Two points A and B lie on opposite sides of a river. Another point C is located on the same side of the river as B at a distance of 240 ft from B. If angle ABC is 100° and angle ACB is 25°, find the distance across the river from A to B. (Round your answer to two decimal places.)

Answer:128.98 feetStep-by-step explanation:By the given information,Suppose ABC is a triangle,According to the question,BC = 240 feet,m∠ABC = 100°,m∠ACB = 25°,∵ m∠ACB + m∠ABC + m∠BAC = 180°,25° + 100° + m∠BAC = 180°,125° + m∠BAC = 180°,⇒ m∠BAC = 55°,Now the law of sine,[tex]\frac{\sin C}{AB}=\frac{\sin A}{BC}[/tex][tex]\frac{\sin 25^{\circ}}{AB}=\frac{\sin 55^{\circ}}{240}[/tex][tex]\implies AB = \frac{250 \sin 25^{\circ}}{\sin 55^{\circ}}\approx 128.98[/tex]Hence, the distance between A and B would be 128.98 ft ( approx )