Gilligan sees a ship coming close to the shore he's standing on. He wants to determine the distance (segment SD) from the ship to the shore. He walks 130 ft along the shore from point D to point C and marks that spot. Then he walks 23 ft further and marks point B. He turns 90°and walks until his location (point A), point C, and point S are collinear.

Answer the following questions making sure to show all your work.

(a) Can Gilligan conclude that triangle ABC and triangle SDC are similar? Why or why not?

(b) Suppose AB = 50 ft. What is the distance from the ship to the shore? Show all your work and round your final answer to the nearest tenth of a foot.

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oeerivona oeerivona · Answer 1 · 2022-02-24T12:01:56+01:00

The two triangles are both right triangles and are both formed by the

intersection of the lines AS and BD and are therefore similar.

Response:

(a) Yes, ΔABC is similar to ΔSDC according AA similarity postulate

(b) The distance from the ship to the shore is approximately 282.6 feet

(a) Yes, Gilligan can conclude that ΔABC and ΔSDC are similar by

Angle-Angle, AA, similarity postulate.

Reason:

Angle ∠SCD in ΔSDC and ∠ACB in ΔABC are vertically opposite angles,

and they are therefore congruent by vertical angles theorem.

∠SDC = 90° and ∠ABC = 90° therefore ∠SDC ≅ ∠ABC because all 90°

are congruent.

Therefore, two angles in ΔABC are congruent to two angles in ΔSDC,

therefore, ΔABC is similar to ΔSDC according to AA similarity postulate

(b) The ratio of corresponding sides in similar triangles are equal,

therefore;

[tex]\dfrac{BC}{DC} = \mathbf{\dfrac{AB}{SD}}[/tex]

Which gives;

[tex]\dfrac{23}{130} = \dfrac{50}{SD}[/tex]

[tex]SD = \dfrac{130}{23} \times 50 \approx 282.6[/tex]

Learn more about triangle similarity postulates here:

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