Advertisements
Advertisements
Question
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
Solution

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
RELATED QUESTIONS
Find the value of 'A', if 2cos 3A = 1
Solve for 'θ': `sec(θ/2 + 10°) = (2)/sqrt(3)`
Find the value of: `sqrt((1 - sin^2 60°)/(1 + sin^2 60°)` If 3 tan2θ - 1 = 0, find the value
a. cosθ
b. sinθ
Find the value 'x', if:
Find the value of 'y' if `sqrt(3)` = 1.723.
Given your answer correct to 2 decimal places.
Find x and y, in each of the following figure:
In the given figure; ∠B = 90°, ∠ADB = 30°, ∠ACB = 45° and AB = 24 m. Find the length of CD.
Evaluate the following: `(5sec68°)/("cosec"22°) + (3sin52° sec38°)/(cot51° cot39°)`
If sin(θ - 15°) = cos(θ - 25°), find the value of θ if (θ-15°) and (θ - 25°) are acute angles.
Prove the following: `(tan(90° - θ)cotθ)/("cosec"^2 θ)` = cos2θ
