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Brilliant_brown [7]

3 years ago

15

Help pl0x, Algebra 1

1 answer:

djyliett [7]3 years ago

3 0

$width=8+x+x=8+2x\\length=9+x+x=9+2x\\\\Area:A=width\cdot length\\\\A=(8+2x)(9+2x)=(8)(9)+(8)(2x)+(2x)(9)+(2x)(2x)\\\\=72+16x+18x+4x^2=4x^2+34x+72\\\\A=75ft^2\\therefore\\\\4x^2+34x+72=75\ \ \ \ |subtract\ 75\ from\ both\ sides\\\boxed{4x^2+34x-3=0}$

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ozzi

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riadik2000 [5.3K]

Answer:

$\frac{(\sin 18^\circ)}{75} = \frac{(\sin 35^\circ)}{x}$

Step-by-step explanation:

Incomplete question:

<em></em> $\angle CBD = 53^\circ$ <em></em>

<em></em> $\angle CAB = 35^\circ$ <em></em>

<em></em> $AB = 75$ <em></em>

<em></em>

<em>See attachment for complete question</em>

Required

Determine the equation to find x

First, is to complete the angles of the triangle (ABC and ACB)

$\angle ABC + \angle CBD = 180$ --- angle on a straight line

$\angle ABC + 53= 180$

Collect like terms

$\angle ABC =- 53+ 180$

$\angle ABC =127^\circ$

$\angle ABC + \angle ACB + \angle CAB = 180$ --- angles in a triangle

$\angle ACB + 127 + 35 = 180$

Collect like terms

$\angle ACB =- 127 - 35 + 180$

$\angle ACB =18$

Apply sine rule

$\frac{\sin A}{a} = \frac{\sin B}{b}$

In this case:

$\frac{\sin ACB}{AB} = \frac{\sin CAB}{x}$

This gives:

$\frac{(\sin 18^\circ)}{75} = \frac{(\sin 35^\circ)}{x}$

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