Trigonometry Question Help :L (1 Viewer)

kevinsta · Jun 5, 2014

1)i)Given A,B and C are angles of a triangle and:
TanA/5 = TanB/6 = TanC/7 = k, show that k= Square-root of 3/35

ii) Hence calculate the smallest of the angles to the nearest minute.

I'm kind of lost in this question, any help would be great thanks!

aDimitri · Jun 5, 2014

kevinsta said:
1)i)Given A,B and C are angles of a triangle and:
TanA/5 = TanB/6 = TanC/7 = k, show that k= Square-root of 3/35

ii) Hence calculate the smallest of the angles to the nearest minute.

I'm kind of lost in this question, any help would be great thanks!

alright so part i was actually pretty tough

$\tan{\frac{A}{5}} = k \\ A = 5\arctan{k} \\ B = 6\arctan{k} \\ C = 7\arctan{k} \\ A+B+C = (5+6+7)\arctan{k} \\ \text{Now since A B and C are the angles of a triangle, } A+B+C = \pi \\ k = \tan{\frac{\pi}{18}} \\ \text{If you evaluate that, you WILL get }\sqrt{\frac{3}{35}} \text { but that's just grinding so I'll leave that to you if you want to do it.}$

braintic · Jun 5, 2014

Solution: View attachment Q.pdf

aDimitri · Jun 5, 2014

i realised that evaluating tan(pi/18) is way out of the scope of the extension course but seriously, i thought the denominators were in the brackets. This makes such a huge difference.....

Squar3root · Jun 6, 2014

kevinsta said:
1)i)Given A,B and C are angles of a triangle and:
TanA/5 = TanB/6 = TanC/7 = k, show that k= Square-root of 3/35

ii) Hence calculate the smallest of the angles to the nearest minute.

I'm kind of lost in this question, any help would be great thanks!

huehuehue

Trigonometry Question Help :L (1 Viewer)

kevinsta

Member

aDimitri

i'm the cook

braintic

Well-Known Member

aDimitri

i'm the cook

Squar3root

realest nigga

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