Let △ABC be a right triangle with m∠C = 90°. Given tan ∠A = 0.25, find tan ∠B.

Rule:

If A+B = 90, then tan(A) = 1/( tan(B) ) and tan(B) = 1/( tan(A) )

Since C = 90, this must mean the other two angles A and B are acute and complementary. So A+B = 90 must be the case. This allows us to use that rule above.

So,

tan(B) = 1/( tan(A) )

tan(B) = 1/( 0.25 )

tan(B) = 4

Here's another way to think about it.

If tan(A) = 0.25, then we could have BC = 1 and AC = 4 as the opposite and adjacent sides respectively. That leads to tan(A) = opposite/adjacent = BC/AC = 1/4 = 0.25

When computing tan(B), the BC and AC sides swap roles in terms of opposite and adjacent,

tan(B) = opposite/adjacent = AC/BC = 4/1 = 4.

For any of these cases, we don't involve the hypotenuse AB.

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step-by-step explanation: