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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)
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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.
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In Δ ABC and Δ PQR,
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Prove that: `(AD)/(BD) = (AE^2)/(BE^2)`.
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Draw seg AB = 6.8 cm and draw perpendicular bisector of it.
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Draw the circumcircle of ΔPMT in which PM = 5.6 cm, ∠P = 60°, ∠M = 70°.
From the information given in the figure, determine whether MP is the bisector of ∠KMN.
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Complete the proof by filling in the boxes.
solution:
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Ray MX is the bisector of ∠PMQ.
∴ `("MP")/("MQ") = square/square` .............(I) [Theorem of angle bisector]
Similarly, in ∆PMR, Ray MY is the bisector of ∠PMR.
∴ `("MP")/("MR") = square/square` .............(II) [Theorem of angle bisector]
But `("MP")/("MQ") = ("MP")/("MR")` .............(III) [As M is the midpoint of QR.]
Hence MQ = MR
∴ `("PX")/square = square/("YR")` .............[From (I), (II) and (III)]
∴ XY || QR .............[Converse of basic proportionality theorem]
