Math, asked by bathing, 1 year ago

In ∆ABC, If r1=8, r2=12, r3=24,Show that a=12,b=16,c=20. (From Properties of Triangles)​

Answers

Answered by mathsdude85
1

Step-by-step explanation:

Given, r₁ = 8, r₂ = 12, r₃ = 24.

∴ Derivative of (1/r) = (1/r₁) + (1/r₂) + (1/r₃)

⇒ (1/r) = (1/8) + (1/12) + (1/24)

⇒ (1/r) = (3 + 2 + 1)/24

⇒ r = 4.

Now,

∴ Δ = √rr₁r₂r₃

⇒ Δ = √4 * 8 * 12 * 24

       = 96

(i)

r = Δ/s

⇒ 4 = 96/s

⇒ 4s = 96

⇒ s = 24.

(ii)

r₁ = Δ/s - a

⇒ 8 = 96/24 - a

⇒ 192 - 8a = 96

⇒ a = 12

(iii)

r₂ = Δ/s - b

⇒ 12 = 96/24 - b

⇒ 288 - 12b = 96

⇒ b = 16

(iv)

r₃ = Δ/s - c

⇒ 24 = 96/24 - c

⇒ 576 - 24c = 96

⇒ 480 = 24c

⇒ c = 20.

Therefore, a = 12, b = 16, c = 20.

Hope it helps!

Answered by OJASVI22
0

Answer:

Step-by-step explanation:

Given lengths represent 'Pythagorean triplet' (18²+24²)=(30)²

a = 18, b = 24, c = 30 gives C = 90 degrees as a^2 + b^2 = c^2 then

cos A = b/c = 24/30 = 4/5 ANSWER

verify trigonometrically

cos A = (b^2 + c^2 -- a^2) / 2bc = (24^2 + 30^2 -- 18^2) / 2*24*30 = 1152/1440 = 4/5

cos B = a/c = 18/30 = 3/5 ANSWER

cos C = cos (90) = 0 ANSWER

sin A = a/c = 18/30 = 3/5 ANSWER

sin B = b/c = 24/30 = 4/5 ANSWER

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