### 2010 credit question 10

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• #### Message 1.

Posted by U15235539 (U15235539) on Tuesday, 1st May 2012

Really struggling on the 2010 credit question 10 , I get so far and then can't do it I was wondering if you could show me please

[Title edited by moderator]

• #### Message 2

, in reply to message 1.

Posted by Beth (U15228510) on Tuesday, 1st May 2012

hiya, this is a mix of trig, the use of our quadrants, and some common sense

we know two sides, and the area, so we can substitute this knowledge into the area of a triangle

Area = 1/2 abSinC (for the half, i will type it as 0.5 because its easier, it doesnt matter if you make a fraction in your calculator or make it a decimal)

plug in what we know

12 = 0.5 x 5 x 6 x SinC

swap the sides so that the SinC is on the left

0.5 x 5 x 6 x SinC = 12

work out and simplify (0.5 x 5 x 6 = 15)

15SinC = 12
SinC = 12/15 (as the 15 was multiplying with the SinC, thats why its easier that its on the left, so that you dnt miss a step)
C = sin ̄ ¹(0.8) <<<< 12 divided by 15
C = 53.13

now, if you look back to the question, it states that "angle QPR is obtuse"
you can see that the angle found is NOT obtuse, but this is where our quadrant comes in. (i will solve this, if you dont understand the quadrant, just reply and i'll explain)

since its sine, we go to the sine quadrant, therefore:

180 - 53.13
= 126.87
= 126.9 (the answer showed it to 1 decimal place, so remember to round!)

you can see, that this is obtuse.

hope this helped, if you have any questions on the quadrants, feel free to ask..

happy studying and best wishes in your exams, you dont need luck to pass x

• #### Message 3

, in reply to message 2.

Posted by Maths Teacher BBC (U15178780) on Wednesday, 2nd May 2012

Great explanation!

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