Question

From the top of a light house, the angles of depression of a ship sailing towards it was found
to be 45°. After 10 minutes, the angle of depression changes to 60°. Assuming that the ship

is sailing at uniform speed, find how much time it will take to reach the light house.

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Answer 1

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Let A and B be the two positions of the ship. Let d be the distance travelled by the ship during the period of observation i.e. AB = d metres.

Let A and B be the two positions of the ship. Let d be the distance travelled by the ship during the period of observation i.e. AB = d metres.Let the observer be at O, the top of the lighthouse PO.

Let A and B be the two positions of the ship. Let d be the distance travelled by the ship during the period of observation i.e. AB = d metres.Let the observer be at O, the top of the lighthouse PO.It is given that PO = 100 m and the angle of depression from O of A and B are 30 ∘ and 45 ∘ respectively.

Let A and B be the two positions of the ship. Let d be the distance travelled by the ship during the period of observation i.e. AB = d metres.Let the observer be at O, the top of the lighthouse PO.It is given that PO = 100 m and the angle of depression from O of A and B are 30 ∘ and 45 ∘ respectively.∴∠OAP=30 ∘

Let A and B be the two positions of the ship. Let d be the distance travelled by the ship during the period of observation i.e. AB = d metres.Let the observer be at O, the top of the lighthouse PO.It is given that PO = 100 m and the angle of depression from O of A and B are 30 ∘ and 45 ∘ respectively.∴∠OAP=30 ∘ and∠OBP=45 ∘

Let A and B be the two positions of the ship. Let d be the distance travelled by the ship during the period of observation i.e. AB = d metres.Let the observer be at O, the top of the lighthouse PO.It is given that PO = 100 m and the angle of depression from O of A and B are 30 ∘ and 45 ∘ respectively.∴∠OAP=30 ∘ and∠OBP=45 ∘ In Δ OPB, we have

Let A and B be the two positions of the ship. Let d be the distance travelled by the ship during the period of observation i.e. AB = d metres.Let the observer be at O, the top of the lighthouse PO.It is given that PO = 100 m and the angle of depression from O of A and B are 30 ∘ and 45 ∘ respectively.∴∠OAP=30 ∘ and∠OBP=45 ∘ In Δ OPB, we havetan45 ∘ = BP OP

OP⇒1= BP 100

100

100 ⇒ BP = 100 m

100 ⇒ BP = 100 mIn Δ OPA, we have

100 ⇒ BP = 100 mIn Δ OPA, we have⇒tan30 ∘ =APOP

100 ⇒ BP = 100 mIn Δ OPA, we have⇒tan30 ∘ =APOP⇒ 31 = d+BP100

100 ⇒ BP = 100 mIn Δ OPA, we have⇒tan30 ∘ =APOP⇒ 31 = d+BP100

100 ⇒ BP = 100 mIn Δ OPA, we have⇒tan30 ∘ =APOP⇒ 31 = d+BP100 ⇒ d + BP = 100 3

100 ⇒ BP = 100 mIn Δ OPA, we have⇒tan30 ∘ =APOP⇒ 31 = d+BP100 ⇒ d + BP = 100 3⇒ d + BP = 100 3

100 ⇒ BP = 100 mIn Δ OPA, we have⇒tan30 ∘ =APOP⇒ 31 = d+BP100 ⇒ d + BP = 100 3⇒ d + BP = 100 3⇒ d = 100 3 - 100⇒ d = 100( 3 1) = 100(1.732 - 1) = 73.2 m.

Hence, the distance travelled by the ship from A to B is 73.2 m.

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Answer 2

Step-by-step explanation:

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