# Thread: Trig angle problem

1. ## Trig angle problem

A woman on an island wishes to reach point R on a straight shore on the mainland from a point P on the island. The point P is 9km from the shore and 15km from the point R. If the woman rows a boat at a rate of 3 km/h to a point Q on land, then walks the rest of the way at a rate of 5km/h, express the total time it takes the woman to reach point R as a function of the indicated angle $\Theta$

2. So this is what I've gotten so far

sin(theta) = 9/PQ so PQ = 9/sin(theta).

Call the point at the right angle S.

RS = sqrt(15^2 + 9^2) = sqrt(225 + 81) = sqrt(306) = 17.5

9/QS = tan(theta)

QS = 9/tan(theta)

QR = 17.5 - 9/tan(theta)
PQ = 9/sin(theta)

she travels 3km/h on PQ and 5 on QR. So it's PQ/3 + QR/5.

Is there a way to find out the lengths of PQ or QR or is what I have the answer?

3. use s=vt to find t.find time taken to travel from P to Q and then add it to the time taken to travel from Q to R. You will get the answer in terms of theta. The lengths PQ and QR can only be found if the angle is given.

4. Hello, OVechkin8!

A woman on an island wishes to reach point $R$ on a straight shore
on the mainland from a point $P$ on the island.
The point $P$ is 9 km from the shore and 15k m from the point $R.$
If the woman rows a boat at a rate of 3 km/h to a point $Q$ on land,
then walks the rest of the way at a rate of 5 km/h,
express the total time it takes to reach point $R$ as a function of the indicated angle $\theta.$
Code:
      : - - - - - 12  - - - - :
S   x   Q      12-x     R
o - - - o - - - - - - - o
|      *             *
|     *           *
|    *         *
9 |   *       * 15
|  *     *
|θ*   *
|* *
o
P
Let $\theta = \angle SPQ$

In right triangle $PSR\!:\;PS = 9,\;PR = 15$
Pythagorus says: . $SR = 12$

Let: $x = SQ$, then $QR = 12-x$

In right triangle $PSQ\!:\;\;\cos\theta \,=\,\frac{9}{PQ} \quad\Rightarrow\quad PQ \,=\,\frac{9}{\cos\theta} \,=\,9\sec\theta$

She rows $9\sec\theta$ km at 3 km/hr.

. . This will take her: . $\frac{9\sec\theta}{3}\:=\:3\sec\theta$ hours.

In right triangle $PSQ\!:\;\;\tan\theta \,=\,\frac{x}{9} \quad\Rightarrow\quad x \,=\,9\tan\theta$
Hence: . $QR \:=\:12-x \:=\:12 - 9\tan\theta$

She walks $(12 - 9\tan\theta)$ km at 5 km/hr.

. . This will take her: . $\frac{12-9\tan\theta}{5}$ hours.

Therefore, her total time is: . $T \;=\;3\sec\theta + \frac{12-9\tan\theta}{5}$ hours.