Answer:

Step-by-step explanation:
In right triangles only, the tangent of an angle is equal to its opposite side divided by its adjacent side.
Therefore, we have the following equation:

Solving, we get:

Also in right triangles only, the cosine of an angle is equal to its adjacent side divided by the hypotenuse of the triangle.

We can also use the Pythagorean theorem now that we've found BC. However, if you choose to do so, make sure you don't use a rounded value for BC, as that could cause a notable deviation in your final answer.
Using the Pythagorean theorem to verify our answers:

(In English), your question would state the following:
<span>Calculate the circumference and circumference of the circle entered in a square 8 cm long.
</span>
So when finding the answer this this question, we know that the circumference would be the complete area of a circle.
So, I believe we would want to find what is the radius and the diameter that would make this circumference 8

.
The radius would then be (1.274).
__________________________________________________________
Now in your own language (Croatian).
<span>Dakle, kad pronađemo odgovor na ovo pitanje, znamo da bi opseg bio cjelokupno područje kruga.
Dakle, vjerujem da bismo željeli otkriti što je radijus i promjer koji bi ovaj opseg mogao učiniti 8cm.
Radijus bi onda bio (1.274).</span>
Answer:
(a) See attachment for tree diagram
(b) 24 possible outcomes
Step-by-step explanation:
Given


Solving (a): A possibility tree
If urn 1 is selected, the following selection exists:
![B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]](https://tex.z-dn.net/?f=B_1%20%5Cto%20%5BR_1%2C%20R_2%2C%20R_3%5D%3B%20R_1%20%5Cto%20%5BB_1%2C%20R_2%2C%20R_3%5D%3B%20R_2%20%5Cto%20%5BB_1%2C%20R_1%2C%20R_3%5D%3B%20R_3%20%5Cto%20%5BB_1%2C%20R_1%2C%20R_2%5D)
If urn 2 is selected, the following selection exists:
![B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]](https://tex.z-dn.net/?f=B_2%20%5Cto%20%5BB_3%2C%20R_4%2C%20R_5%5D%3B%20B_3%20%5Cto%20%5BB_2%2C%20R_4%2C%20R_5%5D%3B%20R_4%20%5Cto%20%5BB_2%2C%20B_3%2C%20R_5%5D%3B%20R_5%20%5Cto%20%5BB_2%2C%20B_3%2C%20R_4%5D)
<em>See attachment for possibility tree</em>
Solving (b): The total number of outcome
<u>For urn 1</u>
There are 4 balls in urn 1

Each of the balls has 3 subsets. i.e.
![B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]](https://tex.z-dn.net/?f=B_1%20%5Cto%20%5BR_1%2C%20R_2%2C%20R_3%5D%3B%20R_1%20%5Cto%20%5BB_1%2C%20R_2%2C%20R_3%5D%3B%20R_2%20%5Cto%20%5BB_1%2C%20R_1%2C%20R_3%5D%3B%20R_3%20%5Cto%20%5BB_1%2C%20R_1%2C%20R_2%5D)
So, the selection is:


<u>For urn 2</u>
There are 4 balls in urn 2

Each of the balls has 3 subsets. i.e.
![B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]](https://tex.z-dn.net/?f=B_2%20%5Cto%20%5BB_3%2C%20R_4%2C%20R_5%5D%3B%20B_3%20%5Cto%20%5BB_2%2C%20R_4%2C%20R_5%5D%3B%20R_4%20%5Cto%20%5BB_2%2C%20B_3%2C%20R_5%5D%3B%20R_5%20%5Cto%20%5BB_2%2C%20B_3%2C%20R_4%5D)
So, the selection is:


Total number of outcomes is:



15+12+6= total number of marbles
33= total
Not clear marbles = 15+12 or 27
27 /33
.818181
82 percent