# Finding The Angle In Radians

• June 17th 2009, 10:36 PM
AlphaRock
Finding The Angle In Radians
1. In a circle with radius 12 cm, an arc of length 20 cm subtends a central angle of theta. Determine the measure of theta in radians.

Answer: 1.67 (Rounded to two decimals)

My work:
(theta in radians) x Radius = arclength
(theta in radians) x 12 = 20
theta in radians = 20/12
Unsure how to continue...

2. Solve: 7tan(x) = -3 0 <= x <= 2pi

(Rounded to two decimals)

My work:
tan(x) = -3/7

3. For the function f(x) = 3sin(bx) + d, where b and d are positive constants, determine an expression for the smallest positive value of x that produces the maximum value of f(x)

How and Why?
• June 17th 2009, 10:43 PM
yeongil
Quote:

Originally Posted by AlphaRock
1. In a circle with radius 12 cm, an arc of length 20 cm subtends a central angle of theta. Determine the measure of theta in radians.

Answer: 1.67 (Rounded to two decimals)

My work:
(theta in radians) x Radius = arclength
(theta in radians) x 12 = 20
theta in radians = 20/12
Unsure how to continue...

$\frac{20}{12} = \frac{5}{3} \approx 1.67$

Quote:

2. Solve: 7tan(x) = -3 0 <= x <= 2pi

(Rounded to two decimals)

My work:
tan(x) = -3/7
$x = \tan^{-1}\left(-\frac{3}{7}\right)$
$x \approx -0.40$

You need answers in $0 \le x \le 2\pi$, but when you take the inverse tangent, you get an answer in $-\pi/2 \le x \le \pi/2$ (because the latter is the range of the inverse tangent). Since our answer is negative, add $\pi$ to it to get $x \approx 2.74$. Then add $\pi$ to it again to get $x \approx 5.88$. Don't add $\pi$ a third time, because you'll then get an answer greater than $2\pi$.

01
• June 17th 2009, 10:49 PM
yeongil
Quote:

Originally Posted by AlphaRock
3. For the function f(x) = 3sin(bx) + d, where b and d are positive constants, determine an expression for the smallest positive value of x that produces the maximum value of f(x)

How and Why?

If you look at the basic sine function
$y = \sin x$,
the maximum value (for y) is 1, so the smallest positive value of x that gives the maximum value is
$x = \frac{\pi}{2}$.

Now, looking at your function
$f(x) = 3\sin (bx) + d$,
since there is no reflection on the x-axis (ie. no negative in front of the 3), and since b and d are positive, the smallest positive value of x that gives the maximum value is when what's inside the sine is $\frac{\pi}{2}$, or
$bx = \frac{\pi}{2}$.
Solve for x to get
$x = \frac{\pi}{2b}$.

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