Question-15212

Question Number 15212 by ajfour last updated on 08/Jun/17

Commented by ajfour last updated on 08/Jun/17

S o l u t i o n t o Q . 15175

Answered by ajfour last updated on 08/Jun/17

C D = \frac{b}{2}, A D = \frac{b}{2}

∠ H D I + ∠ C D I = \frac{π}{2}

∠ H D I + \frac{π}{2} - ∠ C = \frac{π}{2} \Rightarrow ∠ H D I = ∠ C

∠ H D L = \frac{C}{2}

H D = \frac{b}{2} \sin C \cos C

H I = \frac{b}{2} \sin^{2} C

J D = \frac{b}{2} \tan A, J H = J D - H D

J H = \frac{b}{2} (\tan A - \sin C \cos C)

∠ I J H = \boldsymbol α, \tan α = \frac{H I}{J H}

\tan α = \frac{\sin^{2} C}{\tan A - \sin C \cos C} \dots . (i)

∠ K J L = \frac{α}{2}

J K + K D = J D

\frac{\boldsymbol r}{\tan (\boldsymbol α / 2)} + \frac{\boldsymbol r}{\tan (\boldsymbol C / 2)} = \frac{\boldsymbol b}{2} \tan \boldsymbol A

s o,

\boldsymbol r = \frac{\boldsymbol b}{2} \tan \boldsymbol A [\frac{\tan (\boldsymbol α / 2) \tan (\boldsymbol C / 2)}{\tan (\boldsymbol α / 2) + \tan (\boldsymbol C / 2)}]

\dots \dots \dots \dots \dots \dots . \dots \dots \dots

\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} = \frac{225 + 196 - 169}{2 \times 15 \times 14}

\cos A = \frac{3}{5} \Rightarrow \tan A = \frac{4}{3}

\cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b} = \frac{169 + 225 - 196}{2 \times 13 \times 15}

= \frac{33}{65} .

\Rightarrow \sin C = \frac{56}{65} a n d \tan C = \frac{56}{33}

\tan α = \frac{\sin^{2} C}{\tan A - \sin C \cos C}

= \frac{{(56 / 65)}^{2}}{(4 / 3) - (56 / 65) (33 / 65)}

\tan α = 0.8285 \Rightarrow α = 0.692 r a d

C = \tan^{- 1} (56 / 33) = 1.0383

\boldsymbol r = \frac{(\boldsymbol b / 2) \tan \boldsymbol A}{[\frac{1}{\tan (α / 2)} + \frac{1}{\tan (C / 2}]}

= \frac{(\frac{15}{2} \times \frac{4}{3})}{[\frac{1}{\tan (0.346)} + \frac{1}{\tan (0.519)}]}

= \frac{10}{(2.774 + 1.75)} = 2.21 u n i t s ∙

Commented by mrW1 last updated on 08/Jun/17

Tip

when we know the sides of a triangle,

the radius of its inscribed circle r_{i}

can also be calculated like this :

Area of triangle A = \sqrt{s (s - a) (s - b) (s - c)}

but

A = \frac{1}{2} {ar}_{i} + \frac{1}{2} {br}_{i} + \frac{1}{2} {cr}_{i} = \frac{a + b + c}{2} \times r_{i} = {sr}_{i}

\Rightarrow r_{i} = \frac{\sqrt{s (s - a) (s - b) (s - c)}}{s} = \sqrt{\frac{(s - a) (s - b) (s - c)}{s}}

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 08/Jun/17

h e l l o m y f r i e n d : m r A j f o u r .

y o u r s o l u t i o n i s s m a r t a n d s i m p l e .

i l o v e i t s o m u c h . t h a n k s a l o t .

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 09/Jun/17

y e s s i r . w e h a v e : S = p . r .

a l s o : \frac{1}{h_{a}} + \frac{1}{h_{b}} + \frac{1}{h_{c}} = \frac{1}{r}