Question

In the figure given alongside, AB = 6 cm, AC = 10 cm, ∠B = 90° and EC = 21 cm.

Find:

1. AD and AE
2. cos θ + sin θ
3. cot α + cosec α.

Answer 1

Given: △ABC is a right-angled triangle with AB = 6 cm, AC = 10 cm, ∠B = 90° and EC = 21 cm.

Step 1: Find BC using the Pythagorean theorem in △ABC.

AC² = AB² + BC²   ...[Using the Pythagorean theorem]

10² = 6² + BC²

100 = 36 + BC²

BC² = 100 − 36

BC² = 64

BC = `sqrt64`

BC = 8 cm

Step 2:
(i) Find AD

since ADCB forms a rectangle (implied by the right angles at B and D, and AD || BC), AD = BC.

Therefore, AD = 8 

Finding DE

DE = EC − DC

= 21 − 6

= 15

Find AE using the Pythagorean theorem in △ADE 

△ADE is a right-angled triangle with ∠D = 90°

AE² = AD² + DE²

AE² = 8² +15²

AE² = 64 + 225 = 289 

AE = `sqrt289` = 17 CM

AE = 17 cm`

Step 3:

(ii) Find cos θ + sinθ

In △ABC, cos θ = `"adjacent"/"hypotenuse" = (BC)/(AC) = 8/10 = 4/5`

Find sin θ in △ABC 

In △ABC, sin θ = `"opposite"/"hypotenuse" = (AB)/(AC) = 6/10 = 3/5`

cos θ + sin θ = `4/5 + 3/5 = 7/5`    ... [Calculating cos θ + sin θ]

cos θ + sin θ = `7/5` or 1.4

(iii) Find cot α + cosec α.

In △ADE, cot α = `"adjacent"/"opposite" = (DE)/(AD) = (15)/8`    ...[cot α in △ADE]

In △ADE, cosec α = `"hypotenuse"/"opposit" = (AE)/(AD)`   ... [Finding cosec α in △ADE]

= `17/8`

= cot α + cosec α    ... [calculating cot α + cosec α]

= `15/8 + 17/8`

= `32/8`

= 4

cot α + cosec α = 4