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

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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Algebra 2

Complex conjugate theorem

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Q. 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:

Apply Law of Sines: To find the possible values of $\angle X$ , 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: $\newline$ $\frac{\sin(W)}{w} = \frac{\sin(X)}{x}$ $\newline$ Substituting the given values, we get: $\newline$ $\frac{\sin(110°)}{9.5} = \frac{\sin(X)}{2.1}$
Calculate $\sin(X)$ : Now we solve for $\sin(X)$ :
$\sin(X) = (\sin(110°) \times 2.1) / 9.5$
Calculating the value of $\sin(110°)$ and then multiplying by $2.1$ and dividing by $9.5$ , we get:
$\sin(X) \approx (0.9397 \times 2.1) / 9.5$
$\sin(X) \approx 0.2071$
Find Angle X: Next, we find the angle $X$ by taking the inverse sine (arcsin) of $\sin(X)$ : $\newline$ $X \approx \arcsin(0.2071)$ $\newline$ Calculating the arcsin of $0.2071$ , we get: $\newline$ $X \approx 11.9^\circ$
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 $180^\circ$ . We already have one angle, $\angle W$ , which is $110^\circ$ . Adding the smallest possible value for $\angle X$ , which is $11.9^\circ$ , we get: $\newline$ $110^\circ + 11.9^\circ = 121.9^\circ$ $\newline$ This leaves $180^\circ - 121.9^\circ = 58.1^\circ$ for the third angle, which is a valid value for an angle in a triangle. Therefore, both values for $\angle X$ are possible.