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3 years ago

if you draw two right triangles using the line as the hypotenuse, do the triangles have to be similar? Why or why not?

Mathematics

1 answer:

diamong [38]3 years ago

4 0

Answer:

All the right triangles will be similar.

Step-by-step explanation:

See the attached diagram.

Let us assume any two right triangles are drawn on the line OP, and AB and A'B' as the hypotenuse on the same line OP.

Since the slope of the line OP is constant and assume that it makes x° with the positive x-axis.

Hence, $\tan x = \frac{AC}{BC} = \frac{A'C'}{B'C'}$

⇒ $\frac{AC}{A'C'} = \frac{BC}{B'C'}$

Therefore, the sides of the two right triangles are in the same ratio and hence, Δ ABC and Δ A'B'C' are similar.

Hence, all the right triangles will be similar. (Answer)

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4 years ago

What is the equation of a line that passes through the point (6, 1) and is perpendicular to the line whose equation is y=−2x−6 ?

ladessa [460]

Answer:

y=\frac{1}{2}x-2

Step-by-step explanation:

we are given

Line is perpendicular to y=−2x−6

we can compare it with

y=mx+b

where

m is slope

so, m=-2

so, slope is -2

Our line is perpendicular to this slope

so, slope of our line will be

$m'=-\frac{1}{-2}$

$m'=\frac{1}{2}$

now, we are given point as (6,1)

so, $x_1=6,y_1=1$

now, we can use point slope form of line

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we can plug values

$y-1=\frac{1}{2}(x-6)$

now, we can solve for y

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3 years ago

In the figure above, point O is the center of the circle,

snow_tiger [21]

Answer:

16.67 or 16 2/3

Step-by-step explanation:

so, as BC is a tangent, it means the angle OBC is 90 degrees.

therefore, the angle BOC is 180 - 90 - 30 = 60 degrees.

that means the angle AOB is 180 - 60 = 120 degrees.

the triangle ABO must be an isosceles triangle, as the sides are both the radius of the circle. so, both remaining angles are the same : half of 180 - 120 = 60 = 30 degrees.

and so, we know that ABC is an isosceles triangle, because both "side angles" are equal = 30 degrees.

so, AB = BC.

and we know the circumference of the circle (50) and can calculate the radius

50 = 2×pi×r

25 = pi×r

r = 25/pi ≈ 7.96

now we have multiple ways to calculate AB.

AB = BC = tan(60) × r

AB = sqrt(r² + r² - 2×r×r×cos(120))

or ...

the first one might be easiest.

AB = tan(60) × 7.96 ≈ 13.78

oh, I just recognized, we need the arc AB (not the straight line).

even easier.

50 is the full "arc" of the circle for all 360 degrees.

but we only need the part of the angle AOB = 120 degrees.

so, this is then only the 120/360 part of the full circle circumference.

ABarc = 50 × 120/360 = 50 / 3 ≈ 16.67 or 16 2/3

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