Two sides and an angle are given below. Determine whether the given information results in one triangle, two triangles, or no triangle at all. Solve any triangle(s) that results. b equals 6 comma c equals 8 comma Upper B equals 30 degrees

The given information results in a two triangles.

The first one with sides ;

a = 11.4, b = 6 and c = 8

Angle A = 108, B = 30 and C = 42

The second one with sides;

a = 10, b = 6 and c = 8

Angle A = 90, B = 30 and C = 60.

Step-by-step explanation: With the information provided, we can find the size of the angle opposite the third side which we shall call side a, and then calculate the length of side a. We use the sine rule which states that

a/SinA = b/SinB = c/SinC

6/Sin 30 = 8/SinC

By cross multiplication we now have

SinC = (8 x Sin 30)/6

SinC = (8 x 0.5)/6

SinC = 4/6

SinC = 0.6667

C = 41.8128

Approximately, angle C is 42 degrees.

That makes angle A to be

A = 180 - (42 + 30) {Sum of angles in a triangle equals 180}

A = 108

Therefore to calculate length a using the sine rule;

a/SinA = b/SinB

a/Sin 108 = 6/Sin 30

a = (6 x Sin 108)/Sin 30

a = (6 x 0.9510)/0.5

a = 5.706/0.5

a = 11.412

Approximately line a equals 11.4 units

Therefore the sides of the first triangle becomes 6, 8 and 11.4 units while it’s angles are 30, 42 and 108 degrees.

Also, if two sides are given as 6 units and 8 units respectively, then the third side can be calculated by using the Pythagoras theorem and this immediately presumes that it’s a right angled triangle (one of the angles equals 90 degrees). The theorem states that;

AC^2 = AB^2 + BC^2

Where AC is the hypotenuse/longest side. Taking the other two as 6 and 8 units, the formula now becomes;

AC^2 = 6^2 + 8^2

AC^2 = 36 + 64

AC^2 = 100

Add the square root sign to both sides of the equation

AC = 10.

Therefore the sides of the second triangle becomes 6, 8 and 10 units with angles as 30, 60 and 90 degrees.

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step-by-step explanation:

step-by-step explanation:

hans did not square the 40-unit side; he merely multiplied it by 2.   incorrect.