\\\ &=\left( {2,3} \right)\\\\ & \text{length of segment}\\\\ &=\sqrt{(3-1)^2+(1-5)^2}\\\\ &= 2\sqrt{5} \end{aligned}$

$\begin{aligned} \text{(c)}\ & \text{The midpoint between }(-3,-3)\ \text{and }(0,0)\\\\ &=\left( {\displaystyle \frac{{-3+0}}{2},\displaystyle \frac{{-3+0}}{2}} \right)\\\\ &=\left( {-\displaystyle \frac{{3}}{2},-\displaystyle \frac{{3}}{2}} \right)\\\\ & \text{length of segment}\\\\ &=\sqrt{(0+3)^2+(0+3)^2}\\\\ &=\sqrt{18}\\\\ &= 3\sqrt{2} \end{aligned}$

$\begin{aligned} \text{(d)}\ & \text{The midpoint between }(-1,3)\ \text{and }(5,1)\\\\ &=\left( {\displaystyle \frac{{-1+5}}{2},\displaystyle \frac{{3+1}}{2}} \right)\\\\ &=\left( {2,2} \right)\\\\ & \text{length of segment}\\\\ &=\sqrt{(5+1)^2+(1-3)^2}\\\\ &=\sqrt{40}\\\\ &= 2\sqrt{10} \end{aligned}$

$ \begin{aligned} \text{(e)}\ & \text{The midpoint between }(-1,6)\ \text{and }(2,-2)\\\\ &=\left( {\displaystyle \frac{{-1+2}}{2},\displaystyle \frac{{6-2}}{2}} \right)\\\\ &=\left(\displaystyle \displaystyle \frac{1}{2},2 \right)\\\\ &\text{length of segment}\\\\ &=\sqrt{(2+1)^2+(-2-6)^2}\\\\ &=\sqrt{73} \end{aligned}$

$ \begin{aligned} \text{(f)}\ &\text{The midpoint between }(-3,-4)\ \text{and }(3,-1)\\\\ &=\left( {\displaystyle \frac{{-3+3}}{2},\displaystyle \frac{{-4-1}}{2}} \right)\\\\ &=\left( {0,-\displaystyle \frac{{5}}{2}} \right)\\\\ & \text{length of segment}\\\\ &=\sqrt{(3+3)^2+(-1+4)^2}\\\\ &=\sqrt{45}\\\\ &= 3\sqrt{5} \end{aligned}$

4.           If $(1,0)$ is the midpoint of the line passing through the points $A(-5, 2)$ and $B(x, y)$, find the value of $x$ and of $y$.

Show/Hide Solution

Since $(1,0)$ is the midpoint of $A(-5,2)$ and $B(x,y)$, using midpoint formula,

$(1,0) = \left(\displaystyle \frac{-5+x}{2}, \displaystyle \frac{2+y}{2}\right)$

$\therefore\ \displaystyle \frac{-5+x}{2} = 1\ \text{and}\ \displaystyle \frac{2+y}{2} = 0$

$\therefore\ x=7\ \text{and}\ y = -2$

5.           Calculate the perimeter of given polygons correct to one decimal place.

(a) A triangle with vertices $P(-2, 3), Q(5, -4)$ and $R(1, 8)$.

(b) A parallelogram with vertices $A(-10, 1), B(6, -2), C(14, 4)$ and $D(-2, 7)$.

(c) A trapezium with vertices $E(-6, -2), F(1, -2), G(0, 4)$ and $H(-5, 4)$.

Show/Hide Solution

(a) $ P=(-2,3),Q=(5,-4),R=(1,8)$

Since the distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{\left( x_2-x_1 \right)^2 + \left(y_2-y_1 \right)^2},$

$\begin{aligned} PQ&=\sqrt{(5+2)^2+(-4-3)^2}\\\\ &=\sqrt{98}\\\\ &=9.9\\\\ QR &=\sqrt{(1-5)^2+(8+4)^2}\\\\ &=\sqrt{160}\\\\ &=12.7\\\\ PR&=\sqrt{(1+2)^2+(8-3)^2}\\\\ &=\sqrt{34}\\\\ &=5.8 \end{aligned}$

$\therefore\ PQ+QR+PR=28.4$

$\therefore$ the perimeter of $\triangle PQR = 28.4$ units.

(b) $A=(-10, 1), B=(6, -2), C=(14, 4), D=(-2, 7)$

Since the distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{\left( x_2-x_1 \right)^2 + \left(y_2-y_1 \right)^2},$

$\begin{aligned} AB &=\sqrt{(6+10)^2+(-2-1)^2}\\\\ &=\sqrt{265}\\\\ &=16.3\\\\ BC &=\sqrt{(14-6)^2+(4+2)^2}\\\\ &=\sqrt{100}\\\\ &=10.0 \end{aligned}$

Since $ABCD$ is a parallelogram, $CD=AB$ and $AD = BC$

$\therefore\ AB + BC + CD + AD =2(16.3) + 2(10)=52.6$

$\therefore$ the perimeter of triangle parallelogram $ABCD = 52.6$ units.

(c) $E=(-6, -2), F=(1, -2), G=(0, 4), H=(-5, 4)$

Since the distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{\left( x_2-x_1 \right)^2 + \left(y_2-y_1 \right)^2},$

$\begin{aligned} EF &=\sqrt{(1+6)^2+(-2+2)^2}\\\\ &=\sqrt{7^2}\\\\ &=7.0\\\\ FG &=\sqrt{(0-1)^2+(4+2)^2}\\\\ &=\sqrt{37}\\\\ &=6.1\\\\ GH &=\sqrt{(-5-0)^2+(4-4)^2}\\\\ &=\sqrt{5^2}\\\\ &=5.00\\\\ EH &=\sqrt{(-5+6)^2+(4+2)^2}\\\\ &=\sqrt{37}\\\\ &=6.1\\\\ \end{aligned}$

$\therefore\ AB + BC + CD + AD =7.0+6.1+5.0+6.1=24.2$

6.           A circle has centre $(2, 1$). Find the coordinates of the endpoint of a diameter if one endpoint is $(7, 1)$.

Show/Hide Solution


Let the other endpoint be $(x, y)$.

Hence, $(2, 1)$ is the midpoint between $(x, y)$ and $(7,1)$

Using midpoint formula,

$(2, 1) = \left(\frac{x+7}{2}, \frac{y+1}{2}\right)$

$\therefore\ \frac{x+7}{2} = 2\ \text{and}\ \frac{y+1}{2} = 1$

$\therefore\ x = -3\ \text{and}\ y = 1$

Hence the other endpoint is $(-3, 1).$


7.           $\triangle KLM$ has vertices $K(-5,18)$, $L(10,14)$ and $M(-5, -10)$.

(a) Find the length of each side.

(b) Find the perimeter of $\triangle KLM$.

(c) Find the area of $\triangle KLM$.

Show/Hide Solution


Since the distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{\left( x_2-x_1 \right)^2 + \left(y_2-y_1 \right)^2},$

$\begin{aligned} KL &=\sqrt{(10+5)^2+(14-18)^2}\\\\ &=\sqrt{241}\\\\ &=15.5\\\\ LM &=\sqrt{(-5-10)^2+(-10-14)^2}\\\\ &=\sqrt{801}\\\\ &=3\sqrt{89}\\\\ &=28.3\\\\ KM &=\sqrt{(-5+5)^2+(-10-18)^2}\\\\ &=\sqrt{28^2}\\\\ &=28\\\\ \text{The}\ & \text{perimeter of}\ \triangle KLM \\\\ & = KL+LM+KM\\\\ & = 15.5+28.3+28\\\\ &= 71.8\ \text{units} \end{aligned}$

Since $K$ and $M$ have the same $x$-coordinate, $KM$ is a vertical line.

Draw $LN\bot KM$.

Hence the coordinates of $N$ is $(-5,14)$.

$\begin{aligned} \therefore\ LN &=\sqrt{(-5-10)^2+(14-14)^2}\\\\ &=\sqrt{15^2}\\\\ &=15\\\\ \text{The}\ & \text{area of}\ \triangle KLM \\\\ & = \frac{1}{2}\cdot KM\cdot LN\\\\ & = \frac{1}{2}\cdot 28\cdot 15\\\\ & = 210\ \text{sq-units} \end{aligned}$


မြန်မာနိုင်ငံ၏ ပညာရေးစနစ်ကို K+12 စနစ်သစ်သို့ အတန်းလိုက် နှစ်အပိုင်းအခြားအလိုက် ပြောင်းလဲ လာခဲ့ရာ ၂၀၂၀-၂၁ ပညာသင်နှစ်တွင် အထက်တန်းဆင့် ပြောင်းလဲမှုမှာ အသက်ဝင်လာပြီ ဖြစ်သည်။

အတန်းစနစ် ပြောင်းလဲမှုနှင့်အတူ သင်ခန်းစာသင်ရိုးများလဲ ပြောင်းလဲလာခဲ့ရာ Grade 10 Mathematics လည်းပါဝင်လာပါသည်။ Grade 10 Mathematics သည် ယခင် န၀မတန်း သင်ရိုးကို အခြေခံ ပြောင်းလဲထားပြီး နိုင်ငံတကာ သင်ရိုးများနှင့် လိုက်လျောညီထွေ ဖြစ်စေရန် ပြင်ဆင်ပြောင်းလဲမှု အများအပြား ပါဝင်လာပါသည်။

သင်ရိုးအပြောင်းအလဲတွင် သင်ယူလေ့လာသူ ကျောင်းသား၊ ကျောင်းသူများ သင်ခန်းစာဆိုင်ရာ သဘောတရားများ၊ ဝေါဟာရများ၊ လွယ်ကူစွာ လေ့လာနိုင်ရေးအတွက် Target Mathematics မှ သင်ရိုးသစ် သင်္ချာအထောက်ကူပြု စာအုပ်ကို စီစဉ် ထုတ်ဝေလိုက်ပါသည်။ သင်ရိုးသစ် သင်္ချာတွင် ယခင်သင်ရိုးဟောင်း ကဲ့သို့ အခန်း (၁၀) ခန်းပါဝင်ပါသည်။

သင်ရိုးသစ်ပြဌာန်းချက်တွင် ပါဝင်သော သင်ခန်းစာများ


Chapter Title
Chapter (1) Introduction to Coordinate Geometry
Chapter (2) Exponents and Radicals
Chapter (3) Logarithms
Chapter (4) Functions
Chapter (5) Quadratic Functions
Chapter (6) Absolute Value Functions
Chapter (7) Probability
(chapter (8) Similarity
Chapter (9) Circles
Chapter (10) Trigonometry

Target Mathematics သည် အထက်ပါ သင်ခန်းစားများမှ Chapter (1) မှ Chapter (6) အထိကို အတွဲ (၁) ဖြစ် ထုတ်ဝေခဲ့ပြီး Chapter (7) မှ (10) အထိကို အတွဲ(၂) အဖြစ်ဆက်လက် ထုတ်ဝေလိုက်ပါပြီ။ အတွဲ(၁) နှင့် အတွဲ(၂) နှစ်အုပ်လုံးအတွက် အောက်ပါအတိုင်း စီစဉ်ထားပါသည်။

Grade 10 - Vol (1), Price = K 3500 Grade 10 - Vol (2), Price = K 3000

  • သင်ခန်းစာ ရှင်းလင်းချက်
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  • ထပ်ဆောင်းလေ့ကျင့်ခန်းများ
  • သင်ရိုးအကျဉ်းချုပ် စုစည်းမှု (Summary)
  • လေ့ကျင့်ခန်းတိုင်း အတွက် answer key များ ဖြစ်ပါသည်။

Target Mathematics အတွဲ (၁) နှင့် အတွဲ(၂) နှစ်အုပ်လုံးသည် လေ့လာသူအတွက် ...

  • ပြဌာန်းသင်ရိုးစာအုပ် အဖြစ်သော်လည်းကောင်း
  • အထူးထုတ်အသွင် အဖြစ်သော်လည်းကောင်း
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ဘက်စုံအသုံးချနိုင်ရန် ရေးသားပြင်ဆင် ပေးထားသဖြင့် သင်ယူလေ့လာသူ ကျောင်းသား၊ ကျောင်းသူတို့အတွက် များစွာအထောက်အကူ ဖြစ်စေလိမ့်မည် ဟုယုံကြည်မိပါသည်။

ထို့အပြင် သင်ခန်းစာ လေ့ကျင့်ခန်းတိုင်းအတွက် အဖြေအပြည့်အစုံ (Complete Solution) ကိုလည်း ယခု website တွင် တစ်နှစ်ပတ်လုံး စဉ်ဆက်မပြတ် တင်ပေးသွားမည်ဖြစ်ကြောင်း သတင်းကောင်း ပါးအပ်ပါသည်။

မှာယူလိုပါက အောက်ပါ Order Form တွင် ပုံစံဖြည့်၍ မှာယူနိုင်ပါသည်။ ပုံစံဖြည့်၍ Submit လုပ်လိုက်သည်နှင့် မှာယူသူ၏ အမှာစာကို လက်ခံရပြီး ငွေလွှဲရန်အတွက် ပြန်လည်ဆက်သွယ်ပေးမည် ဖြစ်ပါသည်။


1) The Midpoint Formula

If $M(x,y)$ is the midpoint between the two endpoints $A(x_1,y_1)$ and $B(x_2,y_2)$ then $$ \displaystyle M(x,y)=\left( {\frac{{{{x}_{1}}+{{x}_{2}}}}{2}+\frac{{{{y}_{1}}+{{y}_{2}}}}{2}} \right)$$

2) Distance Formula

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $$ \displaystyle \sqrt{{{{{\left( {{{x}_{2}}-x} \right)}}^{2}}+{{{\left( {{{y}_{2}}-{{y}_{1}}} \right)}}^{2}}}}$$

3) The slope of a Line

$$ \displaystyle \text{slope }(m)=\frac{{\text{vertical change}}}{{\text{horizontal change}}}=\frac{{\text{rise}}}{{\text{run}}}$$

4) Slope Formula

If the slope of the line passing through any two points $A(x_1, y_1)$ and $B(x_2, y_2)$ is $m$, then $$m=\displaystyle\frac{y_2-y_1}{x_2-x_1}$$

5) Positive Slope

A line ascending from left to right has a positive slope.

6) Negative Slope

A line descending from left to right has a negaitive slope.

7) Zero Slope

The slope of the horizontal line is zero.

8) Undefined Slope

The slope of the vertical line is undefined.

9) Slope-Intercept Form of a Line

The equation of the form
$$y=mx+c$$ is the equation of a straight line with slope $m$ and $y$-intercept $c$, which is called the slope-intercept form.

10) Point-Slope Form of a Line

The equation of a straight line, with slope $m$, and passes through the point $(x_1, y_1)$ is
$$y-y_1=m(x-x_1)$$ which is called the point-slope form of a straight line.

11) Horizontal Line

  • The $X$-axis and all lines parallel to it are called horizontal lines.
  • The equation of horizontal line intersecting the $Y$-axis at $(0,c)$ is $y=c$.
  • The equation of the $X$-axis is $y=0$.

12) Vertical Line

  • The $Y$-axis and all lines parallel to it are called vertical lines.
  • The equation of vertical line intersecting the $X$-axis at $(a,0)$ is $x=a$.
  • The equation of the $Y$-axis is $x=0$.

13) Parallel Lines and Perpendicular Lines

  • Any two horizontal lines are parallel.
  • Any two vertical lines are parallel.
  • Vertical and horizontal lines are perpendicular to each other.

14) Some Important Properties

  • Two non-vertical lines are parallel if and only if they have the same slope.
  • Two non-vertical lines are perpendicular if and only if the product of their slopes is -1 (i.e., one is the negative reciprocal of the other).
  • On a same straight line all segments have the same slope.
  • Three or more points that lie on the same straight line are said to be collinear.