17sin(θ+61.9)=k

Solución

17 sin (θ + 61. 9^{\circ}) = k

θ = arcsin (\frac{k}{17}) + 36 0^{\circ} n - 61. 9^{\circ}, θ = 18 0^{\circ} + arcsin (- \frac{k}{17}) + 36 0^{\circ} n - 61. 9^{\circ}

Pasos de solución

17 sin (θ + 61. 9^{\circ}) = k

Dividir ambos lados entre

17

sin (θ + 61. 9^{\circ}) = \frac{k}{17}

Aplicar propiedades trigonométricas inversas

θ + 61. 9^{\circ} = arcsin (\frac{k}{17}) + 36 0^{\circ} n, θ + 61. 9^{\circ} = 18 0^{\circ} + arcsin (- \frac{k}{17}) + 36 0^{\circ} n

Resolver

θ + 61. 9^{\circ} = arcsin (\frac{k}{17}) + 36 0^{\circ} n : θ = arcsin (\frac{k}{17}) + 36 0^{\circ} n - 61. 9^{\circ}

Resolver

θ + 61. 9^{\circ} = 18 0^{\circ} + arcsin (- \frac{k}{17}) + 36 0^{\circ} n : θ = 18 0^{\circ} + arcsin (- \frac{k}{17}) + 36 0^{\circ} n - 61. 9^{\circ}

θ = arcsin (\frac{k}{17}) + 36 0^{\circ} n - 61. 9^{\circ}, θ = 18 0^{\circ} + arcsin (- \frac{k}{17}) + 36 0^{\circ} n - 61. 9^{\circ}

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