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Question Number 7142 by Tawakalitu. last updated on 13/Aug/16

T h e s i d e s o f a t r i a n g l e a r e x c m, (x - 4) c m, (x - 8) c m

r e s p e c t i v e l y . i f t h e c o s i n e o t h e l a r g e s t i s \frac{1}{5},

c a l c u l a t e t h e a n g l e s o f t r i a n g l e .

Answered by Rasheed Soomro last updated on 13/Aug/16

∵ x, x - 4, x - 8 > 0 [m e a s u r e s o f t h e s i d e s m u s t b e p o s i t i v e]

∴ x > 8 a n d {i t}^{'} s t h e l a r g e s t .

L e t t h e o p p o s i t e a n g l e t o t h e l a r g e s t s i d e (w h o s e

m e a s u r e i s x) i s α . A c c o r d i n g t o c o s i n e l a w :

x^{2} = {(x - 4)}^{2} + {(x - 8)}^{2} - 2 (x - 4) (x - 8) \cos α

x^{2} = {(x - 4)}^{2} + {(x - 8)}^{2} - 2 (x - 4) (x - 8) (\frac{1}{5}) [∵ \cos α = \frac{1}{5}]

5 x^{2} = 5 x^{2} - 40 x + 80 + 5 x^{2} - 80 x + 320 - 2 x^{2} + 24 x - 64

- 40 x + 80 + 5 x^{2} - 80 x + 320 - 2 x^{2} + 24 x - 64 = 0

3 x^{2} - 96 x + 336 = 0

x^{2} - 32 x + 112 = 0

x = \frac{32 \pm \sqrt{1024 - 448}}{2} = \frac{32 \pm \sqrt{576}}{2} = \frac{32 \pm 24}{2} = 16 \pm 12

x = 28 ∣ x = 4

x = 4 i s d i s c a r d a b l e s i n c e x > 8 [x = 4 m a k e s t h e s i d e s

o f t r i a n g l e 0 a n d n e g a t i v e]

S o, x = 28, x - 4 = 24 a n d x = 20

I - e t h e s i d e s o f t h e t r i a n g l e a r e 28, 24 a n d 20

L e t α, β a n d γ a r e o p p s i t e a n g l e s o f t h e s i d e s h a v i n g

m e a s u r e s 28, 24 a n d 20 r e s p e c t i v e l y .

\cos α = \frac{20^{2} + 24^{2} - 28^{2}}{2 \times 20 \times 24} = \frac{1}{5} \Rightarrow α = \cos^{- 1} \frac{1}{5} \approx 78.46 °

\cos β = \frac{20^{2} + 28^{2} - 24^{2}}{2 \times 20 \times 28} = \frac{19}{35} \Rightarrow β = \cos^{- 1} \frac{19}{35} \approx 57.12 °

\cos β = \frac{24^{2} + 28^{2} - 20^{2}}{2 \times 24 \times 28} = \frac{5}{7} \Rightarrow β = \cos^{- 1} \frac{5}{7} \approx 44.42 °

Commented by Tawakalitu. last updated on 13/Aug/16

T h a n k y o u s o m u c h s i r .