maths bearing question

Consider the △ABC,
length of AB = 2km, length of BC = 3km
∠ABC = (90°- 49°) + (90°-53°) = 78°

By cosine formula:
AC² = AB² + BC² - 2(AB)(BC)cos∠ABC
AC² = 2² + 3² - 2(2)(3)cos78°
AC² = 13 - 12(0.208)
AC² = 10.504
AC = 3.24 km (corrected to 2 decimal places)

By sine formula:
BC/sin∠CAB = AC/sin∠ABC
3/sin∠CAB = 3.24/sin78°
sin∠CAB = 0.9056
∠CAB = 64.90°(corrected to 2 decimal places)
∠CAB = (∠ between C and the north direction of A) + (∠ between B and the north direction of A)
64.90° = (∠ between C and the north direction of A) + 49°
∠ between C and the north direction of A = 15.9°

a. the distance of C west of A
= AC sin(∠ between C and the north direction of A)
= 3.24 sin15.9°
= 0.89 km (corrected to 2 decimal places)

b. the distance of C north of A
= AC cos(∠ between C and the north direction of A)
= 3.24 cos15.9°
= 3.12 km (corrected to 2 decimal places)


2007-04-15 10:04:12 補充：
For diagram and to calculate without using cosine and sin formula,due to limited words spaces, please see the following link for reference :http://www.geocities.com/coidatabase/MathBearing.doc