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2 years ago

Help please!! I will mark brainilist and give a lot of points.

Mathematics

2 answers:

julia-pushkina [17]2 years ago

8 0

Answer:

a = 23.588

Step-by-step explanation:

a^2 = 17 1/3^2 + 16^2

a^2 = 300.4 + 256

a^2 = 556.4

a = 23.588 (I rounded to thousandths but you can round to whatever)

notka56 [123]2 years ago

7 0

Hello there,

I just looked into the question and you are absolutely right whatever you inferred.
I think the question might be wrong.
The value of c should be greater than b.
I guess it will like this:
$b = 16 \: \: \: c = 17 \frac{1}{3}$
The values should be like this.
I am solving the equation by this:
${a}^{2} = {c}^{2} - {b}^{2} \\ = > {a}^{2} = {(17 \frac{1}{3}) }^{2} - {(16)}^{2} \\ = > {a}^{2} = {( \frac{52}{3} )}^{2} - {(16)}^{2} \\ = > {a}^{2} = ( \frac{52}{3} - 16)( \frac{52}{3} + 16) \\ = > {a}^{2} = ( \frac{52 - 48}{3} )( \frac{52 + 48}{3} ) \\ = > {a}^{2} = \frac{4}{3} \times \frac{100}{3} \\ = > {a}^{2} = \frac{400}{9} \\ = > a = \sqrt{ \frac{400}{9} } \\ = > a = \frac{20}{3} \\ = > a = 6 \frac{2}{3}$
So, the value of a is
$a = 6 \frac{2}{3}$

Hope you could get an idea from it.

Doubt clarification - use comment section.

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$A)cos^2\theta=sin^2\theta-1$ $B)sin\theta=\frac{1}{csc\theta}$ $\begin{gathered} C)sec\theta=\frac{1}{cot\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}$

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Solution

Verify each of the given function

$\begin{gathered} cos^2\theta=sin^2\theta-1 \\ Note\text{ that} \\ sin^2\theta+cos^2\theta=1 \\ cos^2\theta=1-sin^2\theta \\ Therefore \\ cos^2\theta sin^2\theta-1,NOT\text{ }IDENTITIES \end{gathered}$

$\begin{gathered} sin\theta=\frac{1}{csc\theta} \\ Note\text{ that} \\ csc\theta=\frac{1}{sin\theta} \\ sin\theta\times csc\theta=1 \\ sin\theta=\frac{1}{csc\theta} \\ Therefore \\ sin\theta=\frac{1}{csc\theta},is\text{ an identities} \end{gathered}$

$\begin{gathered} sec\theta=\frac{1}{cot\theta} \\ note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ tan\theta cot\theta=1 \\ tan\theta=\frac{1}{cot\theta} \\ Therefore, \\ sec\theta\ne\frac{1}{cot\theta},NOT\text{ IDENTITY} \end{gathered}$

$\begin{gathered} cot\theta=\frac{cos\theta}{sin\theta} \\ Note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ cot\theta=1\div tan\theta \\ tan\theta=\frac{sin\theta}{cos\theta} \\ So, \\ cot\theta=1\div\frac{sin\theta}{cos\theta} \\ cot\theta=1\times\frac{cos\theta}{sin\theta} \\ cot\theta=\frac{cos\theta}{sin\theta} \\ Therefore \\ cot\theta=\frac{cos\theta}{sin\theta},is\text{ an Identity} \end{gathered}$

$\begin{gathered} 1+cot^2\theta=csc^2\theta \\ csc^2\theta-cot^2\theta=1 \\ csc^2\theta=\frac{1}{sin^2\theta} \\ cot^2\theta=\frac{cos^2\theta}{sin^2\theta} \\ So, \\ \frac{1}{sin^2\theta}-\frac{cos^2\theta}{sin^2\theta} \\ \frac{1-cos^2\theta}{sin^2\theta} \\ Note, \\ cos^2\theta+sin^2\theta=1 \\ sin^2\theta=1-cos^2\theta \\ So, \\ \frac{1-cos^2\theta}{sin^2\theta}=\frac{sin^2\theta}{sin^2\theta}=1 \\ Therefore \\ 1+cot^2\theta=csc^2\theta,\text{ is an Identity} \end{gathered}$

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