A right spherical triangle ABC of a sphere of radius 50 cm has angle C = 90°, A = 56° and side b = 47°. Find the area of the triangle in square centimeter. A. 941.2 C. 786.3 B. 986.1 D. 652.4 Solution Click here to… Jhun Vert Sat, 06/29/2024 - 21:42 Solution Click here to expand or collapse this section SIN-COOP Rule $\sin \bar{B} = \cos \bar{A} \cos b$ $\sin (90^\circ - B) = \cos (90^\circ - 56^\circ) \cos 47^\circ$ $B = 55.57^\circ$ Spherical excess $E = A + B + C - 180^\circ$ $E = 56^\circ + 55.57^\circ + 90^\circ - 180^\circ$ $E = 21.57^\circ$ $\text{Area} = \dfrac{\pi R^2 E}{180^\circ} = \dfrac{\pi(50^2)(21.57^\circ)}{180^\circ}$ $\text{Area} = 941.17 ~ \text{cm}^2$ Log in or register to post comments Log in or register to post comments
Solution Click here to… Jhun Vert Sat, 06/29/2024 - 21:42 Solution Click here to expand or collapse this section SIN-COOP Rule $\sin \bar{B} = \cos \bar{A} \cos b$ $\sin (90^\circ - B) = \cos (90^\circ - 56^\circ) \cos 47^\circ$ $B = 55.57^\circ$ Spherical excess $E = A + B + C - 180^\circ$ $E = 56^\circ + 55.57^\circ + 90^\circ - 180^\circ$ $E = 21.57^\circ$ $\text{Area} = \dfrac{\pi R^2 E}{180^\circ} = \dfrac{\pi(50^2)(21.57^\circ)}{180^\circ}$ $\text{Area} = 941.17 ~ \text{cm}^2$ Log in or register to post comments
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