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Help with an Obtuse triangle

We've been told that the area is 12.5cm2 and that C is an Obtuse angle:


triangle problem.png

I used 0.5(abSin(C)) = Area

So 12.5 = 0.5(5*6*Sin(C))

Sin(C) = 2(12.5)/30

C = Sin-1(5/6)

C = 56.44...

But this can't be true as it's obtuse, am I missing something here? AQA - C2
Reply 1
Avatar for 2710
2710
18
Original post by MitchHeath
We've been told that the area is 12.5cm2 and that C is an Obtuse angle:


triangle problem.png

I used 0.5(abSin(C)) = Area

So 12.5 = 0.5(5*6*Sin(C))

Sin(C) = 2(12.5)/30

C = Sin-1(5/6)

C = 56.44...

But this can't be true as it's obtuse, am I missing something here? AQA - C2


C = Sin-1(5/6)

So we want an obtuse angle. Sin inverse of 5/6 is 56.44, but it is also an infinite number of other answers. Look at the sin graph and find your correct answer
Reply 2
Avatar for MitchHeath
MitchHeath
OP
Original post by 2710
C = Sin-1(5/6)

So we want an obtuse angle. Sin inverse of 5/6 is 56.44, but it is also an infinite number of other answers. Look at the sin graph and find your correct answer


Thank you for replying, but I don't understand what you mean; in terms of applying it to a Sin graph. Could you possibly expand? Sorry!
Reply 3
Avatar for LaurenB16
LaurenB16
Not too good at explaining - but as the sin graph goes up and down there will be more than one angle for sin@ = 5/6 - so when u get the 56 degrees, you hav to do 180-56 to find the obtuse angle


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Reply 4
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LaurenB16
ImageUploadedByStudent Room1367482439.339319.jpg between 0-180 degrees there r two places on the sin graphs where the line is at height 5/6


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Reply 5
Avatar for 2710
2710
18
Original post by MitchHeath
Thank you for replying, but I don't understand what you mean; in terms of applying it to a Sin graph. Could you possibly expand? Sorry!


As Lauren has explained above. What does sin inverse 5/6 mean? It means, for what angle 'x' does sin(x) give me 5/6? Now look at the sine graph between 0 and 360 degrees. You know one answer, you should be able to find the other answer by symmetry.
Reply 6
Avatar for TenOfThem
TenOfThem
18
Original post by MitchHeath
Thank you for replying, but I don't understand what you mean; in terms of applying it to a Sin graph. Could you possibly expand? Sorry!


When solving Trig equations and finding the "other solutions" do you use

Graphs
Equations
CAST

Whichever you use ... find another value that has the same Sine

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