Answer:
√(4/5)
Step-by-step explanation:
First, let's use reflection property to find tan θ.
tan(-θ) = 1/2
-tan θ = 1/2
tan θ = -1/2
Since tan θ < 0 and sec θ > 0, θ must be in the fourth quadrant.
Now let's look at the problem we need to solve:
sin(5π/2 + θ)
Use angle sum formula:
sin(5π/2) cos θ + sin θ cos(5π/2)
Sine and cosine have periods of 2π, so:
sin(π/2) cos θ + sin θ cos(π/2)
Evaluate:
(1) cos θ + sin θ (0)
cos θ
We need to write this in terms of tan θ. We can use Pythagorean identity:
1 + tan² θ = sec² θ
1 + tan² θ = (1 / cos θ)²
±√(1 + tan² θ) = 1 / cos θ
cos θ = ±1 / √(1 + tan² θ)
Plugging in:
cos θ = ±1 / √(1 + (-1/2)²)
cos θ = ±1 / √(1 + 1/4)
cos θ = ±1 / √(5/4)
cos θ = ±√(4/5)
Since θ is in the fourth quadrant, cos θ > 0. So:
cos θ = √(4/5)
Or, written in proper form:
cos θ = (2√5) / 5
Incorrect info please upload a correct photo of the problem
Answer:
B .-12 and -12
Step-by-step explanation:
it is because when we multiply two -digit the answer is in +
so like in the same -12 * -12=144
And when we add the digit we get -24
Hence, it is proved
Answer:
outliar is23
Step-by-step explanation:
Answer:
The distance between the two airplanes (to the nearest mile) is 1058 miles.
Step-by-step explanation:
An airplane A is at a location 800 miles due west of city X. So AX = 800 miles.
Another airplane is at a distance of 1,200 miles southwest of city X. So BX = 1200 miles.
The angle at city X created by the paths of the two planes moving away from city X measures 60°. So angle ∠AXB = 60°.
In triangle ΔAXB, AX = 800 miles, BX = 1200 miles, ∠AXB = 60°.
Using law of cosines:-
AB² = AX² + BX² - 2 * AX * BX * cos(∠AXB).
AB² = 800² + 1200² - 2 * 800 * 1200 * cos(60°).
AB² = 640000 + 1440000 - 2 * 960000 * 1/2
AB² = 2080000 - 960000
AB² = 1120000
AB = √(1120000) = 1058.300524
Hence, the distance between the two airplanes (to the nearest mile) is 1058 miles.
