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AysviL [449]
3 years ago
5

One leg of a right triangle is 21 inches longer than the other leg, and the hypotenuse is 39 inches.

Mathematics
1 answer:
Valentin [98]3 years ago
4 0

Answer:

One leg - 15 inches

Second leg - 36 inches.

Step-by-step explanation:

One leg - x inches.

Second leg - (x + 21) inches.

Hypotenuse - 39 inches.

Using Pythagorean theorem

x² + (x + 21)² = 39²

x² + x² + 2*21x  + 21² = 39²

2x² + 42x - 39² + 21² = 0

2x² + 42x - 1080 = 0

x² + 21x - 540 = 0

(x - 15)(x + 36) = 0

x = 15, x = - 36

We can use only positive value.

One leg - x inches = 15 inches

Second leg - (x + 21) inches = 15+21=36 inches.

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\tan(\alpha)*\cot^2(\alpha)\\\\

============================================================

Explanation:

Recall that \tan(x) = \frac{\sin(x)}{\cos(x)} and \cot(x) = \frac{\cos(x)}{\sin(x)}. The connection between tangent and cotangent is simply involving the reciprocal

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\tan(\alpha)*\cot^2(\alpha)\\\\\\\frac{\sin(\alpha)}{\cos(\alpha)}*\left(\frac{\cos(\alpha)}{\sin(\alpha)}\right)^2\\\\\\\frac{\sin(\alpha)}{\cos(\alpha)}*\frac{\cos^2(\alpha)}{\sin^2(\alpha)}\\\\\\\frac{\sin(\alpha)*\cos^2(\alpha)}{\cos(\alpha)*\sin^2(\alpha)}\\\\\\\frac{\cos^2(\alpha)}{\cos(\alpha)*\sin(\alpha)}\\\\\\\frac{\cos(\alpha)}{\sin(\alpha)}\\\\

In the second to last step, a pair of sine terms cancel. In the last step, a pair of cosine terms cancel.

All of this shows why \tan(\alpha)*\cot^2(\alpha)\\\\ is identical to \frac{\cos(\alpha)}{\sin(\alpha)}\\\\

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