The hypotenuse of a right triangle is 6 inches. if one of the legs is 2 inches, find the degree measure of each angle

1 answer:

5 0

A and b are the 2 smaller sides, and c = hypotenus; only for right triangles,
Pythagorean Theorem states that:
${c}^{2} = {a}^{2} + {b}^{2}$
given c = 6 in., a = 2 in.
${c}^{2} = {a}^{2} + {b}^{2} \\ {6}^{2} = {2}^{2} + {b}^{2} \\ {b}^{2} = {6}^{2} - {2}^{2} = 36 - 4 \\ \sqrt{ {b}^{2} } = \sqrt{36 - 4} = \sqrt{32}$
$b = \sqrt{32} = \sqrt{16} \times \sqrt{2} = 4 \sqrt{2} \\ b = 5.66 \: inches$
So the three side are 2, 6, and 5.66 inches

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poizon [28]

Answer:

See below

Step-by-step explanation:

It has something to do with the Weierstrass substitution, where we have

$\int\, f(\sin(x), \cos(x))dx = \int\, \dfrac{2}{1+t^2}f\left(\dfrac{2t}{1+t^2}, \dfrac{1-t^2}{1+t^2} \right)dt$

First, consider the double angle formula for tangent:

$\tan(2x)= \dfrac{2\tan(x)}{1-\tan^2(x)}$

Therefore,

$\tan\left(2 \cdot\dfrac{x}{2}\right)= \dfrac{2\tan(x/2)}{1-\tan^2(x/2)} = \tan(x)=\dfrac{2t}{1-t^2}$

Once the double angle identity for sine is

$\sin(2x)= \dfrac{2\tan(x)}{1+\tan^2(x)}$

we know $\sin(x)=\dfrac{2t}{1+t^2}$ , but sure, we can derive this formula considering the double angle identity

$\sin(x)= 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)$

Recall

$\sin \arctan t = \dfrac{t}{\sqrt{1 + t^2}} \text{ and } \cos \arctan t = \dfrac{1}{\sqrt{1 + t^2}}$

Thus,

$\sin(x)= 2 \left(\dfrac{t}{\sqrt{1 + t^2}}\right) \left(\dfrac{1}{\sqrt{1 + t^2}}\right) = \dfrac{2t}{1 + t^2}$

Similarly for cosine, consider the double angle identity

Thus,

$\cos(x)= \left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2- \left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2 = \dfrac{1}{t^2+1}-\dfrac{t^2}{t^2+1} =\dfrac{1-t^2}{1+t^2}$