Two men p and q set off from a base camp r prospecting for o, p moves 40km on a bearing 205 degree and q moves 30km on a bearing of 060 degree a. Find the distance of q from p
b. The bearing of q from p​

(a) By cosine rule, /PQ/2=202+152−2×20×15×cos145°

= 400+225−600cos145°

= 625+600×0.8192

= 491.52+625

/PQ/2=1116.52⟹/PQ/=1116.52−−−−−−√=33.41

≊33km ( to the nearest whole number)

(b) By sine rule,

sin<QPR15=sin14533.41

sin<QPR=15sin14533.41

sin<QPR=15×0.573633.41=0.2575

<QPR=14.92°

∴ Bearing of Q from P = 25° + 14.92° = 39.92°.

≊40° (to nearest whole number)

Step-by-step explanation:

hope it helps

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step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200.   after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes           1800 - 6400 which gets you 11,600.