# Can anyone help me with this question?Watch

#1
Hello. Could anyone please help me with this question from Higher Maths ? The answer guide just says "Proof" which is of no help! Thank you
0
3 weeks ago
#2
pythagorus of the outer triangle will get you how many units C to B, SQROOT2 is (approx) 1.4142, SQROOT6 is (approx) 2.4495,
then you can use the old Cos = opposite/hypotenuse (is that right?, S=O/H, T= O/A, C = A/H) , nope it's Cos = adjacent/hypotenuse

other methods are available
0
#3
Hi thanks for replying.I got that far - I used Pyth to work out that BC was 2 so CD =1. I've tried using addition formulae for cos but can't find a way past not knowing the value of sin theta....
0
3 weeks ago
#4
Okay you have CD as 1

You also can use pythagoras to work out opposite to the right angle which is root 3 as (root 2)^ + 1^ = 3 obviously

So now you can use cos A = b^ + c^ - a^ / 2bc for the non right angled triangle
b^ = 3
c^ = 6 or vice versa doesn't really matter which one is b or c
a^ = 1
so now cos cos A = 3 + 6 - 1 / 2 x rt3 x rt6
cos A = 8 / 2 x (3rt x rt6) then get rid of the 2 and simplify the surd
cos A = 4 / 3rt2 as required
0
#5
Great, thank you very much. This question has been bugging me for ages!! Cosine rule - so obvious - I feel really daft now! I got hung up on thinking that as it was a Higher question it had to be a trig double angle or addition formula question!! :-(
(Original post by fucthesqa)
Okay you have CD as 1

You also can use pythagoras to work out opposite to the right angle which is root 3 as (root 2)^ + 1^ = 3 obviously

So now you can use cos A = b^ + c^ - a^ / 2bc for the non right angled triangle
b^ = 3
c^ = 6 or vice versa doesn't really matter which one is b or c
a^ = 1
so now cos cos A = 3 + 6 - 1 / 2 x rt3 x rt6
cos A = 8 / 2 x (3rt x rt6) then get rid of the 2 and simplify the surd
cos A = 4 / 3rt2 as required
0
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