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Cosine Rule

co.jpg

ABCD is a quadrilateral.
Angle BAC = 37 degrees.
Work out the area of ABCD.

I started by splitting it up into two triangles by drawing a line between AC and adding in the angle it gives you. I tried to use cosine fore the bottom triangle but ending up using Cos 106 in the formula which gives a negative number. I have been trying this for ages now and my exam is tommorow so I wanna get it out the way as soon as possible. Would be very grateful is someone could explain this to me, thanks.
Reply 1
Although cos(106) is negative the cosine rule

a2=b2+c22bcCos(A)a^2 = b^2 + c^2 - 2bcCos(A)

will still be positive since it is minus a minus which is positive. So find the longest side as you have split the triangles and use

Area=12abSin(C) Area = \frac{1}{2}abSin(C)

for the 2 triangles and add the areas together. If this still doesn't work I will give a more detail solution :smile:
(edited 11 years ago)
Reply 2
Original post by Dungey95


I tried to use cosine fore the bottom triangle but ending up using Cos 106 in the formula which gives a negative number.




Gives a negative number where?

This is correct ... the correct method and the correct value

It gives AC
Reply 3
Okay I realised my stupid mistake now but I did this, added the two areas together and got 4420 but in the book the answer is 4210. Is angle DCA also 37 as they are alternate angles?

I did 0.5x113x76xsin37 for the bottom triangle and 0.5x113x54x37 for the top triangle.
Reply 4
Original post by Dungey95
Okay I realised my stupid mistake now but I did this, added the two areas together and got 4420 but in the book the answer is 4210. Is angle DCA also 37 as they are alternate angles?

I did 0.5x113x76xsin37 for the bottom triangle and 0.5x113x54x37 for the top triangle.


They aren't alternate angles, but you have enough to find the area of the bottom triangle without the angle DAC because you have the bottom angle and the sides next to it.

So it would be

12(65)(76)sin(106)+12(113)(54)Sin(37)=4210m2\frac{1}{2}(65)(76)sin(106) + \frac{1}{2}(113)(54)Sin(37) = 4210m^2 to 4 sig fig

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