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
...
Here, let the age of student be x and teacher be 4x after 28 years, the teacher is double the age of student. So the age of teacher will be 4x+28 and student will be 2(x+28)
So,
4x+28=2(x+28)
4x+28=2x+56
4x-2x=56-28
2x=28
x=14
Therefore the age of the teacher in present will be
=4x+28
=4*14+28
=84
And age of student will be:
=x+28
=14+28
=42
Hope this helps
Answer:
no,yes,no,yes,yes
Step-by-step explanation:
Hello,
x=1/2(141-25)
==>x=116/2
==>x=58°
Answer:
The GCF of 32 and 56 are 8.
Step-by-step explanation:
Answer above! Enjoy! :)
