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IrinaVladis [17]
3 years ago
15

The length of the hypotenuse of an isosceles, right triangle is 12 6 . How many degrees are in each angle? What is the length of

each leg?

Mathematics
2 answers:
Fynjy0 [20]3 years ago
8 0
Isosceles right triangle is a 90 45 45 degree triangle
leg = 12.6 / sq rt(2)
leg = 12.6 / <span> <span> <span> 1.4142135624 </span> </span> </span>
leg = <span> <span> <span> 8.909545443 </span> </span> </span>

Amanda [17]3 years ago
7 0

Answer:

Length of each leg = 8.91 units

Length of each angle other than right angle is 45°

Step-by-step explanation:

For better understanding of the solution see the attached figure :

ΔABC is right angled isosceles triangle ⇒ AB = BC

Now, by using Pythagoras theorem :

AC² = AB² + BC²

12.6² = 2×AB²

AB² = 79.38

AB = 8.91 units

So, BC = 8.91 units

Now, Since AB = AC

⇒ ∠BAC = ∠ACB ( equal sides have equal angles opposite to them)

Now, By using angle sum property of a triangle

∠ABC + ∠BAC + ∠ACB = 180°

90° + 2∠BAC = 180°

2∠BAC = 90°

∠BAC = 45°

⇒ ∠ACB = 45°

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Let the three numbers be x, y and z respectively, then
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The above three expressions could be used to represent the numbers.

Solving the three equations, putting (2) and (3) into (1) gives
x + x - 4 + 4x = 62
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A student solves the following equation and
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Answer:

Since the equation is undefined for -2

Therefore, NO SOLUTION for the given equation.

Step-by-step explanation:

Considering the expression

\frac{3}{a+2}-6\cdot \frac{a}{-4+a^2}=\frac{1}{a-2}

\frac{3}{a+2}-\frac{6a}{-4+a^2}=\frac{1}{a-2}

\mathrm{Find\:Least\:Common\:Multiplier\:of\:}a+2,\:-4+a^2,\:a-2:\quad \left(a+2\right)\left(a-2\right)

\mathrm{Multiply\:by\:LCM=}\left(a+2\right)\left(a-2\right)

\frac{3}{a+2}\left(a+2\right)\left(a-2\right)-\frac{6a}{-4+a^2}\left(a+2\right)\left(a-2\right)=\frac{1}{a-2}\left(a+2\right)\left(a-2\right)

as

  • \frac{3}{a+2}\left(a+2\right)\left(a-2\right):\quad 3\left(a-2\right)
  • -\frac{6a}{-4+a^2}\left(a+2\right)\left(a-2\right):\quad -6a
  • \frac{1}{a-2}\left(a+2\right)\left(a-2\right):\quad a+2

so equation becomes

3\left(a-2\right)-6a=a+2  

-3a-6=a+2

-3a-6+6=a+2+6

-4a=8

\mathrm{Divide\:both\:sides\:by\:}-4

\frac{-4a}{-4}=\frac{8}{-4}

a=-2

\mathrm{Verify\:Solutions}

\mathrm{Take\:the\:denominator\left(s\right)\:of\:}\frac{3}{a+2}-6\frac{a}{-4+a^2}-\frac{1}{a-2}\mathrm{\:and\:compare\:to\:zero}

\mathrm{Solve\:}\:a+2=0:\quad a=-2

\mathrm{Solve\:}\:-4+a^2=0:\quad a=2,\:a=-2

\mathrm{Solve\:}\:a-2=0:\quad a=2

So the following points are undefined

a=-2,\:a=2

Since the equation is undefined for -2

Therefore, NO SOLUTION for the given equation.

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