Find the values of the trigonometric functions of the acute angles of
the right triangle ABC, given the other side is a =2 and hypotenuse is
c= 2√5.
Wow, that's a "right" triangle! Let's have some trigonometric fun, shall we?
For angle A, we have the opposite side (a) equal to 2 and the hypotenuse (c) equal to 2√5. So, let's find the sine:
sin(A) = opposite/hypotenuse = 2/(2√5) = 1/√5
Now, let's move on to the cosine. Angle A is between the adjacent side (b) and the hypotenuse (c), so we have:
cos(A) = adjacent/hypotenuse = b/(2√5)
Unfortunately, we don't have the value of b yet. But don't worry, we can use the Pythagorean theorem to find it!
Using Pythagoras, we know that a² + b² = c².
Substituting the given values, we have:
2² + b² = (2√5)²
4 + b² = 20
b² = 16
b = 4
Now with the value of b, we can find cos(A):
cos(A) = 4/(2√5) = 2/√5 = (2/√5) * (√5/√5) = 2√5/5
Alright, let's move on to tangent. Tangent is simply the ratio of sine to cosine:
tan(A) = sin(A)/cos(A) = (1/√5) / (2√5/5) = (1/√5) * (5/2√5) = 1/2
So, to sum it all up:
sin(A) = 1/√5
cos(A) = 2√5/5
tan(A) = 1/2
That's it! I hope you had a laugh and found the values of the trigonometric functions too.
To find the values of the trigonometric functions of the acute angles of the right triangle ABC, we need to identify the missing side of the triangle.
Given that side a = 2, we can use the Pythagorean theorem to find side b:
a^2 + b^2 = c^2
(2)^2 + b^2 = (2√5)^2
4 + b^2 = 20
b^2 = 20 - 4
b^2 = 16
Taking the square root of both sides:
b = √16
b = 4
Now we have all three sides of the triangle: a = 2, b = 4, and c = 2√5.
Next, we can calculate the trigonometric functions of the acute angles.
Let's define angle A as the acute angle adjacent to side a, and angle B as the acute angle adjacent to side b.
- Sine (sin) function:
sin(A) = opposite/hypotenuse = b/c
sin(A) = 4/(2√5)
To rationalize the denominator, multiply both the numerator and denominator by √5:
sin(A) = (4√5)/(2√5 * √5)
sin(A) = 4√5/(2 * 5)
sin(A) = (2/5)√5
Similarly, sin(B) = a/c = 2/(2√5) = (2/2√5) = 1/√5 = √5/5
- Cosine (cos) function:
cos(A) = adjacent/hypotenuse = a/c
cos(A) = 2/(2√5)
cos(A) = 1/√5 = √5/5
Similarly, cos(B) = b/c = 4/(2√5) = 2/√5 = (2√5)/5
- Tangent (tan) function:
tan(A) = opposite/adjacent = b/a
tan(A) = 4/2
tan(A) = 2
Similarly, tan(B) = a/b = 2/4 = 1/2
So, the values of the trigonometric functions of the acute angles of the right triangle ABC are:
sin(A) = (2/5)√5
sin(B) = √5/5
cos(A) = √5/5
cos(B) = (2√5)/5
tan(A) = 2
tan(B) = 1/2
To find the values of the trigonometric functions of the acute angles of the right triangle ABC, we need to use the definitions of these functions.
Let's label the acute angle opposite to side a as angle A, and the acute angle opposite to side b as angle B.
First, we can use the Pythagorean theorem to find the remaining side b:
a^2 + b^2 = c^2
Substituting the given values, we have:
2^2 + b^2 = (2√5)^2
4 + b^2 = 20
b^2 = 20 - 4
b^2 = 16
b = √16
b = 4
Now, let's find the values of the trigonometric functions for angle A:
1. Sine (sin): sin(A) = opposite/hypotenuse = a/c = 2/(2√5) = √5/5
2. Cosine (cos): cos(A) = adjacent/hypotenuse = b/c = 4/(2√5) = 2/√5 = (2/√5) * (√5/√5) = 2√5/5
3. Tangent (tan): tan(A) = opposite/adjacent = a/b = 2/4 = 1/2
Now, let's find the values of the trigonometric functions for angle B:
1. Sine (sin): sin(B) = opposite/hypotenuse = b/c = 4/(2√5) = 2/√5 = (2/√5) * (√5/√5) = 2√5/5
2. Cosine (cos): cos(B) = adjacent/hypotenuse = a/c = 2/(2√5) = √5/5
3. Tangent (tan): tan(B) = opposite/adjacent = b/a = 4/2 = 2
Therefore, the values of the trigonometric functions for the acute angles of the right triangle ABC are:
sin(A) = sin(B) = √5/5
cos(A) = cos(B) = 2√5/5
tan(A) = 1/2
tan(B) = 2
c^2 = 20
a^2 = 4
b^2 = 16
sinA = cosB = a/c
sinB = cosA = b/c
tanA = cotB = a/b