Pythagorean theorem A=8 B=? C=17

2 answers:

7 0

$\text{Solve for b:}\\\\\text{Pythagorean Theorem:}\\\\a^2+b^2=c^2\\\\\text{Plug in and solve:}\\\\8^2+b^2=17^2\\\\64+b^2=17^2\\\\64+b^2=289\\\\\text{Subtract 64 from both sides}\\\\b^2=225\\\\\text{Square root both sides}\\\\\sqrt{b^2}=\sqrt{225}\\\\\boxed{b=15}$

DiKsa [7]4 years ago

5 0

Answer:

b=15

Step-by-step explanation:

a^2+b^2=c^2

8^2+b^2=17^2

64+b^2=289

**subtract**

b^2=225

**square root 225**

b=15

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Verify the identity tan(???? − ????) = tan(????)−tan(????) / 1+tan (????) tan(????) for all (???? − ????) ≠ ???? / 2 + ????n.

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Answer:

See the proof below.

Step-by-step explanation:

For this case we need to proof the following identity:

$tan(x-y) = \frac{tan x - tan y}{1 + tan x tan y}$

So we need to begin with the definition of tan, we know that $tan x = \frac{sin x}{cos x}$ and we have this:

$tan (x-y) = \frac{sin(x-y)}{cos(x-y)}$ (1)

We also have the following identities:

$sin (a-b) = sin a cos b - sin b cos a$

$cos(a-b) = sin a sin b + cos a cos b$

And if we apply those identities into equation (1) we got:

$tan(x-y) = \frac{sin x cos y - sin y cos x}{sin x sin y + cos x cos y}$ (2)

We can divide numerator and denominator from expression (2) by $\frac{1}{cos x cos y}$ like this:

$tan(x-y) = \frac{\frac{sin x cos y}{cos x cos y} - \frac{sin y cos x}{cos x cos y}}{\frac{sin x sin y}{cos x cos y} + \frac{cos x cos y}{cos x cos y}}$