Hi, I'm really stuck on Q6 of this link - https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FMaths%2FA-level%2FPapers%2FAQA%2FAS-Paper-1%2FMA%2FSpecimen%20MA.pdf

I understand how Sina = 8/9 but at that point I tried to get a by using inverse sin. This gave me a= 62.73.... which i then put into tan to get 1.94... whi h was wrong and for some reason the solutions find two possible ones at first. where have I gone wrong?

I understand how Sina = 8/9 but at that point I tried to get a by using inverse sin. This gave me a= 62.73.... which i then put into tan to get 1.94... whi h was wrong and for some reason the solutions find two possible ones at first. where have I gone wrong?

Original post by mitostudent

Hi, I'm really stuck on Q6 of this link - https://www.physicsandmathstutor.com/pdf-pages/?pdf=https%3A%2F%2Fpmt.physicsandmathstutor.com%2Fdownload%2FMaths%2FA-level%2FPapers%2FAQA%2FAS-Paper-1%2FMA%2FSpecimen%20MA.pdf

I understand how Sina = 8/9 but at that point I tried to get a by using inverse sin. This gave me a= 62.73.... which i then put into tan to get 1.94... whi h was wrong and for some reason the solutions find two possible ones at first. where have I gone wrong?

I understand how Sina = 8/9 but at that point I tried to get a by using inverse sin. This gave me a= 62.73.... which i then put into tan to get 1.94... whi h was wrong and for some reason the solutions find two possible ones at first. where have I gone wrong?

It is clear from the diagram that alpha is an obtuse angle > 90. You need to know that there are two angles between 0-180 degrees that produce the same value of the sine function. In this case approx 62.7 and 180-62.7 = approx 117.3. HOWEVER The question asks for the EXACT value of tan alpha. The mark scheme you link literally explains how to do it. You cannot usually do that in an exam question by using inverse trig functions. Once you have an exact value for one trig ratio, you can work out exact values of the others by sketching a right angle triangle and using pythagoras. You should also be aware of what angles result in negative trig values either from a CAST diagram or a sketch of the graph of each function.

(edited 3 months ago)

Original post by gdunne42

It is clear from the diagram that alpha is an obtuse angle > 90. You need to know that there are two angles between 0-180 degrees that produce the same value of the sine function. In this case approx 62.7 and 180-62.7 = approx 117.3. HOWEVER The question asks for the EXACT value of tan alpha. The mark scheme you link literally explains how to do it. You cannot usually do that in an exam question by using inverse trig functions. Once you have an exact value for one trig ratio, you can work out exact values of the others by sketching a right angle triangle and using pythagoras. You should also be aware of what angles result in negative trig values either from a CAST diagram or a sketch of the graph of each function.

nice, thanks!

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