Tak matma, ale może ktoś lubi :P Zadanka w załaczniku :).... Question from @kasiaf120

October 2018 1 28 Report

Tak matma, ale może ktoś lubi :P Zadanka w załaczniku :)

1.
$\sqrt{cos^2\alpha-sin^2\alpha}=\sqrt{cos^2\alpha-(1-cos^2\alpha)}= \\ \sqrt{cos^2\alpha-1+cos^2\alpha}= \sqrt{2cos^2\alpha-1}=\\ \sqrt{2\cdot ( \frac{12}{13})^2-1}=\sqrt{2\cdot \frac{144}{169}-1}=\\ \sqrt{\frac{288}{169}-1}= \sqrt{\frac{119}{169}}= \frac{\sqrt{119}}{13}$

D
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2.
$tg\alpha= \frac{sin20^o}{sin70^o}=\frac{sin20^o}{sin(90^o-20^o)}=\\ \frac{sin20^o}{cos 20^o}=tg20^o$

$\alpha=20^o$

A
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3.
$tg(90^o-\alpha)=\frac{1}{2}$

$ctg\alpha=\frac{1}{2}$

$\frac{cos\alpha}{sin\alpha}=\frac{1}{2}$

$cos\alpha=\frac{1}{2}sin\alpha$

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

$sin^2\alpha+(\frac{1}{2}sin\alpha)^2=1$

$sin^2\alpha+\frac{1}{4}sin^2\alpha=1$

$\frac{5}{4}sin^2\alpha=1\ /\cdot \frac{4}{5}$

$sin^2\alpha= \frac{4}{5}$

$sin\alpha= \sqrt{\frac{4}{5}}$

$sin\alpha=\frac{2}{\sqrt{5}}$

$sin\alpha=\frac{2 \sqrt{5}}{5}$

A
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4.
$cos\alpha\cdot \frac{tg\alpha}{sin\alpha}= \frac{cos\alpha}{sin\alpha}\cdot tg\alpha=ctg\alpha\cdot tg\alpha=$

$1= \sqrt{2}\cdot \frac{\sqrt{2}}{2}= \sqrt{2}sin45^o$

B
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5.
$cos\alpha= \sqrt{3}cos(90^o-\alpha)$

$cos\alpha= \sqrt{3}sin\alpha \ /:sin\alpha$

$\frac{cos\alpha}{sin\alpha}= \sqrt{3}$

$ctg\alpha= \sqrt{3}$

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$\frac{1}{\tg^2\alpha}=( \frac{1}{tg\alpha} )^2=(ctg\alpha)^2=(\sqrt{3})^2=3$

D
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6.
$cos\alpha\cdot tg\alpha=cos\alpha\cdot \frac{sin\alpha}{cos\alpha}=sin\alpha$

$sin\alpha= \frac{\sqrt{7,5^2-7,2^2}}{7,5}=\frac{\sqrt{56,25-51,84}}{7,5}=$

$\frac{\sqrt{4.41}}{7,5}= \frac{2,1}{7,5}=0,28$

C
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7.
$tg\alpha\cdot ( \frac{sin\beta}{sin\alpha} +tg\beta)=tg\alpha\cdot ( \frac{cos\alpha}{sin\alpha} +ctg\alpha)=$

$tg\alpha\cdot ( ctg\alpha+ctg\alpha)=tg\alpha\cdot 2ctg\alpha=2\cdot 1=2= \sqrt[3]{2^3}= \sqrt[3]{8}$

A
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8.
$P= \frac{bc sin\alpha}{2}$

$P= \frac{2 \sqrt{2}\cdot 6\cdot \frac{\sqrt{3}}{2}}{2}$

$P= \frac{6\sqrt{6}}{2}$

$P=3 \sqrt{6} cm^2$

D
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9.
Obliczam $sin\alpha$
$sin\alpha= \sqrt{1-cos^2\alpha}$

$sin\alpha= \sqrt{1-( \frac{7}{25})^2}$

$sin\alpha= \sqrt{1-\frac{49}{625}}$

$sin\alpha= \sqrt{\frac{576}{625}}$

$sin\alpha= \frac{24}{25}$

Obliczam $P$
$P=a^2sin\alpha$

$P=10^2\cdot \frac{24}{25}$

$P=100\cdot \frac{24}{25}$

$P=96cm^2$

B