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Random trigonometry question help please.

Hi all,

Can anyone shed some light on how to go about solving this question?

(Question 7, for clarity).

Thank you!
Attachment-1.pdf285.4KB
Reply 1
Avatar for Zacken
Zacken
22
Original post by londoncricket
Hi all,

Can anyone shed some light on how to go about solving this question?

(Question 7, for clarity).

Thank you!


It's right angled, you should write down: a2+b2=c2a^2 + b^2 = c^2 immediately.

Furthermore, you can see that cos⁡θ=ac\cos \theta = \frac{a}{c} and tan⁡θ=ba\tan \theta = \frac{b}{a}.

Can you take it from here?
(edited 8 years ago)
What level maths is this? (GCSE, C1, C2?
Reply 3
Avatar for Zacken
Zacken
22
Original post by Oblivion99
What level maths is this? (GCSE, C1, C2?


Anybody doing GCSE should be able to do it.
Reply 4
Avatar for okey
okey
15
Original post by Zacken
Anybody doing GCSE should be able to do it.


I don't get how tan of theta is a/b i thought it would be b/a because b is the opposite
Reply 5
Avatar for Zacken
Zacken
22
Original post by okey
I don't get how tan of theta is a/b i thought it would be b/a because b is the opposite


Yep, that's correct. Thanks for pointing it out, must have gotten confused because of all the neck craning I was doing. :lol:
Original post by Zacken
It's right angled, you should write down: a2+b2=c2a^2 + b^2 = c^2 immediately.

Furthermore, you can see that cos⁡θ=ac\cos \theta = \frac{a}{c} and tan⁡θ=ba\tan \theta = \frac{b}{a}.

Can you take it from here?


Got it. Thanks!

Original post by Oblivion99
What level maths is this? (GCSE, C1, C2?


Additional Maths at GCSE level.
Reply 7
Avatar for Zacken
Zacken
22
Original post by Oblivion99
Solved it. Its actually quite easy, yes. As someone stated above, just take note of the trigonometric identities (i.e cos 0 = a/b tan = b/c) and then you should be able to take it from here :smile:


These are both wrong.
Original post by Zacken
These are both wrong.


Can you provide me a step to step method onto solving this Q. Yes, ive actually flopped it :tongue:
Reply 9
Avatar for Zacken
Zacken
22
Original post by Oblivion99
Can you provide me a step to step method onto solving this Q. Yes, ive actually flopped it :tongue:


Tell you what, I'll let the OP do it - it'll be good practice for him/her. :biggrin:


Original post by londoncricket
...


If he doesn't reply, then tag me in a bit and I will. :yep:
Reply 10
Avatar for TeeEm
TeeEm
19
Original post by Zacken
Anybody doing GCSE should be able to do it.


hmmm ...
I11.JPG38.1KB
Original post by Zacken
Tell you what, I'll let the OP do it - it'll be good practice for him/her. :biggrin:




If he doesn't reply, then tag me in a bit and I will. :yep:


Its been an hour so :smile:
Reply 12
Avatar for Zacken
Zacken
22
Original post by Oblivion99
Its been an hour so :smile:


If you look at the triangle and select the angle theta, then the adjacent over hypotenuse (i.e: cos) is: cos⁡θ=ac\cos \theta = \frac{a}{c}

The tangent is: tan⁡θba\displaystyle \tan \theta \frac{b}{a} (opposite over adjacent)

We also know it's a right angled triangle, so: a2+b2=c2a^2 + b^2 = c^2.

Squaring and taking the reciprocal of cosine: 1cos⁡2θ=caa2\frac{1}{\cos^2 \theta} = \frac{c^a}{a^2}

And also: 1+tan⁡2θ=1+b2a2=a2+b2a21 + \tan^2 \theta = 1 + \frac{b^2}{a^2} = \frac{a^2 + b^2}{a^2}

But since a2+b2=c2a^2 +b^2 = c^2 then, 1+tan⁡2θ=c2a2=1cos⁡2θ1 + \tan^2 \theta = \frac{c^2}{a^2} = \frac{1}{\cos^2 \theta} as required.

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