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2 years ago

Special Right Trianglws

Mathematics

1 answer:

Masteriza [31]2 years ago

3 0

Answer:

x = $2\sqrt7$

y = $2\sqrt7$

Step-by-step explanation:

Hello!

This is a right isosceles triangle (45°-45°-90°). You can solve for the missing angle by simply subtracting 45° and 90° from 180° (sum of angles in a triangle equals 180°)

Since the triangle is isosceles, x and y are congruent because they are the legs opposite to the congruent angles.

In a right isosceles triangle, the hypotenuse is always the measure of one leg multiplied by $\sqrt2$ .

So, let's use leg y, $y\sqrt2 = 2\sqrt{14}$

Solve for y:

$y\sqrt2 = 2\sqrt{14}\\\\y = $\frac{2\sqrt{14}}{\sqrt2}$\\\\y = 2\sqrt7$

y = $2\sqrt7$ , which means that x is also $2\sqrt7$

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2 years ago

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Consider the first four terms of an arithmetic sequence below.

VMariaS [17]

7, 13, 19 and 25 have a common difference: 6.
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3 years ago

Find the measures of the angles of the triangle whose vertices are A = (-3,0) , B = (1,3) , and C = (1,-3).A.) The measure of ∠A

alekssr [168]

Answer:

$\theta_{CAB}=128.316$

$\theta_{ABC}=25.842$

$\theta_{BCA}=25.842$

Step-by-step explanation:

A = (-3,0) , B = (1,3) , and C = (1,-3)

We're going to use the distance formula to find the length of the sides:

$r= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

$AB= \sqrt{(-3-1)^2+(0-3)^2}=5$

$BC= \sqrt{(1-1)^2+(3-(-3))^2}=9$

$CA= \sqrt{(1-(-3))^2+(-3-0)^2}=5$

we can use the cosine law to find the angle:

it is to be noted that:

the angle CAB is opposite to the BC.

the angle ABC is opposite to the AC.

the angle BCA is opposite to the AB.

to find the CAB, we'll use:

$BC^2 = AB^2+CA^2-(AB)(CA)\cos{\theta_{CAB}}$

$\dfrac{BC^2-(AB^2+CA^2)}{-2(AB)(CA)} =\cos{\theta_{CAB}}$

$\cos{\theta_{CAB}}=\dfrac{9^2-(5^2+5^2)}{-2(5)(5)}$

$\theta_{CAB}=\arccos{-\dfrac{0.62}}$

$\theta_{CAB}=128.316$

Although we can use the same cosine law to find the other angles. but we can use sine law now too since we have one angle!

To find the angle ABC

$\dfrac{\sin{\theta_{ABC}}}{AC}=\dfrac{\sin{CAB}}{BC}$

$\sin{\theta_{ABC}}=AC\left(\dfrac{\sin{CAB}}{BC}\right)$

$\sin{\theta_{ABC}}=5\left(\dfrac{\sin{128.316}}{9}\right)$

$\theta_{ABC}=\arcsin{0.4359}\right)$

$\theta_{ABC}=25.842$

finally, we've seen that the triangle has two equal sides, AB = CA, this is an isosceles triangle. hence the angles ABC and BCA would also be the same.

$\theta_{BCA}=25.842$

this can also be checked using the fact the sum of all angles inside a triangle is 180

$\theta_{ABC}+\theta_{BCA}+\theta_{CAB}=180$