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arsen [322]
3 years ago
11

Find the value of x and y.

Mathematics
1 answer:
zysi [14]3 years ago
8 0

Answer:

<em>x = 9</em>

<em>y = 36</em>

Step-by-step explanation:

<u>Lines and Angles</u>

Triangle CDE is isosceles. This means the two angles of the base DE are congruent (have the same measure):

7x + 1 = 4x + 28

Subtracting 4x + 1:

3x = 27

Dividing by 3:

x = 9

Substituting in the expression for the angles:

7x + 1 = 7*9 + 1 = 64°

The angles are 64° and 64°. The other internal angle at vertex C is 180°-64°-64°=52°. This angle is congruent with its vertical angle in the triangle ABC. We are given another angle of 43°. Thus the measure of angle A is 180°-52°-43°=85°

This last angle is equal to the expression of y:

-2(3 - y) + 19 = 85

Subtracting 19:

-2(3 - y) = 66

Removing the parentheses:

-6 + 2y = 66

Adding 6:

2y = 72

Dividing by 2:

y = 36

Final answer:

x = 9

y = 36

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Ronch [10]

Answer: m<KHL = 37 degrees.

m<GHK = 53 degrees

Step-by-step explanation:

<CHD and <KHL are vertical angles. They are equal in measure. m<CHD is 37 degrees, so m<KHL is also 37 degrees.

Notice <GHL is a 90 degree angle (because the lines that form it are perpendicular) and <GHL is made up of two angles: <GHK and <KHL.

m<GHK + m<KHL (37) = m< GHL (90 degrees)

Subtract 37 from both sides.

m<GHK = 53 degrees

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3 years ago
A sector of a circle makes a 127° angle at its centre. If the arc of the sector has length 36 mm, find
Veseljchak [2.6K]

Answer:

Approximately 68.5\; \rm mm.

Step-by-step explanation:

Convert the angle of this sector to radians:

\begin{aligned}\theta &= 127^{\circ} \\ &= 127^{\circ} \times \frac{2\pi}{360^{\circ}} \\ &\approx 2.22\end{aligned}.

The formula s = r\, \theta relates the arc length s of a sector of angle \theta (in radians) to the radius r of this sector.

In this question, it is given that the arc length of this sector is s = 36\; \rm mm. It was found that \theta = 2.22 radians. Rearrange the equation s = r\, \theta to find the radius r of this sector:

\begin{aligned} r&= \frac{s}{\theta} \\ &\approx \frac{36\; \rm mm}{2.22} \\ &\approx 16.2\; \rm mm\end{aligned}.

The perimeter of this sector would be:

\begin{aligned}& 2\, r + s \\ =\; & 2 \times 16.2\; {\rm mm} + 36\; {\rm mm} \\ =\; & 68.5\; \rm mm\end{aligned}.

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