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Alla [95]
2 years ago
15

37.Where is the incenter of any given triangle ?

Mathematics
1 answer:
Vladimir [108]2 years ago
5 0

Answer:

Part 1) Option The point of concurrency of the angle bisectors of the triangle

Part 2) Option A segment drawn  from a vertex to the midpoint of the opposite side

Step-by-step explanation:

Part 1)

we know that

The <u>incenter</u> is the point where the angle bisectors intersect. The incenter is also the center of the triangle's incircle.

therefore

Is the point of concurrency of the angle bisectors of the triangle

Part 2)

we know that

A <u>median</u> of a triangle is a line segment joining a vertex to the midpoint of the opposite side

therefore

Is a segment drawn  from a vertex to the midpoint of the opposite side

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Step-by-step explanation:

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3 years ago
What is the length of seg AZ if A is between Y and Z, YA = 24, AZ = 15x, and<br> YZ = 189 ?
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YA+AZ=YZ

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Step-by-step explanation:

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2 years ago
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3 years ago
Can you please answer this question
tamaranim1 [39]
3tan^{2} \theta +7sec\theta=3
First I converted the equation terms into sine and cosine.
tan^{2}\theta = \frac{sin^{2}\theta}{cos^{2}\theta} and sec\theta= \frac{1}{cos\theta}
Substitution:
\frac{3sin^2\theta}{cos^2\theta} + \frac{7}{cos\theta} =3
Common Denominator Created:
\frac{3sin^2\theta}{cos^2\theta} + \frac{7cos\theta}{cos^2\theta} =3
Multiply each term by the LCD:
3sin^2\theta+7cos\theta=3cos^2\theta
Substitution: Recall ⇒sin^2\theta =1-cos^2\theta 
3(1-cos^2\theta)+7cos\theta=3cos^2\theta
Distribute and collect all terms on one side:
6cos^2\theta-7cos\theta-3=0
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2cos\theta-3=0⇒theta=cos^{-1} \frac{3}{2}
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The 2nd factor provides only possible answer 109.5 degrees

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3 years ago
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