30.
the longest side of an acute triangle measures 30 inches. the two remaining sides are congruent, but their length is
unknown
what is the smallest possible perimeter of the triangle, rounded to the nearest tenth?
vx
0
41.0 in.
51.2 in.
72.4 in.
81.2 in.
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The smallest possible perimeter of the triangle, rounded to the nearest tenth is 72.4 in

Step-by-step explanation:

The Triangle Inequality Theorem states that the sum of any 2 sides of a triangle must be greater than the measure of the third side

Let

x > the length of  the remaining side

Applying the triangle inequality theorem

1) x+x > 30

2x > 30

x > 15 in

The perimeter is equal to

P=30+2x

Verify each case

1) For P=41.0 in

substitute in the formula of perimeter and solve for x

41.0=30+2x

2x=41.0-30

x=5.5 in

Is not a solution because the value of x must be greater than 15 inches

2) For P=51.2 in

substitute in the formula of perimeter and solve for x

51.2=30+2x

2x=51.2-30

x=10.6 in

Is not a solution because the value of x must be greater than 15 inches

3) For P=72.4 in

substitute in the formula of perimeter and solve for x

72.4=30+2x

2x=72.4-30

x=21.2 in

Could be a solution because the value of x is greater than 15 inches

4) For P=81.2 in

substitute in the formula of perimeter and solve for x

81.2=30+2x

2x=81.2-30

x=25.6 in

Could be a solution because the value of x is greater than 15 inches

therefore

The smallest possible perimeter of the triangle, rounded to the nearest tenth is 72.4 in

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47