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gladu [14]
3 years ago
12

A triangle with vertices labeled as X, Y, and Z. Side X Z is base. Side Y X contains a midpoint M. A line segment drawn from Z t

o M bisects angle Y Z X into two parts labeled as Y Z M and X Z M. Angles Y Z M and X Z M are marked with single arc. Side Y Z is labeled 7. Base X Z is labeled 11. Side Y M is labeled 3.5.

Mathematics
1 answer:
MakcuM [25]3 years ago
3 0

Answer:

The value of MX is 5.5 units.

Step-by-step explanation:

As the complete question is not given, the complete question is found online and is attached herewith

brainly.com/question/11944039

The diagram is attached with the solution.

As the line ZM is dividing the angle in two parts thus as the two angles are equal thus

The sides are given as

XZ=11

YM=3.5

YZ=7

Now by the theorem of the angles, the bisected angles are such that sin(<YZM)=sin(<XZM)

so

sin(

So the value of MX is 5.5 units.

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Step-by-step explanation:

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x = 9.5

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What is the length of BC , rounded to the nearest tenth?
Arte-miy333 [17]

Step 1

In the right triangle ADB

<u>Find the length of the segment AB</u>

Applying the Pythagorean Theorem

AB^{2} =AD^{2}+BD^{2}

we have

AD=5\ units\\BD=12\ units

substitute the values

AB^{2}=5^{2}+12^{2}

AB^{2}=169

AB=13\ units

Step 2

In the right triangle ADB

<u>Find the cosine of the angle BAD</u>

we know that

cos(BAD)=\frac{adjacent\ side }{hypotenuse}=\frac{AD}{AB}=\frac{5}{13}

Step 3

In the right triangle ABC

<u>Find the length of the segment AC</u>

we know that

cos(BAC)=cos (BAD)=\frac{5}{13}

cos(BAC)=\frac{adjacent\ side }{hypotenuse}=\frac{AB}{AC}

\frac{5}{13}=\frac{AB}{AC}

\frac{5}{13}=\frac{13}{AC}

solve for AC

AC=(13*13)/5=33.8\ units

Step 4

<u>Find the length of the segment DC</u>

we know that

DC=AC-AD

we have

AC=33.8\ units

AD=5\ units

substitute the values

DC=33.8\ units-5\ units

DC=28.8\ units

Step 5

<u>Find the length of the segment BC</u>

In the right triangle BDC

Applying the Pythagorean Theorem

BC^{2} =BD^{2}+DC^{2}

we have

BD=12\ units\\DC=28.8\ units

substitute the values

BC^{2}=12^{2}+28.8^{2}

BC^{2}=973.44

BC=31.2\ units

therefore

<u>the answer is</u>

BC=31.2\ units

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