A surveyor leaves her base camp and drives 43km on a bearing of 032°.She then drives 28km on a bearing of 154°.How far is she then from her base camp and what is her bearing from it

36.84km ; 252.13°

Step-by-step explanation:

Using the cosine rule :

a² = b² + c² - 2bcCosA

a² = 28² + 43² - (2*28*43)*Cos58

a² = 2633 - 1276.0455

a² = 1356.9544

a = sqrt(1356.9544)

a = 36.8368

a = 36.84 km

For the bearing,

Obtain the angle at B using the Sine rule

a/SinA = b/SinB

36.84/sin58 = 28/sinB

36.84*SinB = 28 * Sin58

SinB = 23.745346 / 36.84

SinB = 0.6445533

B = sin^-1(0.6445533)

B = 40.13°

180 + (32 + 40.13)

180 - 72.13

= 252.13°

answer: no idea

step-by-step explanation: i just want points

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)