Question

Trigonometry: Angles and Length Escape Room (level 3)

so i need help bc i just can’t figure the letters out!!

Answer 1

Answer:

A -) E -) I -) F -) C -) H -) K -) Escape!

Step-by-step explanation:

I calculated it, so shouldn't be wrong but tell me if it is.

Answer 2

Answer:

The steps for your escape are:

A-> E -> I -> F -> C -> H -> K -> ESCAPE

Step-by-step explanation:

For solving this question, you need to apply the right triangle properties and trigonometric ratios in each triangle.

1. Triangle A

Here the question gives the adjacent side (52) from the angle x and the hypotenuse (56) of the right triangle. Therefore, you can find x from the trigonometric ratio of cos :

$cos(x)=\frac{adj}{hyp}= \frac{52}{56}=\frac{13}{14}$

After that, you should calculate the arccos(x).

$\arccos \left(\frac{13}{14}\right) = 22^{\circ \:}$

Let's to triangle E.

2. Triangle E

The question gives two angles (90° and 43°) and the hypotenuse (22) of the right triangle. Therefore, you can find x from the trigonometric ratio of sin (43°):

$\sin \left(43^{\circ \:}\right)=\frac{opp}{hyp} =\frac{x}{22}\\\\\sin \left(43^{\circ \:}\right)=0.68199\dots \\\\Then,\\\\0.68199=\frac{x}{22}\\ 0.68199\cdot \:22=15\\x=15^{\circ \:}$

Let's to triangle I.

3. Triangle I

The question gives two sides (34  and 42). Therefore, you can find x from the trigonometric ratio of tan(43°):

$tan(x)=\frac{opp}{adj} =\frac{34}{42} =\frac{17}{21}\\ \\Then,\\\\arctan (\frac{17}{21})=39^{\circ \:}$

Let's to triangle F.

4. Triangle F

Again the question gives two sides (54 and 51). Therefore, you can find x applying the trigonometric ratio of sin (x):

$sin(x)=\frac{opp}{hyp}= \frac{51}{54} =\frac{17}{18}\\ \\Then,\\\\arcsin (\frac{17}{18})=71^{\circ \:}$

Let's to triangle C.

5. Triangle C

The question gives an angle (38°) and the opposite side (20) from the angle 38°  of the right triangle. Therefore, you can find x from the trigonometric ratio of tan (38°):

$tan(38^{\circ \:})=\frac{opp}{adj} =\frac{20}{x}\\\\\tan \left(38^{\circ \:}\right)=0.78128\\\\Then,\\\\0.78128=\frac{20}{x}\\x=25.6^{\circ \:}$

Let's to triangle H.

6. Triangle H

The question gives an angle (40°) and the hypotenuse (20). Therefore, you can find x from the trigonometric ratio of sin (40°):

$\sin \left(40^{\circ \:}\right)=\frac{opp}{hyp} =\frac{x}{20}\\\\\sin \left(40^{\circ \:}\right)=0.64278\dots \\\\Then,\\\\0.64278=\frac{x}{20}\\ 0.64278\cdot \:20=12.9\\x=12.9^{\circ \:}$

Let's to triangle K.

7. Triangle K

the question gives the adjacent side (26) from the angle x and the hypotenuse (45) of the right triangle. Therefore, you can find x from the trigonometric ratio of cos :

$cos(x)=\frac{adj}{hyp}= \frac{26}{45}$

After that, you should calculate the arccos(x).

$\arccos \left(\frac{26}{45}\right) = 55^{\circ \:}$

Finally...ESCAPE!

Learn more about trigonometric ratio here:

https://brainly.com/question/11967894