Question

Vic is standing on the ground at a point directly south of the base of the CN Tower and can see the top when looking at an angle of elevation of 61°. Dan is standing on the ground at a point directly west of the base of the tower and must look up at an angle of elevation of 72° in order to see the top. If the CN Tower is 553.3 m tall,how far apart are Vic and Dan to the nearest meter? Include a well-labeled diagram as part of your solution.

Answer 1

The angle of elevation of 61° and 72° with the height of the tower being 553.3 m. gives Vic's distance from Dan as approximately 356 meters.

How can the distance between Vic and Dan be calculated?

Location of Vic relative to the tower = South

Vic's sight angle of elevation to the top of the tower = 61°

Dan's location with respect to the tower = West

Dan's angle of elevation in order to see the top of the tower = 72°

Height of the tower = 553.3m

$tan( \theta) = \frac{opposite}{adjacent}$

$tan( 61 ^{ \circ}) = \frac{553.3}{ Vic' s\: distance \: to \: tower}$

$distance = \frac{553.3}{tan ( {61}^{ \circ} )} = 306.7$

• Vic's distance from the tower ≈ 306.7 m

Similarly, we have;

$Dan's distance = \frac{553.3}{tan ( {72}^{ \circ} )} = 179.8$

• Dan's distance from the tower ≈ 179.8 m

Given that Vic and Dan are at right angles relative to the tower (Vic is on the south of the tower while Dan is at the west), by Pythagorean theorem, the distance between Vic and Dan d is found as follows;

• d = √(306.7² + 179.8²) ≈ 356

Therefore;

• Vic is approximately 356 meters from Dan

Learn more about Pythagorean theorem here:

brainly.com/question/343682

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