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Kobotan [32]
3 years ago
15

Find a side of an equilateral triangle with a 12 inch altitude

Mathematics
1 answer:
saveliy_v [14]3 years ago
8 0
Since the triangle is an equilateral triangle we know all of it's sides must be the same length, with that in mind the angles that make up the triangle must be equal as well. Knowing that a triangle's three interior angles make up 180 degrees we know that the size of each angle must be one third of this (as each angle must be equal).

180/3 = 60


then we may split the triangle along it's altitude into two special right triangles
more specifically two 30-60-90 triangles.

this means that the side with 30 degrees will be some value "x" where the side for 60 degrees will be related as it is "x*sqrt(3)" and the hypotenuse (which would be the side of the triangle) would be proportionally "2x"

this would mean that the altitude is the side associated with the 60 degree angle as such we can solve for "x" using this.

12= x*sqrt(3)
12/sqrt(3)=x
4sqrt(3)=x  (simplifying the radical we get "x" equals 4 square root 3)

now we may solve for the side length of the triangle which is "2x"

2*4sqrt(3) ->   8sqrt (3)  

eight square root of three is the answer.
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I'm thinking of a fraction that is equivalent to . The numerator is less than the​ denominator." What fraction is thinking​ of?
Vesna [10]

Answer:

Proper fraction

Step-by-step explanation:

If the numerator is less than the denominator , then it is proper fraction.

Example : \frac{5}{8},  \frac{124}{1000}

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3 years ago
5 PIONTS+BRAINLYESS
Lady_Fox [76]
The first one is the answer 0.9
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3 years ago
Read 2 more answers
Find the solution of this system of equations.
beks73 [17]
<span>x= -8 + y
-9x + 6y= 9

substitute </span>x= -8 + y into -9x + 6y= 9

-9x + 6y= 9
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<span>x= -8 + y
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7 0
3 years ago
Question 20 not yet answered marked out of 1.00 not flaggedflag question question text luna created a trash can in the shape of
Alex Ar [27]

Consider the triangle with sides a=1.6, b=2.3 and c=1.2 feet. You can use cosine theorem twice to find two unknown angles:

1.

b^2=a^2+c^2-2ac\cos \angle B,\\ (2.3)^2=(1.6)^2+(1.2)^2-2\cdot 1.6\cdot 1.2\cos \angle B,\\ 5.29=2.56+1.44-3.84\cos \angle B,\\ 3.84\cos \angle B=4-5.29=-1.29,\\\cos \angle B=-\dfrac{1.29}{3.84} =-0.336

Then you can conclude that ∠B is obtuse (because \cos \angle B) and m∠B=110°.

2.

c^2=a^2+b^2-2ab\cos \angle C,\\ (1.2)^2=(1.6)^2+(2.3)^2-2\cdot 1.6\cdot 2.3\cos \angle C,\\ 1.44=2.56+5.29-7.36\cos \angle C,\\ 7.36\cos \angle C=7.85-1.44=6.41,\\\cos \angle C=\dfrac{6.41}{7.36} =0.871.

Then you can conclude that ∠C is acute (because \cos \angle C>0) and m∠C=29°.

3. m∠A+m∠B+m∠C=180°, then m∠A=180°-110°-29°=41°.

Answer: m∠A=41° (∠A is opposite to the side a=1.6), m∠B=110° (∠B is opposite to the side b=2.3) and m∠C=29° (∠C is opposite to the side c=1.2)

3 0
3 years ago
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