Trigonometry problem (multiple choice)

[ZyzzBrah]

1. The problem statement, all variables and given/known data
A car moves from point A northward for 35 minutes, then eastward for 1 hour, then changed course N35W. How far is the car directly from A after 2 hours if it is traveling at a uniform speed of 45kph.
a. 53.12km
b. 54.22km
c. 55.12km
d. 56.36km


2. Relevant equations
google law of sine and cosine




3. The attempt at a solution
Let
AB - north
BC - eastward
CD - N35W at (t = 2hours - 35minutes - 1hours = 0.4167hrs)

Required DA = ?

AB = (35/60)*45kph = 26.25km
BC = (45km)
CD = 0.4167(45kph) = 18.75km

Now i drew a line from point D to B and apply the law of cosine
DB^2 = CD^2 + BC^2 - 2(CD)(BC)cos(90-35)
DB = 37.532km

through sine law (getting angle DBC)
sin55/37.532 = sin(<DBC)/18,75
<DBC = 24.1563

<DBA = 24.15 + 90
<DBA = 114.1563

law of cosine(from point D to A )
DA ^2 = (DB)^2 + (BA)^2 - 2(DB)(BA)cos(114.1563)
DA = 53.893km

not in the choices
help....

 

[soroban]

Hello, ZyzzBrah!

I solved it with "simple" trig and Pythagorus.

A car moves from point A northward for 35 minutes,
then eastward for 1 hour, then changed course N35W.
How far is the car directly from A after 2 hours if it is traveling at a uniform speed of 45 kph?

. . (a) 53.12 km . . (b) 54.22 km . . (c) 55.12 km . . (d) 56.36 km . . Really bad choices!

      :   34.25   D 10.75 :
    P + - - - - - * - - - + Q
      :            \      :
      :             \     :
      :              \    : 15.36
15.36 :         18.75 \   :
      :                \35:
      :                 \ :
      :                  \:
    B * - - - - - - - - - * C
      |         45
26.25 |
      |
    A *

The car drove from A to B for 35 minutes at 45 kph.
. . \(\displaystyle AB = 26.25\)

Then it drove from B to C for an hour at 45 kph.
. . \(\displaystyle BC = 45\)

Then it drove from C to D for 25 minutes at 45 kph.
. . \(\displaystyle CD = 18.75\)

Draw line segment \(\displaystyle AD.\)


In right triangle \(\displaystyle CQD:\:\cos35^o \:=\:\dfrac{CQ}{CD} \quad\Rightarrow\quad CQ = CD\cos35^o \)

. . Hence: .\(\displaystyle CQ = 18.75\cos35^o \:\approx\:15.36 \:=\:BP\)


In right triangle \(\displaystyle CQD:\:\sin35^o \:=\:\dfrac{DQ}{CD} \quad\Rightarrow\quad DQ \:=\:CD\sin35^o\)

. . Hence: .\(\displaystyle DQ \:=\:18.75\sin35^o \:\approx\:10.75\)

Then: .\(\displaystyle PD \:=\:45 - 10.75 \:=\:34.25\)


In right triangle \(\displaystyle DPA\), we have two legs: .\(\displaystyle PD = 34.25,\;PA = 15.36 + 26.25 \,=\,41.61\)

Therefore: .\(\displaystyle AD \;=\;\sqrt{34.25^2 + 41.61^2} \;=\;\sqrt{2904.4546} \;=\;53.89299212\)


Our answer is closest to choice (b)
. . but we should not be forced to make that judgment.