Advertisements
Advertisements
प्रश्न
In the given figure, AB and EC are parallel to each other. Sides AD and BC are 1.5 cm each and are perpendicular to AB. Given that ∠AED = 45° and ∠ACD = 30°. Find:
a. AB
b. AC
c. AE
Advertisements
उत्तर

a. In right ΔADC,
tan30° = `"AD"/"DC"`
⇒ `(1)/sqrt(3) = (1.5)/"DC"`
⇒ DC = `1.5sqrt(3)`
Since AB || DC and AD ⊥ EC, ABCD is a parallelogram and hence opposite sides are equal.
⇒ AB
= DC
= `1.5sqrt(3)"cm"`.
b. In right ΔADC,
sin30° = `"AD"/"AC"`
⇒ `(1)/(2) = (1.5)/"AC"`
⇒ AC
= 2 x 1.5
= 3cm.
c. In right ΔADE,
sin45° = `"AD"/"AE"`
⇒ `(1)/sqrt(2) = (1.5)/"AE"`
⇒ AE = `1.5sqrt(2)`.
APPEARS IN
संबंधित प्रश्न
From the given figure,
find:
(i) cos x°
(ii) x°
(iii) `(1)/(tan^2 xx°) – (1)/(sin^2xx°)`
(iv) Use tan xo, to find the value of y.
Solve the following equation for A, if 2 sin A = 1
Solve for x : 2 cos (3x − 15°) = 1
Solve for x : cos `(x/(2)+10°) = (sqrt3)/(2)`
Find the value of 'A', if 2cos 3A = 1
Solve for 'θ': `sec(θ/2 + 10°) = (2)/sqrt(3)`
If θ < 90°, find the value of: `tan^2θ - (1)/cos^2θ`
Find the value 'x', if:
If tan x° = `(5)/(12) . tan y° = (3)/(4)` and AB = 48m; find the length CD.
If sin(θ - 15°) = cos(θ - 25°), find the value of θ if (θ-15°) and (θ - 25°) are acute angles.
