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3 years ago

In triangle ABC, angle A is equal to angle C, BA=x+20,CA=4x-30, and BC=3x+14. What is the length of BC?

Mathematics

1 answer:

Anna [14]3 years ago

3 0

Answer:

BC = 23

Step-by-step explanation:

Since ∠ A = ∠ C then the triangle is isosceles with AB = BC, that is

3x + 14 = x + 20 ( subtract x from both sides )

2x + 14 = 20 ( subtract 14 from both sides )

2x = 6 ( divide both sides by 2 )

x = 3

Thus

BC = 3x + 14 = 3(3) + 14 = 9 + 14 = 23

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2 years ago

Given points A (1, 2/3), B (x, -4/5), and C (-1/2, 4) determine the value of x such that all three points are collinear

AlladinOne [14]

Answer:

$x=\frac{83}{50}$

Step-by-step explanation:

we know that

If the three points are collinear

then

$m_A_B=m_A_C$

we have

A (1, 2/3), B (x, -4/5), and C (-1/2, 4)

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

step 1

Find the slope AB

we have

$A(1,\frac{2}{3}),B(x,-\frac{4}{5})$

substitute in the formula

$m_A_B=\frac{-\frac{4}{5}-\frac{2}{3}}{x-1}$

$m_A_B=\frac{\frac{-12-10}{15}}{x-1}$

$m_A_B=-\frac{22}{15(x-1)}$

step 2

Find the slope AC

we have

$A(1,\frac{2}{3}),C(-\frac{1}{2},4)$

substitute in the formula

$m_A_C=\frac{4-\frac{2}{3}}{-\frac{1}{2}-1}$

$m_A_C=\frac{\frac{10}{3}}{-\frac{3}{2}}$

$m_A_C=-\frac{20}{9}$

step 3

Equate the slopes

$m_A_B=m_A_C$

$-\frac{22}{15(x-1)}=-\frac{20}{9}$

solve for x

$15(x-1)20=22(9)$

$300x-300=198$

$300x=198+300$

$300x=498$

$x=\frac{498}{300}$

simplify

$x=\frac{83}{50}$

8 0

3 years ago

In triangle ABC, angle A is equal to angle C, BA=x+20,CA=4x-30, and BC=3x+14. What is the length of BC?​

In triangle ABC, angle A is equal to angle C, BA=x+20,CA=4x-30, and BC=3x+14. What is the length of BC?