tan θ = tan 1 5 ∘ tan 2 5 ∘ tan 3 5 ∘
If θ is a positive acute angle satisfying the equation above, find the measure of θ in degrees.
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You could simplify your working by a bit. Hint: 2 5 ∘ + 3 5 ∘ = 6 0 ∘ .
Challenge Master:
With tan A tan ( 6 0 − A ) tan ( 6 0 + A ) = tan ( 3 A ) , set A = 2 5 ∘ .
tan 2 5 ∘ × tan 3 5 ∘ × tan 8 5 ∘ tan 2 5 ∘ × tan 3 5 ∘ × cot 5 ∘ tan 2 5 ∘ × tan 3 5 ∘ × tan 1 5 ∘ = = = tan 7 5 ∘ cot 1 5 ∘ tan 5 ∘
how come we get tan 15*tan 75=tan15 * tan45 * tan 75??? plzz.. I need to know it..!!
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What does tan 45 equal? I believe you can multiply by 1 all you like, no matter what form it comes in.
Where did you get that formula? I see how it is derived, but I've never seen it before, it would have been nice to know.
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tan 1 5 ∘ tan 2 5 ∘ tan 3 5 ∘
= tan 1 5 ∘ ⋅ tan 8 5 ∘ tan 2 5 ∘ tan 3 5 ∘ tan 8 5 ∘
= tan 1 5 ∘ ⋅ tan 8 5 ∘ tan 7 5 ∘
= tan 4 5 ∘ tan 8 5 ∘ tan 1 5 ∘ tan 4 5 ∘ tan 7 5 ∘
= tan 4 5 ∘ tan 8 5 ∘ tan 4 5 ∘ = tan 8 5 ∘ 1 = tan 5 ∘ ⟹ θ = 5 ∘
Formula used is :
tan x tan ( 6 0 ∘ − x ) tan ( 6 0 ∘ + x ) = tan 3 x