Bearing and Triangles

[IloveManUtd]

The diagram shows a point A which lies 9 km south of point B. The points C and D are both 6 km from B and the bearing of C from A is 036.3 degrees. The points A, C, D and E all lie on a straight line. Calculate angle ACB and the shortest distance from B to the line ACD.

In the second part of the question are they asking for a perpendicular line of B? Thanks.

 

[mmm4444bot]

In the second part of the question are they asking for a perpendicular line of B? Two points define a line, not one.

The shortest distance from B to the line AE is along the line perpendicular to AE passing through B.

In other words, pick a point between C and D, and call it F. Then, BF is shortest when it is perpendicular to CD.

 

[soroban]

Hello, IloveManUtd!

The diagram shows a point $\displaystyle A$ which lies 9 km south of point $\displaystyle B$.
The points $\displaystyle C$ and $\displaystyle D$ are both 6 km from $\displaystyle B$
and the bearing of $\displaystyle C$ from $\displaystyle A$ is 036.3 degrees.
The points $\displaystyle A , C, D, E$ are collinear.

Calculate: .(a) $\displaystyle \angle ACB$
. . . . . . . (b) the shortest distance from $\displaystyle B$ to the line $\displaystyle AE.$

    B o
      |* *                  E
      | *   *   6         *
      |  *     *        *
      |   *       *   *
      |  6 *        o
    9 |     *     *   D
      |      *  *
      |       o
      |36.3 *   C
      |   *
      | *
    A o

$\displaystyle \text{(a) Use the Law of Sines on }\Delta ACB.$

. . . $\displaystyle \frac{\sin(\angle ACB)}{9} \:=\:\frac{\sin36.3^o}{6} \quad\Rightarrow\quad \sin(\angle ACB) \:=\:\frac{9\sin36.3^o}{6} \:=\:0.888019768$

. . $\displaystyle \text{Hence: }\:\angle ACB \;=\;62.625457^O,\;117.374543^O$

. . $\displaystyle \text{Since }\angle ACB\text{ is obviously obtuse: }\:\angle ACB \;\approx\;117.4^o$



$\displaystyle \text{(b) Let }h = BF\text{ be the shortest distance from }B\text{ to }AE.$

. . $\displaystyle \text{Then: }\:BF \perp AE.$

$\displaystyle \text{In right triangle }BFA\!:\;\;\sin 36.3^o \:=\:\frac{h}{9}$

. . $\displaystyle \text{Therefore: }\;h \;=\;9\sin36.3^o \;=\;5.328118609 \;\approx\;5.3\text{ km.}$