Question

Let ABC be a right-angled triangle with ∠B = 90◦.Let I be the incentre of ABC. Draw a line perpendicular to AI at I. Let it intersect the line CB at D. Prove that CI is perpendicular to AD and prove that ID =Sqrt(b(b − a)) ,where BC = a and CA = b

Answer 1

Answer:

This is the required answer...

Answer 2

Answer:

Given, 12 * (sin Ѳ) = 5 * (cos Ѳ) = a (say).

So, cot Ѳ = (cos Ѳ) / (sin Ѳ) = (12 / 5).

Both sin Ѳ and cos Ѳ will have same sign.

sin Ѳ = (a / 12) and cos Ѳ = (a / 5).

Now, sin^2 Ѳ + cos^2 Ѳ = 1

(a^2) * (169) / (144 * 25) = 1

(a^2) = (144 * 25) / (169)

a = ± (12 * 5) / 13

So, either sin Ѳ = (5 / 13) and cos Ѳ = (12 / 13)

or, sin Ѳ = (-5 / 13) and cos Ѳ = (-12 / 13).

Therefore, sin Ѳ + cos Ѳ - cot Ѳ

= (5 / 13) + (12 / 13) - (12 / 5) or, (-5 / 13) + (-12 / 13) - (12 / 5)

= (17 / 13) - (12 / 5) or, (-17 / 13) - (12 / 5)

= (-71 / 65) or, (-241 / 65).