Angles, Degrees, and Arcs

**juangohan9** · August 7th 2006, 06:23 AM

Hey there, I've got a few questions.

I have no idea how to figure out these kind of word problems. I'm taking an online class and I've got no source of help but here. So, is there some sort of formula I must follow with these? Draw a diagram? What does the "bearing" of 42.26 do for me?

A fire is sighted due west of lookout A. The bearing of the fire from lookout B, 8.6 miles due south of A, is N 42°26'W. How far is the fire from B (to the nearest tenth of a mile)?

Oh, and this one is probably easy, just simple algebra and such, but I dont know what to do... sin A = cos 3A

Thanks in advance, Juan.

**Quick** · August 7th 2006, 06:35 AM

Originally Posted by juangohan9

Hey there, I've got a few questions.

I have no idea how to figure out these kind of word problems. I'm taking an online class and I've got no source of help but here. So, is there some sort of formula I must follow with these? Draw a diagram? What does the "bearing" of 42.26 do for me?

A fire is sighted due west of lookout A. The bearing of the fire from lookout B, 8.6 miles due south of A, is N 42°26'W. How far is the fire from B (to the nearest tenth of a mile)?

$42^{\circ}26'=42^{\circ}+\frac{26}{60}^{\circ}=<br /> \frac{<br /> 1260+13}{30}$

a right triangle is formed when you draw a line from A to B to the fire, the distance from the Fire to base B is the hypotenuse, the distances from A to B and A to the Fire are the bases of the right triangle. do you know secant?

$\sec\left(\frac{1273}{30}\right)=\frac{\overline{B F}}{\overline{AB}}$

therefore:

$\sec\left(\frac{1273}{30}\right)=\frac{\overline{B F}}{8.6}$

thus:

$8.6\sec\left(\frac{1273}{30}\right)=\overline{BF}$

$11.7\approx\overline{BF}$

so the distance is 11.7 miles.

note: $\sec A=\cos^{-1}A=\frac{hypotenuse}{adjacent}$

there is also another way of finding it, using tan and the pythagoreum theorum, would you like to see it?

**ThePerfectHacker** · August 7th 2006, 07:49 AM

Originally Posted by juangohan9

sin A = cos 3A

Thanks in advance, Juan.

Thus,
$\sin x - \cos 3x=0$
Use co-function identity,
$\sin x - \sin\left( \frac{\pi}{2} - 3x \right)=0$
---
For reference use formula,
$\sin x-\sin y=2\cos \frac{x+y}{2} \sin \frac{x-y}{2}$
---
Thus,
$2\cos \left(\frac{\pi}{4}-x\right)\sin \left( \frac{\pi}{4} +2x \right) =0$
Divide by two, thus,
$\cos \left(\frac{\pi}{4}-x\right)\sin \left( \frac{\pi}{4} +2x \right) =0$
Set each factor equal to zero thus,
$\left\{ \begin{array}{c} \cos \left(\frac{\pi}{4}-x\right)=0\\\sin \left( \frac{\pi}{4} +2x \right) =0$
From the first one you have,
$\frac{\pi}{4}-x=\frac{\pi}{2}+\pi k$
Thus,
$x=-\frac{\pi}{4}-\pi k$
To make your answer look cooler sine $-k$ is an integer too is does not matter if you write,
$x=-\frac{\pi}{4}+\pi k$
Since the number being constantly added and subtracted is $\pi$ it does not matter if you write,
$x=-\frac{\pi}{4}+\pi+\pi k$
Simplify,
$x=\frac{3\pi}{4}+\pi k$

For the second one you have,
$\sin \left(\frac{\pi}{4}+2x \right)=0$
Thus,
$\frac{\pi}{4}+2x=\pi k$
Thus,
$2x=-\frac{\pi}{4}+\pi k$
Since the number being constantly subtracted and added is $\pi$ it does not matter if,
$2x=\pi-\frac{\pi}{4}+\pi k$
Thus, simplify,
$2x=\frac{3\pi}{4}+\pi k$
Thus,
$x=\frac{3\pi}{8}+\frac{\pi}{2} k$

**Soroban** · August 7th 2006, 08:17 AM

Hello, Juan!

I'll explain $N\,42^o26'\,W$
. . The $N$ means we start by facing North.
. . Then $42^o26'\,W$ means we turn $42^o26'$ to the west.
And that is the direction indicated by $N\,42^o26'\,W$ .

Similarly, $S30^oE$ means: face South, turn 30° to the east.

A fire is sighted due west of lookout $A$ .
The bearing of the fire from lookout $B$ , 8.6 miles due south of $A$ , is $N\,42^o26'\,W$
How far is the fire from $B$ to the nearest tenth of a mile?

Code:

    F * - - - - - - - * A
        \             |
          \           |
            \         |
            d \       | 8.6
                \     |
                  \ θ |
                    \ |
                      * B

The fire is at $F$ . .Angle $\theta \,= \,42^o26' \,\approx \,42.4333^o$ . .Let $d = BF.$

In right triangle $FAB$ , we have: . $\cos\theta\,=\,\frac{8.6}{d}\quad\Rightarrow\quad d\,=\,\frac{8.6}{\cos\theta}$

Therefore: . $d \:= \:\frac{8.6}{\cos42.4333^o} \:= \:11.6521...\:\approx\:11.7$ miles.

Solve for $A:\;\;\sin A \:= \:\cos 3A$

Here's a back-door approach to this problem . . .

If $\sin A \,= \,\cos B$ , then $A$ and $B$ are complementary: . $A + B \,=\,90^o$

Look at this right triangle.

Code:

                  * B
           c   *  *
            *     * a
         *        *
    A *  *  *  *  * C
            b

We have: . $\sin A = \frac{a}{c}$ .and . $\cos B = \frac{a}{c}$

See? . . . Angles $A$ and $B$ are from the same right triangle.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Back to the problem: . $\sin A \:= \:\cos3A$

Since $A$ and $3A$ are complementary: . $A + 3A \:= \:90^o\quad\Rightarrow\quad 4A \:=\:90^o$

Therefore: . $\boxed{A\:=\:22.5^o}$

Note: TPHacker's solution is complete.

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