Question

Isosceles ΔABC has a right angle at C. Point ‘P’ is inside ΔABC, such that PA = 11, PB = 7, and PC = 6. Legs AC and BC have length S = a + b 2, where a and b are positive integers. What is a + b ?. (Clue: Let D, E and F be the reflections of P about AB , BC , and CA respectively,….).

Answer 1

a + b = 127

Step-by-step explanation:

Given that,

PA = 11, PB = 7, and PC = 6

by using the cosines' law on ΔPBC;

PB^2 = BC^2 + PC^2 - 2 * BC * PC * cos(α)

⇒ 49 = 36 + s^2 - 12s cos(α)

⇒ cos(α) = (s^2 - 13)/12s

Using the same law for ΔPAC,

PA^2 = AC^2 + PC^2 - 2 * AC * PC * cos(90° - α)

⇒ 121 = 36 + s^2 - 12s sin(α)

⇒ sin(α)  = (s^2 - 85)/12s

Now, by using sin^2(α) + cos^2 (α) = 1

sin^2(α) + cos^2 (α) = 1

⇒ (s^4 - 26s^2 + 169)/144s^2 + (s^4 - 170s^2 + 7225)/144s^2 = 1

⇒ 2s^4 - 340s^2 + 7394 = 0

⇒ s^2 = (170 ± 84$\sqrt{2}$)/2 = 85 ± 42$\sqrt{2}$

∵ since s^2  > 85, while sin(α) > 0, a + b = 85 + 42 = 127.

Learn more: Find the ∠ABC

brainly.in/question/28563626

Answer 2

Answer:

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