11. In \( \triangle V W X, v=85 \mathrm{~cm}, x=58 \mathrm{~cm} \) and \( \angle X=107^{\circ} \). Find all possible values of \( \angle V \), to the nearest degree.
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Beyond the Answer
To solve for angle \( V \) in triangle \( VWX \), we can use the Law of Cosines. The formula states: \[ c^2 = a^2 + b^2 - 2ab \cdot \cos(C) \] In this case: - \( a = v = 85 \, \text{cm} \) - \( b = x = 58 \, \text{cm} \) - \( C = \angle X = 107^{\circ} \) First, let's find the length of side \( w \) (the side opposite angle \( V \)): \[ w^2 = v^2 + x^2 - 2 \cdot v \cdot x \cdot \cos(X) \] \[ w^2 = 85^2 + 58^2 - 2 \cdot 85 \cdot 58 \cdot \cos(107^{\circ}) \] \[ w^2 = 7225 + 3364 - 2 \cdot 85 \cdot 58 \cdot (-0.2924) \quad (\text{using } \cos(107^{\circ}) \approx -0.2924) \] \[ w^2 = 10589 + 85 \cdot 58 \cdot 0.5848 \] Calculating \( 85 \cdot 58 \cdot 0.5848 \): \[ 85 \cdot 58 \approx 4930 \] \[ 4930 \cdot 0.5848 \approx 2882.5 \] Now adding to calculate \( w^2 \): \[ w^2 \approx 10589 + 2882.5 \approx 13471.5 \implies w \approx 116 \, \text{cm} \] Now, let's find angle \( V \) using the Law of Sines: \[ \frac{v}{\sin(V)} = \frac{w}{\sin(X)} \] Rearranging gives: \[ \sin(V) = \frac{v \cdot \sin(X)}{w} \] Substituting the values: \[ \sin(V) = \frac{85 \cdot \sin(107^{\circ})}{116} \] Calculating \( \sin(107^{\circ}) \approx 0.9511 \): \[ \sin(V) = \frac{85 \cdot 0.9511}{116} \approx \frac{80.85}{116} \approx 0.696 \implies V \approx \arcsin(0.696) \approx 44^{\circ} \] However, since \( \sin \) can have two solutions, we must also consider the supplementary angle: \[ V' = 180^{\circ} - 44^{\circ} \approx 136^{\circ} \] Thus, the possible values for \( \angle V \) are \( 44^{\circ} \) and \( 136^{\circ} \).
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