Question-82729

Question Number 82729 by TawaTawa last updated on 23/Feb/20

Commented by Kunal12588 last updated on 23/Feb/20

I n t h e △ A B C, l e t G b e t h e c e n t r o i d, a n d l e t I

b e t h e c e n t e r o f t h e i n s c r i b e d c i r c l e . L e t α

a n d β b e t h e a n g l e s a t t h e v e r t i c e s A a n d B

r e s p e c t i v e l y . S u p p o s e t h a t t h e s e g m e n t I G ∥ A B

a n d t h a t β = 2 {t a n}^{- 1} (\frac{1}{3}) . F i n d α .

Commented by TawaTawa last updated on 23/Feb/20

In the triangle Δ ABC, let G be the centroid, and let I

be the centre of the inscribed circle . Let α and β be the

angles at the vertices A and B respectively . Suppose that

the segment IG is parallel to AB and that β = 2 \tan^{- 1} (\frac{1}{3}) .

Find α .

Answered by mr W last updated on 24/Feb/20

Commented by mr W last updated on 24/Feb/20

β = 2 \tan^{- 1} \frac{1}{3}

\tan \frac{β}{2} = \frac{1}{3}

\tan β = \frac{2 \times \frac{1}{3}}{1 - {(\frac{1}{3})}^{2}} = \frac{3}{4}

l e t r a d i u s o f i n c i r c l e = r = 1

B F = \frac{I F}{\tan \frac{β}{2}} = \frac{r}{\frac{1}{3}} = 3 r = 3

K F = \frac{I F}{\tan β} = \frac{r}{\frac{3}{4}} = \frac{4 r}{3} = \frac{4}{3}

B K = B F - K F = 3 - \frac{4}{3} = \frac{5}{3}

\frac{C K}{B K} = \frac{C G}{E G} = \frac{2}{1}

\Rightarrow C K = 2 B K = \frac{10}{3}

B C = B K + C K = \frac{5}{3} + \frac{10}{3} = 5

F C = B C - B F = 5 - 3 = 2

\tan \frac{γ}{2} = \frac{I F}{F C} = \frac{1}{2}

\tan γ = \frac{2 \times \frac{1}{2}}{1 - {(\frac{1}{2})}^{2}} = \frac{4}{3} = \frac{1}{\tan β}

\tan β \times \tan γ = 1

\Rightarrow β + γ = 90 °

α = 180 ° - β - γ

\Rightarrow α = 90 °

Commented by TawaTawa last updated on 24/Feb/20

Wow, God bless you sir . I really appreciate your time .

Commented by TawaTawa last updated on 24/Feb/20

Sir from the diagram . How is IK = BE

Commented by mr W last updated on 24/Feb/20

w h y d o y o u t h i n k I K = B E ?

Commented by mr W last updated on 24/Feb/20

y o u c a n c a l c u l a t e I K = \sqrt{1^{2} + {(\frac{4}{3})}^{2}} = \frac{5}{3}

a n d B E = \frac{A B}{2} = \frac{4}{2} = 2 . t h e r e f o r e I K \neq B E .

Commented by TawaTawa1 last updated on 24/Feb/20

sorry sir, i mean parallel to BE .

Commented by mr W last updated on 24/Feb/20

I t i s g i v e n : I G i s p a r a l l e l t o A B!

Commented by TawaTawa1 last updated on 24/Feb/20

Oh, thank you sir, i appreciate .