if u have time please ty this question which i am stuck on

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#1
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Go to

http://www.gcsemathspastpapers.com/i...p6j2000q16.htm
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ickle_katy
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what are u stuck on?
i did it and got
(A) 134m
(B) 4180 m(squared)
(C) 6160 m(squared)


but its a long time since i did gcse maths.

hope that helps
love Katy***
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ickle_katy
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oh, forgot to say, if u want any explanations of what i did, say and i can explain.


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Expression
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I'm going to do these one at a time so I don't confuse myself !

1) Construct a triangle, by joining P and R by a straight line:

We are looking at triangle PQR.



Using cosine rule:

c^2 = a^2 + b^2 - 2ab cos C

c^2 = 110^2 + 76^2 - ((2x110x76)cos 98)

c^2 = 12100 + 5776 - 16720 cos98

c^2 = 17876 - (-2326.974248)

c^2 = 20202.97425

c = 142.137167m

So to the nearest metre, c = PR = 142 m.

2) Area = 1/2 ab sin C

Area = (1/2 x 110 x 76)sin 98

Area = 4180 sin 98

Area = 4139.320527 m^2

So to the nearest square metre area PQR = 4139 m^2

3) For the whole quadrilateral, Total Area = Area PQR + Area PRS

Area PQR = 4139 m^2

Area PRS = 1/2 x r x p sin S

PRS = 1/2 x 132 x p sin 77

p is currently unknown, and is required.

However, given that we don't know angle P, we can't calculate side p using the sine rule.

However, because we know side r (132), side s (142), and angle S (77deg), we can find angle R using the sine rule, and substract the sum of angles R and S from 180degs to find angle P; after which side p, will be findable.

So sin R sin S
------- = --------
r s

==> sin R sin 77
-------- = ---------
132 142

So, sin R = 132sin 77 / 142 ==> R = 64.9 degs.

P = 180 - (S + R) = 180 - 141.9 ==> 38.1 degs.

Now,

p s
------ = ------
sin P sin S

===> p / sin 38.1 = 142 / sin 77
===> p = 142sin 38.1 / sin 77
===> p = 89.9 m (1 dp).

Now, Area Triangle PRS = 1/2 r p sin S
==> (1/2 x 132 x 89.9) sin 77 = 5781 m^2 (nearest metre)

So area of quadrilateral = 4139 + 5781 = 9920m^2
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