A tower that is 105 feet tall casts a shadow 110 feet long. Find the angle of elevation of the sun to the nearest degree.

Answer 1

Hi Quinn!

A diagram is very helpful in solving this problem.

\ /

-- O --

/ \

||\

|| \

105ft || \

tower || \

|| \

||________Θ_\

(shadow)

110ft

opp 105

TanΘ = ------ = ------ ≈ .9545

adj 110

TanΘ ≈ .9545

Θ ≈ arctan (.9545) ≈ 43.67^o

Answer 2

You should set up this problem by first drawing a sketch. This building and it's shadow will form a right triangle, with the vertical side being 105 ft, the horizontal side (the shadow on the ground) being 110 ft, and the hypotenuse being unknown at first. The angle we need to find is the angle between the hypotenuse and the horizontal side, representing the ground.

Once you have your sketch, I would start with finding the length of the hypotenuse. Since we know this is a right triangle, and we know 2 of the 3 side lengths, we can use Pythagorean Theorem to find the 3rd side.

a² + b² = c² Where c is the hypotenuse.

105² + 110² = c²

c = √(105² + 110²) = √(11,025 + 12,100) = √23,125

Now that we know all 3 sides of the triangle, we can use Law of Cosines to find the angle we need. Please note: the sides a, b, and c below DO NOT correspond with the sides represented by a, b, and c in the above section.

c² = a² + b² - 2ab·cos(C)

Make sure that the angle C is the angle opposite the triangle from side c. Since we are looking for the angle between the hypotenuse and the ground, side c in the equation should be the side representing the building.

c² = a² + b² - 2ab·cos(C)

105² = 110² + √23,125² - 2(110)(√23,125)·cos(C)

105² - 110² - 23,125 = -2(110)(√23,125)·cos(C)

[105² - 110² - 23,125]/[-2(110)(√23,125)] = cos(C)

C = cos^-1( [105² - 110² - 23,125] / [-2(110)(√23,125)] )

And you can just plug this into a calculator from here.

C ≈ 0.762 radians ≈ 43.67°