# Question 3C & 9A (1 Viewer)

#### Teresa_2004

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

#### rckl

##### Member
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)

(ii)

#### bevstarrunner

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

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

#### acmilan

##### I'll stab ya
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

#### Teresa_2004

##### New Member
Thanks for the help!

#### rckl

##### Member
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? #### gerardk

##### New Member
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

##### Member
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

##### Maths Nut
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

#### Li0n

##### spiKu
oh wf did you have to show working
FFS