Advertisements
Advertisements
प्रश्न
From the top of a light house, an abserver looking at a boat makes an angle of depression of 600. If the height of the lighthouse is 90 m then find how far is the boat from the lighthouse. (3 = 1.73)
Advertisements
उत्तर
Let AB be the light house.
The boat is at C and observer is at A.
∠ MAC is the angle of depression.
∠ MAC = ∠ ACB = 60° .....(Alternate angle)
AB = 90 m.
From the figure, tan60° `= (AB)/(BC)`
`sqrt3 = (90)/(BC)`
`BC = (90)/(sqrt3) = (90 xx sqrt3)/(sqrt3 xxsqrt3) = (90sqrt3)/3 = 30sqrt3`
∴ BC = 30 × 1.73
∴ BC = 51.90
∴ The boat is at a distance of 51.90m from the light house.
APPEARS IN
संबंधित प्रश्न
Given below is the triangle and length of line segments. Identify in the given figure, ray PM is the bisector of ∠QPR.
Given below is the triangle and length of line segments. Identify in the given figure, ray PM is the bisector of ∠QPR.
In ∆MNP, NQ is a bisector of ∠N. If MN = 5, PN = 7 MQ = 2.5 then find QP.
Measures of some angles in the figure are given. Prove that `"AP"/"PB" = "AQ"/"QC"`.
Find QP using given information in the figure.
In the given figure, if AB || CD || FE then find x and AE.
In ∆LMN, ray MT bisects ∠LMN If LM = 6, MN = 10, TN = 8, then Find LT.
In ∆ABC, seg BD bisects ∠ABC. If AB = x, BC = x + 5, AD = x – 2, DC = x + 2, then find the value of x.
In the given fig, bisectors of ∠B and ∠C of ∆ABC intersect each other in point X. Line AX intersects side BC in point Y. AB = 5, AC = 4, BC = 6 then find `"AX"/"XY"`.
In ▢ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that `"AP"/"PD" = "PC"/"BP"`.
In Δ ABC and Δ PQR,
∠ ABC ≅ ∠ PQR, seg BD and
seg QS are angle bisector.
`If (l(AD))/(l(PS)) = (l(DC))/(l(SR))`
Prove that : Δ ABC ∼ Δ PQR
Seg NQ is the bisector of ∠ N
of Δ MNP. If MN= 5, PN =7,
MQ = 2.5 then find QP.
In ΔABC, ray BD bisects ∠ABC.
If A – D – C, A – E – B and seg ED || side BC, then prove that:
`("AB")/("BC") = ("AE")/("EB")`
Proof :
In ΔABC, ray BD bisects ∠ABC.
∴ `("AB")/("BC") = (......)/(......)` ......(i) (By angle bisector theorem)
In ΔABC, seg DE || side BC
∴ `("AE")/("EB") = ("AD")/("DC")` ....(ii) `square`
∴ `("AB")/square = square/("EB")` [from (i) and (ii)]
In the figure, ray YM is the bisector of ∠XYZ, where seg XY ≅ seg YZ, find the relation between XM and MZ.
In the following figure, ray PT is the bisector of ∠QPR Find the value of x and perimeter of ∠QPR.
Draw the circumcircle of ΔPMT in which PM = 5.6 cm, ∠P = 60°, ∠M = 70°.
In ΔABC, ray BD bisects ∠ABC, A – D – C, seg DE || side BC, A – E – B, then for showing `("AB")/("BC") = ("AE")/("EB")`, complete the following activity:
Proof :
In ΔABC, ray BD bisects ∠B.
∴ `square/("BC") = ("AD")/("DC")` ...(I) (`square`)
ΔABC, DE || BC
∴ `(square)/("EB") = ("AD")/("DC")` ...(II) (`square`)
∴ `("AB")/square = square/("EB")` ...[from (I) and (II)]
