Roger is going cliff diving and is standing on the edge of a 46 ft high cliff. he can ee his two dogs on the beach below. from the same spot, the angle of depression to his dogs are 34 degrees and 48 degrees, how far apart are the dogs?

the distance between two dogs is 26.77921851 feet.

Step-by-step explanation:

Suppose Roget is standing on the edge of a 46 ft high cliff (AX). So AX = 46 feet.

He can see his two dogs on the beach below. From the same spot, the angle of depression to his dogs are 34 degrees (at point B) and 48 degrees (at point C).

It means ∠ABX = 34° and ∠ACX = 48°.

Considering Right triangle ΔAXB, cot(B) = XB / AX.

XB = AX*cot(B) = 46*cot(34°)

Considering Right triangle ΔAXC, cot(C) = XC / AX.

XC = AX*cot(C) = 46*cot(48°)

We know BC = XB-XC.

BC = 46*cot(34°) - 46*cot(48°)

BC = 46 * {cot(34°) - cot(48°)}

BC = 46 * 0.582156924

BC = 26.77921851 feet.

Hence, the distance between two dogs is 26.77921851 feet.

it is only things thain can be deafeated

step-by-step explanation:

cant see photo

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