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Vesna [10]
3 years ago
5

The shorter leg of a 30°- 60°- 90° right triangle is 12.5 inches. How long are the longer leg and the hypotenuse?

Mathematics
2 answers:
Elodia [21]3 years ago
8 0
A right angle with angles 30° - 60° - 90° is cut from an equilateral triangle, for example as shown below in triangle ABC

The shorter leg would be the side BD (or CD) since D is the midpoint of side BC

We are to find the length of side AD (which is the height of the triangle ABC) and the hypotenuse AB (or AC ⇒ depends on which right angle triangle that you'd want to work with)

Since we know the three angles in the right angle triangle ABD, we can use the trigonometry ratio

Side AD is the opposite of the angle 60° and we know the length of side that is adjacent to 60° so we'll use the tangent ratio

tan (x°) = opposite ÷ adjacent
tan (60°) = AD ÷ 12.5
√3 = AD ÷ 12.5
AD = 12.5 × √3
AD = 21.65

We can then use the Pythagoras theorem to work out AB

AB² = AD² + BD²
AB² = 12.5² + 21.65²
AB² = 624.9725
AB = √624.9725
AB = 25 (rounded to 2 significant figures)



dezoksy [38]3 years ago
4 0
If u know the short leg...which we do, it is 12.5......to find the hypotenuse, double it......so the hypotenuse is : 12.5 * 2 = 25 inches.....and to find the longer leg, multiply shorter leg by square root of 3.....12.5 * sqrt 3 = 21.65 (thats rounded)
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\boxed{x = 7, y = 9, z = 68}

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I will use these three:

\begin{array}{lrcll}(1) & 8x + 13y +7 & = & 180 & \\(2)& 9x - 7 + 13y +7 & = & 180 & \\(3)& 8x + 5y - 11 + z & = & 180 &\text{We can rearrange these to get:}\\(4)& 8x + 13y & = & 173 &\\(5) & 9x + 13y & = & 180 & \\(6)& 8x + 5y + z & = & 169 & \\(7)& x & = & \mathbf{7} & \text{Subtracted (4) from (5)} \\\end{array}

\begin{array}{lrcll}& 8(7) + 13y & = & 173 & \text{Substituted (7) into (4)} \\& 56 + 13y & = & 173 & \text{Simplified} \\& 13y & = & 117 & \text{Subtracted 56 from each side} \\(8)& y & =& \mathbf{9}&\text{Divided each side by 13}\\& 8(7) + 5(9) + z & = & 169 & \text{Substituted (8) and (7) into (6)} \\& 56 + 45 + z& = & 169 & \text{Simplified} \\& 101 + z& = & 169 & \text{Simplified} \\&z& = & \mathbf{68} & \text{Subtracted 101 from each side}\\\end{array}

\boxed{\mathbf{ x = 7, y = 9, z = 68}}

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