Question

The Daredevil cliffs rise vertically from the beach. The beach slopes gently at an angle of 3 feet from the horizontal. Laying at the water's edge, 50 feet from the base of the cliff, you determine your position, and a straight line to the top of the cliff creates an angle of 70 degrees with the line to the base of the cliff. How high is the daredevil cliff?

Answer 1

Answer:

160.70 ft. ( approx )

Step-by-step explanation:

Let AB represents the cliff, C represents Scott's position,

Then according to the question,

m∠ABC = 93°, ( because beach slopes gently at an angle of 3 feet from the horizontal )

m∠ACB = 70°,

BC = 50 ft,

∵ The sum of all interior angles of a triangle is 180°,

i.e. m∠ABC + m∠ACB + m∠BAC = 180°,

⇒ 93° + 70° + m∠BAC = 180°,

⇒ m∠BAC = 180° - 163° = 17°,

By the law of sine,

$\frac{\sin A}{BC}=\frac{\sin C}{AB}$

$\frac{\sin 17^{\circ}}{50}=\frac{\sin 70^{\circ}}{AB}$

$\implies AB = \frac{50 sin 70^{\circ}}{sin 17^{\circ}}\approx 160.70$

Hence, daredevil cliff is approximately 160.70 ft.