Advertisements
Advertisements
प्रश्न
Find x and y, in each of the following figure:
Advertisements
उत्तर
In right ΔABC,
tan30° = `"BC"/"AB"`
⇒ `(1)/sqrt(3) = x/(24 + y)` ....(i)
In right ΔDBC,
tan60° = `"BC"/"DB"`
⇒ `sqrt(3) = x/y`
⇒ x = `sqrt(3)y`
Substituting the value of x in (i), we get
`(1)/sqrt(3) = sqrt(3)/(24 + y)`
⇒ 24 + y = 3y
⇒ 2y = 24
⇒ y = 12cm
⇒ x = `sqrt(3) xx 12 = 12sqrt(3)"cm"`.
APPEARS IN
संबंधित प्रश्न
Solve for x : tan2 (x - 5°) = 3
Find the value of 'A', if 2 sin 2A = 1
Find the value of 'A', if `sqrt(3)cot"A"` = 1
Find the value of 'A', if (1 - cosec A)(2 - sec A) = 0
Find the value of 'A', if (2 - cosec 2A) cos 3A = 0
In a rectangle ABCD, AB = 20cm, ∠BAC = 60°, calculate side BC and diagonals AC and BD.
Evaluate the following: `(sec32° cot26°)/(tan64° "cosec"58°)`
Evaluate the following: sin(35° + θ) - cos(55° - θ) - tan(42° + θ) + cot(48° - θ)
Evaluate the following: `(3sin^2 40°)/(4cos^2 50°) - ("cosec"^2 28°)/(4sec^2 62°) + (cos10° cos25° cos45° "cosec"80°)/(2sin15° sin25° sin45° sin65° sec75°)`
Prove the following: sin58° sec32° + cos58° cosec32° = 2
