Math  /  Trigonometry

QuestionTrigonomet'y Homework
A flagpole [GH][G H], shown in the diagram, is vertical and the ground is inclined at an angle of 55^{\circ} to the horizontal between EE and GG. The angles of elevation from EE and FF to the top of the pole are 3535^{\circ} and 5252^{\circ} respectively. The distance from EE to FF along the incline is 6 m . Find how far FF is from the base of the pole (G)(G) along the incline. Give your answer correct to two decimal places.

Studdy Solution

STEP 1

1. The flagpole is vertical.
2. The ground is inclined at an angle of 55^\circ to the horizontal.
3. The angles of elevation from points EE and FF to the top of the pole are 3535^\circ and 5252^\circ respectively.
4. The distance from EE to FF along the incline is 66 meters.

STEP 2

1. Use trigonometry to find the height of the flagpole.
2. Use the Law of Sines to find the distance from FF to GG.

STEP 3

Use the angle of elevation from EE to find the height of the flagpole. The angle of elevation is 3535^\circ, and the ground is inclined at 55^\circ.
The effective angle of elevation from the horizontal is:
θ=355=30 \theta = 35^\circ - 5^\circ = 30^\circ
Using the tangent function:
tan(30)=Height of flagpoleDistance from E to G \tan(30^\circ) = \frac{\text{Height of flagpole}}{\text{Distance from } E \text{ to } G}
Let h h be the height of the flagpole and d d be the distance from E E to G G .
h=dtan(30) h = d \cdot \tan(30^\circ)
STEP_1.1: Use the angle of elevation from FF to find another expression for the height of the flagpole. The angle of elevation is 5252^\circ, and the ground is inclined at 55^\circ.
The effective angle of elevation from the horizontal is:
θ=525=47 \theta = 52^\circ - 5^\circ = 47^\circ
Using the tangent function:
tan(47)=hDistance from F to G \tan(47^\circ) = \frac{h}{\text{Distance from } F \text{ to } G}
Let x x be the distance from F F to G G .
h=xtan(47) h = x \cdot \tan(47^\circ)

STEP 4

Use the Law of Sines to find the distance from F F to G G .
The triangle EFG \triangle EFG has angles EFG=1804730=103 \angle EFG = 180^\circ - 47^\circ - 30^\circ = 103^\circ .
Using the Law of Sines:
6sin(103)=xsin(30) \frac{6}{\sin(103^\circ)} = \frac{x}{\sin(30^\circ)}
Solve for x x :
x=6sin(30)sin(103) x = \frac{6 \cdot \sin(30^\circ)}{\sin(103^\circ)}
Calculate x x :
x60.50.974373.08 x \approx \frac{6 \cdot 0.5}{0.97437} \approx 3.08
The distance from F F to G G along the incline is approximately:
3.08 meters \boxed{3.08 \text{ meters}}

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