What angles could we find sine and cosine for using information for π/4 and π/6?

1 answer:

7 0

Had to do a bit of guesswork to figure out what you're asking here. If you're given the 2 angles pi/4 and pi/6, you can find any trig function of either angle.

The sine of pi/4 is 1 / sqrt(2); the cosine is the same.

The cosine of pi/6 is sqrt(3) / 2; the sine of pi/6 is 1/2.

If this isn't quite what you were looking for, try paraphrasing your question to make it clearer.

You might be interested in

ONLY ANSWER IF U KNOW! Find the limit of the function algebraically.

blsea [12.9K]

Answer:

lim [(x(2) + 3) × 1/ x(4) ] = 3 × ( + ○○) = + ○○

8 0

3 years ago

What is the solution to 2x-8<12 ? A. X<2 B. X<8 C. X<10 D. X<40

Maurinko [17]

2x - 8 < 12
2x < 12 + 8
2x < 20
x < 20/2
x < 10

3 0

3 years ago

12.20x=372.10 Pls write exactly how to do it not just answer

ANEK [815]

To find x, you divide both sides of the equation by 12.20. This would look like this:
12.20x/12.20 =372.10/12.20 . Then, the answer you get is 30.5, so. X=30.5

6 0

3 years ago

Read 2 more answers

Solve this for Brainliest! See the attachment [Level - GCSE]

madreJ [45]

Answer:

$x = 3 + 4\sqrt{3}$

Step-by-step explanation:

OPQ is a right angle triangle.

Using Pythagoras

$x^{2} + (x+5)^{2} = (x+8)^{2}\\x^{2} + x^{2} + 10x +25 = x^{2} + 16x + 64\\x^{2} + x^{2} - x^{2} + 10x - 16x + 25 - 64 = 0\\x^{2} - 6x -39 = 0$

Using quadratic formula

$x = \frac{-b \± \sqrt{b^{2}-4ac}}{2a}\\x = \frac{-(-6) \± \sqrt{(-6)^{2}-4(1)(-39)}}{2(1)}\\x = \frac{6}{2} \± \frac{\sqrt{192}}{2}\\x = 3 \± \frac{8\sqrt{3}}{2}\\x = 3 \± 4\sqrt{3}$