In triangle $A B C$ $ABC$ , $A B = x$ $AB = x$ , $B C = 5 cm$ $BC = 5" cm"$ , $A C = (10 - x) cm$ $AC = (10-x)" cm"$ , and $cos (∠ A B C) = - \frac{1}{7}$ $cos (angle ABC) = -1/7$ . What is the value of $x$ $x$ ?

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In triangle $A B C$ $ABC$ , $A B = x$ $AB = x$ , $B C = 5 cm$ $BC = 5" cm"$ , $A C = (10 - x) cm$ $AC = (10-x)" cm"$ , and $cos (∠ A B C) = - \frac{1}{7}$ $cos (angle ABC) = -1/7$ . What is the value of $x$ $x$ ?

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Answer 1 · 2017-11-06T18:25:22

$x = 3.5 cm$ $x = 3.5" cm"$

Explanation:

Given: $A B = x cm$ $AB = x" cm"$ , $B C = 5 cm$ $BC = 5" cm"$ , $A C = (10 - x) cm$ $AC = (10-x)" cm"$ , and $cos (∠ A B C) = - \frac{1}{7}$ $cos(angle ABC) = -1/7$

I will switch to the notation where the angle is represented by an uppercase letter and the opposite side is represented by the corresponding lowercase letter:

$c = x cm$ $c = x" cm"$ , $a = 5 cm$ $a = 5" cm"$ , $b = (10 - x) cm$ $b = (10-x)" cm"$ , and $cos (B) = - \frac{1}{7}$ $cos(B) = -1/7$

We may use the Law of Cosines to derive an equation that has x as the only variable:

$b^{2} = a^{2} + c^{2} - 2 (a) (c) cos (B)$ $b^2 = a^2+c^2-2(a)(c)cos(B)$

Substitute in the known values:

${((10 - x) cm)}^{2} = {(5 cm)}^{2} + x^{2} - 2 (5 cm) (x) (- \frac{1}{7})$ $((10-x)" cm")^2 = (5" cm")^2+x^2-2(5" cm")(x)(-1/7)$

Expand the square and write both sides in standard form:

$x^{2} - 20 x + 100 = x^{2} + \frac{10 x}{7} + 25$ $x^2 - 20 x + 100 = x^2 + (10 x)/7 + 25$

Solve for x:

$- 20 x + 100 = \frac{10 x}{7} + 25$ $-20 x + 100 = (10 x)/7 + 25$

$- \frac{150}{7} x = - 75$ $-150/7x=-75$

$x = 3.5 cm$ $x = 3.5" cm"$

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In triangle $A B C$ $ABC$ , $A B = x$ $AB = x$ , $B C = 5 cm$ $BC = 5" cm"$ , $A C = (10 - x) cm$ $AC = (10-x)" cm"$ , and $cos (∠ A B C) = - \frac{1}{7}$ $cos (angle ABC) = -1/7$ . What is the value of $x$ $x$ ?

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Explanation:

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In triangle ABCABC, AB=xAB = x, BC=5 cmBC = 5" cm", AC=(10−x) cmAC = (10-x)" cm", and cos(∠ABC)=−17cos (angle ABC) = -1/7. What is the value of xx?

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Explanation:

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In triangle $A B C$ $ABC$ , $A B = x$ $AB = x$ , $B C = 5 cm$ $BC = 5" cm"$ , $A C = (10 - x) cm$ $AC = (10-x)" cm"$ , and $cos (∠ A B C) = - \frac{1}{7}$ $cos (angle ABC) = -1/7$ . What is the value of $x$ $x$ ?