Mathematics, 21.10.2019 21:00 Blackhawk1881

Week 8 trigonometric ! me is a final grade for school!

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Answer from: dee1334

Part 1) The six trigonometric functions in the procedure

Part 2) The six trigonometric functions in the procedure

Part 3) The six trigonometric functions in the procedure

Part 4) The value of x is $x=7\sqrt{2}\ units$ and the value of y is $y=7\ units$

Part 5) The value of x is $x=5\ units$ and the value of y is $y=5\sqrt{3}\ units$

Part 6) The value of x is $x=2\sqrt{3}\ units$ and the value of y is $y=\sqrt{3}\ units$

Part 7) $cos(27\°)=0.8910$

Part 8) $tan(5\°)=0.0875$

Part 9) $sin(48\°)=0.7431$

Part 10) $cot(81\°)=0.1584$

Part 11) $csc(23\°)=2.5593$

Part 12) $sec(66\°)=2.4586$

Part 13) $cot(13\°)=4.3315$

Part 14) $sin(32\°)=0.5299$

Step-by-step explanation:

Note The complete answers in the attached file

Part 1) In the right triangle of the figure find the hypotenuse

Applying Pythagoras theorem

$c^{2} =8^{2}+15^{2}\\c^{2}=289\\c=17\ units$

1) Find the $sin(\theta)$

$sin(\theta)=\frac{8}{17}$ ----> opposite side angle $\theta$ divided by the hypotenuse

2) Find the $cos(\theta)$

$cos(\theta)=\frac{15}{17}$ ----> adjacent side angle $\theta$ divided by the hypotenuse

3) Find the $tan(\theta)$

$tan(\theta)=\frac{8}{15}$ ----> opposite side angle $\theta$ divided by the adjacent side angle $\theta$

4) Find the $cot(\theta)$

$cot(\theta)=\frac{15}{8}$ ----> adjacent side angle $\theta$ divided by the opposite side angle $\theta$

5) Find the $sec(\theta)$

$sec(\theta)=\frac{1}{cos(\theta)}$

$sec(\theta)=\frac{17}{15}$ ----> hypotenuse divided by the adjacent side angle $\theta$

6) Find the $csc(\theta)$

$csc(\theta)=\frac{1}{sin(\theta)}$

$csc(\theta)=\frac{17}{8}$ ----> hypotenuse divided by the opposite side angle $\theta$

Part 2) In the right triangle of the figure find the adjacent side angle $\theta$

Applying Pythagoras theorem

$5^{2} =2^{2}+a^{2}\\ a^{2}=5^{2}-2^{2}\\a^{2}=21\\a=\sqrt{21}\ units$

1) Find the $sin(\theta)$

$sin(\theta)=\frac{2}{5}$ ----> opposite side angle $\theta$ divided by the hypotenuse

2) Find the $cos(\theta)$

$cos(\theta)=\frac{\sqrt{21}}{5}$ ----> adjacent side angle $\theta$ divided by the hypotenuse

3) Find the $tan(\theta)$

$tan(\theta)=\frac{2}{\sqrt{21}}$ ----> opposite side angle $\theta$ divided by the adjacent side angle $\theta$

4) Find the $cot(\theta)$

$cot(\theta)=\frac{\sqrt{21}}{2}$ ----> adjacent side angle $\theta$ divided by the opposite side angle $\theta$

5) Find the $sec(\theta)$

$sec(\theta)=\frac{1}{cos(\theta)}$

$sec(\theta)=\frac{5}{\sqrt{21}}$ ----> hypotenuse divided by the adjacent side angle $\theta$

6) Find the $csc(\theta)$

$csc(\theta)=\frac{1}{sin(\theta)}$

$csc(\theta)=\frac{5}{2}$ ----> hypotenuse divided by the opposite side angle $\theta$

Part 3) In the right triangle of the figure find the opposite side angle $\theta$

Applying Pythagoras theorem

$3^{2} =1^{2}+b^{2}\\ b^{2}=3^{2}-1^{2}\\b^{2}=8\\b=\sqrt{8}\ units$

1) Find the $sin(\theta)$

$sin(\theta)=\frac{\sqrt{8}}{3}$ ----> opposite side angle $\theta$ divided by the hypotenuse

2) Find the $cos(\theta)$

$cos(\theta)=\frac{\sqrt{1}}{3}$ ----> adjacent side angle $\theta$ divided by the hypotenuse

3) Find the $tan(\theta)$

$tan(\theta)=\frac{\sqrt{8}}{1}$ ----> opposite side angle $\theta$ divided by the adjacent side angle $\theta$

4) Find the $cot(\theta)$

$cot(\theta)=\frac{1}{\sqrt{8}}$ ----> adjacent side angle $\theta$ divided by the opposite side angle $\theta$

5) Find the $sec(\theta)$

$sec(\theta)=\frac{1}{cos(\theta)}$

$sec(\theta)=\frac{3}{1}$ ----> hypotenuse divided by the adjacent side angle $\theta$

6) Find the $csc(\theta)$

$csc(\theta)=\frac{1}{sin(\theta)}$

$csc(\theta)=\frac{3}{\sqrt{8}}$ ----> hypotenuse divided by the opposite side angle $\theta$

Part 4) In the right triangle of the figure

a) Find the value of x

we know that

$sin(45\°)=\frac{7}{x}$

$x=\frac{7}{sin(45\°)}$

Remember that

$sin(45\°)=\frac{\sqrt{2}}{2}$

substitute

$x=\frac{7}{\frac{\sqrt{2}}{2}}$

$x=\frac{14}{\sqrt{2}}$

$x=7\sqrt{2}\ units$

b) Find the value of y

The value of $y=7\ units$ ----> by triangle 45°-90°-45° measures

Note The complete answers in the attached file

Answer from: Quest

828. all there is to it!

here, if you ever need with rounding (or other forms of arithmetic/math) again, use this:

thats the rounding one, but if you just click the top left, it shows you all the calculators you may need!

Answer from: Quest

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

u also answered this so i think u would know but i am here to

put in or use mine