In /_\VWX,x=2.1cm,w=9.5cm and /_W=110^(@). Find all possible values of /_X, to the nearest 1oth of a degree. Answer:

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In  /_\VWX,x=2.1cm,w=9.5cm and  /_W=110^(@). Find all possible values of  /_X, to the nearest 1oth of a degree. Answer:

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Apply Law of Sines: To find the possible values of /_X, we can use the Law of Sines, which relates the sides of a triangle to the sines of its opposite angles. The Law of Sines states that for any triangle ABC with sides a, b, and c opposite angles A, B, and C respectively, the following ratio holds true: (sin A)/a = (sin B)/b = (sin C)/c. We will apply this to triangle VWX.Find Side Length: First, we need to find the length of the side opposite to angle W, which is side v. We can use the Law of Sines to find this. We have: sin(W)/w = sin(X)/x Substituting the given values, we get: sin(110°)/9.5 = sin(X)/2.1Calculate sin(X): Now we solve for sin(X): sin(X) = (sin(110°) * 2.1) / 9.5 Calculating the value of sin(110°) and then multiplying by 2.1 and dividing by 9.5, we get: sin(X) ≈ (0.9397 * 2.1) / 9.5 sin(X) ≈ 0.2071Find Angle X: Next, we find the angle X by taking the inverse sine (arcsin) of sin(X): X ≈ arcsin(0.2071) Calculating the arcsin of 0.2071, we get: X ≈ 11.9°Find Second X Value: However, since the sine function is positive in both the first and second quadrants, there is another possible value for angle X in the second quadrant. To find this, we subtract the first quadrant angle from 180°: 180° - 11.9° = 168.1°Check Validity: We must check if this second possible value for angle X is valid in the context of a triangle. The sum of angles in any triangle is 180°. We already have one angle, /_W, which is 110°. Adding the smallest possible value for /_X, which is 11.9°, we get: 110° + 11.9° = 121.9° This leaves 180° - 121.9° = 58.1° for the third angle, which is a valid value for an angle in a triangle. Therefore, both values for /_X are possible.

In $\triangle \mathrm{VWX}, x=2.1 \mathrm{~cm}, w=9.5 \mathrm{~cm}$ and $\angle \mathrm{W}=110^{\circ}$ . Find all possible values of $\angle \mathrm{X}$ , to the nearest $10$ th of a degree. $\newline$ Answer:

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In $\triangle \mathrm{VWX}, x=2.1 \mathrm{~cm}, w=9.5 \mathrm{~cm}$ and $\angle \mathrm{W}=110^{\circ}$ . Find all possible values of $\angle \mathrm{X}$ , to the nearest $10$ th of a degree. $\newline$ Answer: