Answer to Question #177204 in Trigonometry for Rainbow_Sparkles

Question

From A, B lies 24 km away on a bearing of 057 and C lies 15km away on a bearing of 347 Find a)The distance between B and C. b)The bearing of B from C.

Answer 1

a).

If you lay the points on a circle with origin at A, you will get a triangle with an interior angle of 70 deg at A, which is the sum of the bearing of 57 deg between A and B, and 360 deg - 347 deg = 13 deg, the bearing between points A and C. Let 'a' be the side opposite to angle A at point A, you know the distance between A and B call that side c=24km opposite to angle C and side b=15km opposite to angle B.

Using the law of cosines:

$a^2=b^2+c^2-2bc(cosA\degree)$

$a^2=15^2+24^2-2(15)(24)cos70\degree$

$a^2=554.745$

$a=23.55km$

Thus, the distance between B and C is BC=23.55km.

b).

From the Law of Sines:

$\frac{23.55}{sin70\degree}=\frac{24}{sinC\degree}$

$sinC\degree=\frac{24*sin70\degree}{23.55}=\frac{24*(0.939692621)}{23.55}=0.957648531$

$C=sin^{-1}{(0.957648531)}= 73.265\degree$

The bearing of B from C is the angle formed by the line joining C and B and rotating about C.

By geometry this angle is :

$180\degree-(C\degree +13\degree)=180\degree-(73.265\degree+13\degree)=93.734\degree$

Thus, the bearing of B from C is 93.734 deg.