A shopper is standing on level ground 800 feet from the base of a 250 foot tall department store the shopper looks up and sees a
flag on the stores roof to the nearest degree what is the angle of elevation to the top of the building from the point on the ground where the shopper is standing?
1 answer:
Answer:
The elevation to the top of the building from the point on the ground where the shopper is standing is = 17.35°
Step-by-step explanation:
From the ΔABC
AB = height of the building = 250 feet
BC = 800 ft
Angle of elevation =
= 0.3125
°
Therefore the elevation to the top of the building from the point on the ground where the shopper is standing is = 17.35°
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(C) 6 + 3√3
<u>Explanation:</u>
Area of the square = 3
a X a = 3
a² = 3
a = √3
Therefore, QR, RS, SP, PQ = √3
ΔBAC ≅ ΔBQR
Therefore,
In ΔBAC, BA = AC = BC because the triangle is equilateral
So,
BQ = √3
So, BQ, QR, BR = √3 (equilateral triangle)
Let AP and SC be a
So, AQ and RC will be 2a
In ΔAPQ,
(AP)² + (QP)² = (AQ)²
(a)² + (√3)² = (2a)²
a² + 3 = 4a²
3 = 3a²
a = 1
Similarly, in ΔRSC
(SC)² + (RS)² = (RC)²
(a)² + (√3)² = (2a)²
a² + 3 = 4a²
3 = 3a²
a = 1
So, AP and SC = 1
and AQ and RC = 2 X 1 = 2
Therefore, perimeter of the triangle = BQ + QA + AP + PS + SC + RC + BR
Perimeter = √3 + 2 + 1 + √3 + 1 + 2 + √3
Perimeter = 6 + 3√3
Therefore, the perimeter of the triangle is 6 + 3√3