Advertisements
Advertisements
प्रश्न
Find AB and BC, if:
Advertisements
उत्तर
Let BC = x m
BD = BC + CD = (x + 20) cm
In ΔABD,
tan 45° = `"AB"/"BD"`
1 = `"AB"/(x + 20)`
x + 20 = AB ...(1)
In ΔABC
tan 60° = `"AB"/"BC"`
`sqrt(3) = "AB"/x`
x = `"AB"/sqrt(3)` ...(2)
From (1)
`"AB"/sqrt(3) + 20 = "AB"`
AB + `20sqrt(3) = sqrt(3)"AB"`
AB`(sqrt(3) -1) = 20sqrt(3)`
AB = `(20sqrt(3))/((sqrt(3) - 1)`
AB = `(20sqrt(3))/((sqrt(3) - 1)) xx ((sqrt(3)+1))/((sqrt(3)+1)`
AB = `(20sqrt(3)(sqrt(3) + 1))/(3 - 1)`
AB = 47.32 cm
From (2)
x = `"AB"/sqrt(3)`
x = `(47.32)/sqrt(3)`
x = 27.32 cm
∴ BC = x = 27.32 cm
Therefore, AB = 47.32 cm, BC = 27.32 cm.
APPEARS IN
संबंधित प्रश्न
Find 'x', if :
Find angle 'A' if :
Find angle 'A' if :
Find AD, if :
Find the length of AD.
Given: ∠ABC = 60o.
∠DBC = 45o
and BC = 40 cm.
In trapezium ABCD, as shown, AB // DC, AD = DC = BC = 20 cm and A = 60°. Find: distance between AB and DC.
Use the information given to find the length of AB.
In the given figure, AB and EC are parallel to each other. Sides AD and BC are 2 cm each and are perpendicular to AB.
Given that ∠ AED = 60° and ∠ ACD = 45°. Calculate: AE.
Find: AD
In right-angled triangle ABC; ∠B = 90°. Find the magnitude of angle A, if:
BC is `sqrt(3)` times of AB.
