Answer:
(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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Answer:
.
Step-by-step explanation:
Given:
The figure of a triangle LMN.
P is the centroid of triangle LMN.
To find:
14. Find the value of PN if QN=30.
15. Find the value of PN if QN=9.
Solution:
We know that the centroid in the intersection of medians of a triangle and centroid divides each median in 2:1.
Since P is the centroid it means NQ is the median from vertex N. It means P divides the median NQ in 2:1. So, PN:PQ=2:1.
14. We have QN=30.




Therefore, the value of PN is 20 when QN=30.
15. We have QN=9.




Therefore, the value of PN is 6 when QN=9.
Answer:
A
15:20
Step-by-step explanation: