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Alecsey [184]
3 years ago
14

Right triangle ABC and right triangle ACD overlap as shown below. Angle DAC measures 20 degrees and angel BCA measures 30 degree

s. What are the values of X and Y?

Mathematics
1 answer:
alexdok [17]3 years ago
6 0

The image of the question is attached below.

Given:

m∠DAC = 20° and m∠BCA = 30°

m∠ABC = 90° and m∠CDA = 90°

To find:

The value of x and y.

Solution:

In right triangle ABC,

m∠BAC = x° + 20°

Sum of the interior angles of a triangle = 180°

m∠BAC +m∠ABC + m∠BCA = 180°

x° + 20° + 90° + 30° = 180°

x° + 140° = 180°

Subtract 140° from both sides.

x° + 140° - 140° = 180° - 140°

x° = 40°

In right triangle ADC,

m∠ACD = y° + 30°

Sum of the interior angles of a triangle = 180°

m∠ACD +m∠CDA + m∠DAC = 180°

y° + 30° + 90° + 20° = 180°

y° + 140° = 180°

Subtract 140° from both sides.

y° + 140° - 140° = 180° - 140°

y° = 40°

The value of x is 40 and y is 40.

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3 years ago
The vertices A, B, C of a triangle are (2,1),(5,2),and (3,4) respectively . Find the coordinate of the circumcentre and also the
BlackZzzverrR [31]

Answer:

Circumcenter = (4,0)

Circumcircle = √5

Step-by-step explanation:

The circumcentre is the point of intersection of the perpendicular bisectors of a triangle. The vertices of the triangle are equidistant to the circumcentre.

Let us assume the coordinate of the circumcentre is at O(x, y). Therefore the distance between the cirmcumcenter and the vertices are:

AO=\sqrt{(x-2)^2+(y-1)^2} =\sqrt{x^2-4x+4+(y^2-2y+1)}\\=\sqrt{x^2+y^2-4x-2y+5}  \\\\BO=\sqrt{(x-5)^2+(y-2)^2} =\sqrt{x^2-10x+25+(y^2-4y+4)}\\=\sqrt{x^2+y^2-10x-4y+29}  \\\\CO=\sqrt{(x-3)^2+(y-4)^2} \\=\sqrt{x^2-9x+9+(y^2-8y+16)}=\sqrt{x^2+y^2-9x-8y+25}  \\

AO = BO, therefore

√(x² + y²-4x-2y+5) = √(x² + y² - 10x - 4y + 29)

x² + y²-4x-2y+5 = x² + y² - 10x - 4y + 29

6x + 2y = 24       (1)

BO = CO

√(x² + y² - 10x - 4y + 29) = √(x² + y² - 9x - 8y + 25)

x² + y² - 10x - 4y + 29 = x² + y² - 9x - 8y + 25

-x + 4y = -4         (2)

Multiply equation 2 by 6 and add to equation 1:

26y = 0

y=0

Put y = 0 in -x + 4y = -4

-x + 4(0) = -4

x = 4

The cicumcenter is at (4, 0)

The radius of the circumcircle = AO = BO = CO. Therefore:

Radius=AO=\sqrt{(4-2)^2+(0-1)^2} =\sqrt{4+1}=\sqrt{5}

7 0
3 years ago
X^2 + 18x + 50 = 9<br><br>Solve by completing the square
dsp73
Step 1: both sides subtract 50: x²+18x=-41
step 2: both sides add (1/2 of 18)², which is 81: x²+18x+81=-41+81
step 3: make the square: (x+9)²=40
step 4: square root both sides: x+9=√40 or x+9=-√40
simplify: x=-9+2√10 or x=-9-2√10
8 0
3 years ago
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