Name: Trigonometric Ratios for the Angle of 30 Degrees
Uploaded: 2019-07-19
Duration: 3 min 17 s
Description: What is the angular bisector of 60 degrees? It is 30 degree. Wondering how? Let us learn the trigonometric ratios of 30 degree in this interesting video!

Question

When data is sent electronically, for example when you download a song, you don't want parts of it to be missing. Computer programmers, who imagine data as points in space, use Pythagoras' theorem to make sure that the data is where it should be. Given an original point and its final location, determine whether or not a piece of data is where it should be (a distanced from its original location). Copy and complete this table.

Original Location	Final Location	Actual distance travelled	Distance data was supposed to travel (d)	Did the data end up where it was supposed to? (Y/N)
$(2, 6)$	$(8, 14)$		$10$
$(- 1, 3)$	$(2, - 1)$		$6$
$(1, - 1)$	$(- 4, 5)$		$7$
$(0, 7)$	$(- 5, - 5)$		$13$

Accepted Answer

The shortest distance between $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is $d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$ units

Let the original location $(2, 6)$ be $(x_{1}, y_{1})$ and the final location $(8, 14)$ be $(x_{2}, y_{2})$ then

$d = \sqrt{{(8 - 2)}^{2} + {(14 - 6)}^{2}}$

$\Rightarrow d = \sqrt{6^{2} + 8^{2}}$

$\Rightarrow d = \sqrt{36 + 64}$

$\Rightarrow d = \sqrt{100}$

$\Rightarrow d = 10$

Therefore, the actual distance travelled is $10$ units.

Let the original location $(- 1, 3)$ be $(x_{1}, y_{1})$ and the final location $(2, - 1)$ be $(x_{2}, y_{2})$ then

$d = \sqrt{{(2 + 1)}^{2} + {(- 1 - 3)}^{2}}$

$\Rightarrow d = \sqrt{3^{2} + {(- 4)}^{2}}$

$\Rightarrow d = \sqrt{9 + 16}$

$\Rightarrow d = \sqrt{25}$

$\Rightarrow d = 5$

Therefore, the actual distance travelled is $5$ units.

Let the original location $(1, - 1)$ be $(x_{1}, y_{1})$ and the final location $(- 4, 5)$ be $(x_{2}, y_{2})$ then

$d = \sqrt{{(- 4 - 1)}^{2} + {(5 + 1)}^{2}}$

$\Rightarrow d = \sqrt{{(- 5)}^{2} + 6^{2}}$

$\Rightarrow d = \sqrt{25 + 36}$

$\Rightarrow d = \sqrt{61}$

$\Rightarrow d = 7.81$

Therefore, the actual distance travelled is $7.81$ units.

Let the original location $(0, 7)$ be $(x_{1}, y_{1})$ and the final location $(- 5, - 5)$ be $(x_{2}, y_{2})$ then

$d = \sqrt{{(- 5 - 0)}^{2} + {(- 5 - 7)}^{2}}$

$\Rightarrow d = \sqrt{{(- 5)}^{2} + {(- 12)}^{2}}$

$\Rightarrow d = \sqrt{25 + 144}$

$\Rightarrow d = \sqrt{169}$

$\Rightarrow d = 13$

Therefore, the actual distance travelled is $13$ units.

Now let us complete the table.

Original Location

Final Location

Actual distance travelled
Distance data was supposed to travel (d) Did the data end up where it was supposed to? (Y/N)

$(2, 6)$ $(8, 14)$ $10$ $10$ Y

$(- 1, 3)$ $(2, - 1)$ $5$ $6$ N

$(1, - 1)$ $(- 4, 5)$ $7.81$ $7$ N

$(0, 7)$ $(- 5, - 5)$ $13$ $13$ Y

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