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chubhunter [2.5K]
3 years ago
13

Find x. A. 21√2 B. 7 C. 21√3 over 2 D. 21√2 over 2

Mathematics
2 answers:
kkurt [141]3 years ago
5 0

Answer:

D

Step-by-step explanation:

<em>for </em><em>you </em><em>to </em><em>find </em><em>x </em><em>you </em><em>first </em><em>have </em><em>to </em><em>find </em><em>the </em><em>adjacent</em><em> </em><em>of </em><em>the </em><em>4</em><em>5</em><em>°</em><em> </em><em>angle </em><em>you </em><em>can </em><em>do </em><em>that </em><em>by </em><em>using</em><em> </em><em>the </em><em>other </em><em>triangle.</em><em>u</em><em>s</em><em>i</em><em>n</em><em>g</em><em> </em><em>the </em><em>sin </em><em>ratio</em>

<em>sin60=</em><em>opposite</em><em>/</em><em>hypotenuse</em>

<em>sin60=</em><em>a/</em><em>7</em><em>√</em><em>3</em>

<em>a</em><em>=</em><em>1</em><em>0</em><em>.</em><em>5</em>

<em>then </em><em>after </em><em>you </em><em>have </em><em>found</em><em> </em><em>the </em><em>adjacent</em><em> </em><em>you </em><em>can </em><em>use </em><em>the </em><em>cos </em><em>ratio</em>

<em>cos45=</em><em>adjacent/</em><em>hypotenuse</em>

<em>cos45=</em><em>1</em><em>0</em><em>.</em><em>5</em><em>/</em><em>x</em>

<em>cos45x/</em><em>cos45=</em><em>1</em><em>0</em><em>.</em><em>5</em><em>/</em><em>cos45</em>

<em>x=</em><em>1</em><em>4</em><em>.</em><em>8</em><em>4</em><em>9</em>

<em>which </em><em>is </em><em>the </em><em>same </em><em>as </em><em>2</em><em>1</em><em>√</em><em>2</em><em> </em><em>over </em><em>2</em>

<em>I </em><em>hope</em><em> this</em><em> helps</em>

Evgesh-ka [11]3 years ago
4 0

Answer:

D

Step-by-step explanation:

Using sine ratio in left right angled triangle to find the altitude a of the large triangle which is common to both right triangles and the exact value

sin60° = \frac{\sqrt{3} }{2} , then

sin60° = \frac{opposite}{hypotenuse} = \frac{a}{7\sqrt{3} } = \frac{\sqrt{3} }{2} ( cross- multiply )

2a = 21 ( divide both sides by 2 )

a = \frac{21}{2}

Using the cosine ratio in the right side triangle and the exact value

cos45° = \frac{1}{\sqrt{2} } , then

cos45° = \frac{adjacent}{hypotenuse} = \frac{a}{x} = \frac{1}{\sqrt{2} } ( cross- multiply )

x = \sqrt{2} a = \sqrt{2} × \frac{21}{2} = \frac{21\sqrt{2} }{2} → D

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