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Question 3C & 9A (1 Viewer)

Teresa_2004

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Could someone please tell me the answer that they got for these questions?
 

rckl

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for question 3, C:

(i)

biscally, u just need to use the COSINE RULE to find the disance of R from Q:

a^2 = b^2 + c^2 - 2bc(cos A)

Answer is: 26.14km

(ii)

answer is : 290
 

bevstarrunner

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errr...3 b i) RQ = 19.17m
ii) bearing = 282.36º

(NOTE: those were rough/quick re-estimates, they could be wrong...)
 

acmilan

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9(a)(i)

limiting sum
= a/(1-r)
= 1/(1+tan²x)
= 1/sec²x (since 1 +tan²x = sec²x)
= cos²x

9(a)(ii) limiting sum exists for -pi/4 < theta < pi/4
 

rckl

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bevstarrunner said:
errr...3 b i) RQ = 19.17m
ii) bearing = 282.36º

(NOTE: those were rough/quick re-estimates, they could be wrong...)
u making me crying now.. r u sure? :p
 

gerardk

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acmilan said:
9(a)(i)

limiting sum
= a/(1-r)
= 1/(1+tan²x)
= 1/sec²x (since 1 +tan²x = sec²x)
= cos²x

9(a)(ii) limiting sum exists for -pi/4 < theta < pi/4

acmilan do you mind explaining (ii) to me please..i still dont understand. thanks mate
 

rckl

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bevstarrunner said:
errr...3 b i) RQ = 19.17m
ii) bearing = 282.36º

(NOTE: those were rough/quick re-estimates, they could be wrong...)
u r correct....
 

bevstarrunner

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gerardk said:
acmilan do you mind explaining (ii) to me please..i still dont understand. thanks mate
limiting sum exists when |r| < 1
r = - tan ² x

- tan ² x < 1
tan x > - 1
x > - pi/4

tan ² x < 1
tan x < 1
x < pi/4

.'. - pi/4 < x < pi/4
 

bevstarrunner

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so did I, but that is the working should u have needed it

It was worth 2 marks, so i think you might have needed it
 

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