4 years ago

Another geometry question! Please help! ASAP!

1 answer:

7 0

$D = \sqrt{( y_{2}-y_{1})^{2} +(x_{2}-x_{1})^{2}} = \sqrt{(5-(-6))^{2}+(-9-7)^{2}} =

= \sqrt{11^{2}+16^{2}} = \sqrt{377} = 19

Distance = 19$

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Look at the triangle show on the right. The Pythagorean Theorem states that the sum of the squares of the legs of a right triang

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Answer:

$cos^2\theta + sin^2\theta = 1$

Step-by-step explanation:

Given

$(\frac{b}{r})^2 + (\frac{a}{r})^2$

Required

Use the expression to prove a trigonometry identity

The given expression is not complete until it is written as:

$(\frac{b}{r})^2 + (\frac{a}{r})^2 = (\frac{r}{r})^2$

Going by the Pythagoras theorem, we can assume the following.

a = Opposite
b = Adjacent
r = Hypothenuse

So, we have:

$Sin\theta = \frac{a}{r}$

$Cos\theta = \frac{b}{r}$

Having said that:

The expression can be further simplified as:

$(\frac{b}{r})^2 + (\frac{a}{r})^2 = 1$

Substitute values for sin and cos

$(\frac{b}{r})^2 + (\frac{a}{r})^2 = 1$ becomes

$cos^2\theta + sin^2\theta = 1$

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Exact form:
40/9
mixed form:

4 4/9

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3 years ago

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