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2 years ago

What is the area of the trapezoid to the nearest tenth? PICTURE BELOW ⬇⬇⬇⬇

Mathematics

1 answer:

Nastasia [14]2 years ago

6 0

Answer:

62.4 ft²

Step-by-step explanation:

The unmarked horizontal dimension at the bottom of the triangle is ...

(8 ft)sin(30°) = 4 ft

The unmarked vertical dimension of the triangle (the height of the trapezoid) is ...

(8 ft)cos(30°) ≈ 6.93 ft

Then the area of the trapezoid is given by the formula ...

A = (1/2)(b1 +b2)h

A = (1/2)((4 ft+7 ft) +(7 ft))(6.93 ft) ≈ 62.4 ft²

_____

The mnemonic SOH CAH TOA can remind you of the relationships between right triangle dimensions and angles.

Sin = Opposite/Hypotenuse ⇒ Hypotenuse×Sin = Opposite

Cos = Adjacent/Hypotenuse ⇒ Hypotenuse×Cos = Adjacent

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Answer:

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Step-by-step explanation:

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2 years ago

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Please solve 5 f <br> (Trigonometric Equations)<br> #salute u if u solved it

Zanzabum

Answer:

$\beta=45\degree\:\:or\:\:\beta=135\degree$

Step-by-step explanation:

We want to solve $\tan \beta \sec \beta=\sqrt{2}$ , where $0\le \beta \le360\degree$ .

We rewrite in terms of sine and cosine.

$\frac{\sin \beta}{\cos \beta} \cdot \frac{1}{\cos \beta} =\sqrt{2}$

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Use the Pythagorean identity: $\cos^2\beta=1-\sin^2\beta$ .

$\frac{\sin \beta}{1-\sin^2\beta}=\sqrt{2}$

$\implies \sin \beta=\sqrt{2}(1-\sin^2\beta)$

$\implies \sin \beta=\sqrt{2}-\sqrt{2}\sin^2\beta$

$\implies \sqrt{2}\sin^2\beta+\sin \beta- \sqrt{2}=0$

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By the quadratic formula, we have:

$\sin \beta=\frac{-1\pm \sqrt{1^2-4(\sqrt{2})(-\sqrt{2} ) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{1^2+4(2) } }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm \sqrt{9} }{2\cdot \sqrt{2} }$

$\sin \beta=\frac{-1\pm3}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{2}{2\cdot \sqrt{2} }$ or $\sin \beta=\frac{-4}{2\cdot \sqrt{2} }$

$\sin \beta=\frac{1}{\sqrt{2} }$ or $\sin \beta=-\frac{2}{\sqrt{2} }$

$\sin \beta=\frac{\sqrt{2}}{2}$ or $\sin \beta=-\sqrt{2}$

When $\sin \beta=\frac{\sqrt{2}}{2}$ , $\beta=\sin ^{-1}(\frac{\sqrt{2} }{2} )$

$\implies \beta=45\degree\:\:or\:\:\beta=135\degree$ on the interval $0\le \beta \le360\degree$ .

When $\sin \beta=-\sqrt{2}$ , $\beta$ is not defined because $-1\le \sin \beta \le1$

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3 years ago