Advertisements
Advertisements
प्रश्न
Find x and y, in each of the following figure:
Advertisements
उत्तर
In right ΔABC,
tan45° = `"AB"/"BC"`
⇒ 1 = `x/(15 + y)`
⇒ x = 15 + y ....(i)
In right ΔABD,
tan60° = `"AB"/"BD"`
⇒ `sqrt(3) = x/y`
⇒ `sqrt(3) = (15 + y)/y` ....[From (i)]
⇒ `sqrt(3)y` = 15 + y
⇒ `sqrt(3)y - y` = 15
⇒ `y(sqrt(3) - 1)` = 15
⇒ y = `(15)/(sqrt(3) - 1)`
⇒ y = `(15)/(sqrt(3) - 1) xx (sqrt(3) + 1)/(sqrt(3) + 1`
= `(15(sqrt(3) + 1))/(3 - 1)`
= `(15(sqrt(3) + 1))/(2)"cm"`
⇒ x = `15 + (15(sqrt(3) + 1))/(2)`
= `(30 + 15(sqrt(3) + 1))/(2)`
= `(15(2 + sqrt(3) + 1))/(2)`
= `(15(3 + sqrt(3)))/(2)`
= `(15sqrt(3)(sqrt(3) + 1))/(2)`.
APPEARS IN
संबंधित प्रश्न
In ΔABC, ∠B = 90° , AB = y units, BC = `(sqrt3)` units, AC = 2 units and angle A = x°, find:
- sin x°
- x°
- tan x°
- use cos x° to find the value of y.
If sin 3A = 1 and 0 < A < 90°, find cos 2A
If sin 3A = 1 and 0 < A < 90°, find `tan^2A - (1)/(cos^2 "A")`
Solve for x : sin2 60° + cos2 (3x- 9°) = 1
In the given figure, PQ = 6 cm, RQ = x cm and RP = 10 cm, find
a. cosθ
b. sin2θ- cos2θ
c. Use tanθ to find the value of RQ
Find the value of 'x' in each of the following:
Find the length of EC.
In the given figure, ∠B = 60°, ∠C = 30°, AB = 8 cm and BC = 24 cm. Find:
a. BE
b. AC
If sec2θ = cosec3θ, find the value of θ if it is known that both 2θ and 3θ are acute angles.
Prove the following: `(tan(90° - θ)cotθ)/("cosec"^2 θ)` = cos2θ
