Fragen 213
Antworten 20


We can find cos(C) using the Law of Cosines. Here's how:


Law of Cosines:


The Law of Cosines relates the sides and angles of a triangle. For triangle ABC, it states:


c^2 = a^2 + b^2 - 2ab * cos(C)




c is the side opposite angle C (in this case, side AB)


a is the side adjacent to angle A (in this case, side BC)


b is the side adjacent to angle B (in this case, side AC)


Given Information:


We are given:


cos(A) = sqrt(7/10)


cos(B) = sqrt(3/10)


We need to find cos(C).


Sides are Unknown:


We don't have information about the side lengths (a, b, or c). However, we can express them using cos(A) and cos(B) for further calculations.


Expressing Sides:


Using the trigonometric identity cos^2(x) + sin^2(x) = 1, we can find sin^2(A) and sin^2(B):


sin^2(A) = 1 - cos^2(A) = 1 - (sqrt(7/10))^2 = 3/10


sin^2(B) = 1 - cos^2(B) = 1 - (sqrt(3/10))^2 = 7/10


Relating Sides to Cosines:


Since we are looking for cos(C), let's focus on side c (AB). Using cosine definitions in right triangles:


cos(A) = (adjacent side) / (hypotenuse) = a / c


cos(B) = (adjacent side) / (hypotenuse) = b / c




a = c * cos(A) = c * sqrt(7/10)


b = c * cos(B) = c * sqrt(3/10)


Applying the Law of Cosines:


Substitute the expressions for a and b in the Law of Cosines equation:


c^2 = (c * sqrt(7/10))^2 + (c * sqrt(3/10))^2 - 2 * c * sqrt(7/10) * c * sqrt(3/10) * cos(C)


Simplifying the equation:


c^2 = (7/10)c^2 + (3/10)c^2 - (6/10)c^2 * cos(C)


Combining like terms:


c^2 = (7 + 3 - 6) / 10 * c^2 - (6/10)c^2 * cos(C)


4c^2 / 10 = -(6/10)c^2 * cos(C)


Solving for cos(C):


Isolate cos(C):


cos(C) = -(4c^2 / 10) / -(6/10)c^2


cos(C) = 4/6 (Since both c^2 terms are negative, the negative signs cancel out)


cos(C) = 2/3


Therefore, cos(C) = 2/3.


This problem can be solved by considering the complementary probability (the probability of the opposite event happening) and subtracting it from 1.


Event: The coin is tossed no more than 10 times (either heads or tails appear three times in a row within 10 tosses).


Complementary Event: The coin is tossed more than 10 times (neither heads nor tails appear three times in a row within 10 tosses).


Probability of Complementary Event:


There are two possibilities for the complementary event:


HHH appears within the first 10 tosses: There are 10 possibilities for the location of the first H (tosses 1 through 10). For each of these possibilities, there are two remaining tosses that must be Hs (total of 2 * 10 = 20 outcomes).


TTT appears within the first 10 tosses: Similar to case 1, there are 20 outcomes (10 possibilities for the first T and two remaining tosses must be Ts).


The total number of successful outcomes for the complementary event is 20 (HHH) + 20 (TTT) = 40.


Since each outcome (sequence of tosses) has an equal probability (either H or T on each toss), the probability of any specific outcome is 1/2 raised to the power of the number of tosses (2^-# of tosses).


The total number of possible outcomes for 10 tosses is 2^10 (either H or T on each toss).


Therefore, the probability of the complementary event is:


Probability (complementary event) = (Favorable outcomes) / (Total possible outcomes) Probability (complementary event) = 40 / (2^10)


Probability of the Event:


The probability of the event we're interested in (tossing more than 10 times) is the complement of the complementary event.


Probability (event) = 1 - Probability (complementary event)


Since all outcomes are equally likely, the sum of the probabilities of all possible events must be 1.


Probability (event) = 1 - 40 / (2^10)


Simplifying the Fraction:


We can rewrite 2^10 as 1024.


Probability (event) = 1 - (40 / 1024)

Probability (event) = 1 - (5 / 128)

Probability (event) = 123/128


Therefore, the answer is 123/128.


Here's another way to solve this problem, focusing on proportionality within the parallelogram:


Proportions in Similar Triangles:


Since ABCD is a parallelogram, lines AD and BC are parallel. When line DE intersects line BC at point F, it creates two transversal lines. Because of this, corresponding angles on alternate sides of DE are congruent (alternate interior angles)


Therefore, triangles EAD and EBF are similar by Angle-Angle Similarity (AA).


Since the triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding side lengths:


Area(EBF) / Area(EAD) = (length(BF) / length(AD))^2


Given Information:


We are given that the area of triangle EBF is 4 and the area of triangle EAD is 9. Plugging these values into the equation:


4 / 9 = (length(BF) / length(AD))^2


Proportionality in a Parallelogram:


In a parallelogram, opposite sides have the same length. Therefore, length(BF) = length(AE) and length(AD) = length(BC). Substituting these equalities into the equation from step 2:


4 / 9 = (length(AE) / length(BC))^2


Area of Parallelogram:


The area of a parallelogram is equal to the base times the height. Since AD is parallel to BC, the height of parallelogram ABCD is the same as the height of triangle EAD (considering side AE as the base). We can express this using proportionality:


Area(ABCD) ∝ length(BC) * height(EAD)


From step 3, we know the ratio between the base of triangle EAD (which is also a base of the parallelogram) and side BC of the parallelogram:

length(AE) / length(BC) = √(4/9) (taking the square root of both sides)


Combining Proportions:


Since the area of the parallelogram is proportional to the product of its base and height, and we know the ratio between the base and a side of the parallelogram, we can combine these proportions:


Area(ABCD) ∝ (√(4/9)) * height(EAD)


Note: We don't need to find the actual height of triangle EAD since it cancels out when solving for the relative area of the parallelogram.


Relative Area of Parallelogram:


Since the area of triangle EAD is a constant value (9) and the other factor in the proportion is a constant resulting from the given information, the area of parallelogram ABCD is also proportional to a constant value.


Therefore, relative to the area of triangle EAD, the area of parallelogram ABCD is:


Area(ABCD) ∝ √(4/9) * Area(EAD) = √(4/9) * 9 = 4


Scaling to Actual Area:


While the previous step gives us the relative area compared to triangle EAD, we can find the actual area by considering that the area of triangle EBF is given as 4.


Since triangles EAD and EBF are similar, the ratio of their areas is the square of the ratio of their corresponding side lengths (as established in step 1). Therefore, the area of triangle EAD is 4 times the area of triangle EBF:


Area(EAD) = 4 * Area(EBF) = 4 * 4 = 16


Now, we can use the relative area we found in step 6 to solve for the actual area of the parallelogram:


Area(ABCD) = √(4/9) * Area(EAD) = √(4/9) * 16


Area(ABCD) = 4 * 4 = 16​.