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3 years ago

AB is a diameter of a circle, center O. C is a point on the circumference of the circle, such that

Mathematics

1 answer:

Solnce55 [7]3 years ago

7 0

Given:

AB is the diameter of a circle.

m∠CAB = 26°

To find:

The measure of m∠CBA.

Solution:

Angle formed in the diameter of a circle is always 90°.

⇒ m∠ACB = 90°

In triangle ACB,

Sum of the angles in the triangle = 180°

m∠CAB + m∠ACB + m∠CBA = 180°

26° + 90° + m∠CBA = 180°

116° + m∠CBA = 180°

Subtract 116° from both sides.

116° + m∠CBA - 116° = 180° - 116°

m∠CBA = 64°

The measure of m∠CBA is 64°.

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1 year ago

How do i solve 3x² - 8x + 2 = 0 using quadratic formula?

DENIUS [597]

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2 years ago

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Pachacha [2.7K]

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2 years ago

I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur

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$\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx$

Setting $x=5\sin\theta$ , you have $\mathrm dx=5\cos\theta\,\mathrm d\theta$ . Then the integral becomes

$\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta$
$\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

Now, $\sqrt{x^2}=|x|$ in general. But since we want our substitution $x=5\sin\theta$ to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means $\theta=\sin^{-1}\dfrac x5$ , which implies that $\left|\dfrac x5\right|\le1$ , or equivalently that $|\theta|\le\dfrac\pi2$ . Over this domain, $\cos\theta\ge0$ , so $\sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta$ .

Long story short, this allows us to go from

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to

$\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
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Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

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and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

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