Math Question and Answers


Question and Answers Forum

All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
Question Number 11915 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 04/Apr/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 04/Apr/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 04/Apr/17

$s e e d i a g r a m ⇑⇑⇑ . (n o t i n s c a l e)$
	Answered by mrW1 last updated on 05/Apr/17

	$∡ B = 180 - 62 - 46 = 72$ $w i t h A C a s x - a x i s w e h a v e$ $A (0, 0)$ $C (10, 0)$ $E q . o f l i n e A B : y = (\tan 62) x$ $E q . o f l i n e C B : y = (- \tan 46) (x - 10)$ $(\tan 62) x_{B} = (- \tan 46) (x_{B} - 10)$ $\Rightarrow x_{B} = \frac{(\tan 46) \times 10}{\tan 62 + \tan 46} = 3.551$ $\Rightarrow y_{B} = (\tan 62) \times 3.551 = 6.678$ $E q . o f l i n e A D : y = (\tan 31) x$ $E q . o f l i n e B D : y - y_{B} = (- \tan 70) (x - x_{B})$ $(\tan 31) x_{D} - y_{B} = (- \tan 70) (x_{D} - x_{B})$ $\Rightarrow x_{D} = \frac{(\tan 70) x_{B} + y_{B}}{\tan 31 + \tan 70}$ $= \frac{(\tan 70) \times 3.551 + 6.678}{\tan 31 + \tan 70} = 4.908$ $\Rightarrow y_{D} = (\tan 31) \times 4.908 = 2.949$ $E q . o f l i n e C E : y = (- \tan 23) (x - 10)$ $E q . o f l i n e B E : y - y_{B} = (\tan 86) (x - x_{B})$ $(- \tan 23) (x_{E} - 10) - y_{B} = (\tan 86) (x_{E} - x_{B})$ $\Rightarrow x_{E} = \frac{(\tan 23) \times 10 - y_{B} + (\tan 86) x_{B}}{\tan 86 + \tan 23}$ $= \frac{(\tan 23) \times 10 - 6.678 + (\tan 86) \times 3.551}{\tan 86 + \tan 23} = 3.283$ $\Rightarrow y_{E} = (- \tan 23) (3.283 - 10) = 2.851$ $D E = \sqrt{{(x_{D} - x_{E})}^{2} + {(y_{D} - y_{E})}^{2}}$ $= \sqrt{{(4.908 - 3.283)}^{2} + {(2.949 - 2.851)}^{2}}$ $= 1.628$
	Commented by mrW1 last updated on 05/Apr/17

	Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 06/Apr/17

	$i n Δ A D B : \frac{B D}{s i n \frac{A}{2}} = \frac{A B}{s i n (180 - (\frac{A}{2} + \frac{2 B}{3}))} \Rightarrow$ $\Rightarrow B D = A B . \frac{s i n \frac{A}{2}}{s i n (\frac{A}{2} + \frac{2 B}{3})}$ $i n Δ C E B : \frac{B E}{s i n \frac{C}{2}} = \frac{C B}{s i n (180 - (\frac{C}{2} + \frac{2 B}{3}))} \Rightarrow$ $\Rightarrow B E = C B . \frac{s i n \frac{C}{2}}{s i n (\frac{C}{2} + \frac{2 B}{3})}$ $i n Δ A B C : \frac{A B}{s i n 46} = \frac{B C}{s i n 62} = \frac{A C}{s i n (180 - 46 - 62)} \Rightarrow$ $\Rightarrow A B = 10 \times \frac{s i n 46}{s i n 72} = 7.56, B C = 10 \times \frac{s i n 62}{s i n 72} = 9.28$ $\Rightarrow B D = 7.56 \times \frac{s i n 31}{s i n (31 + 48)} = 3.97$ $\Rightarrow B E = 9.28 \times \frac{s i n 23}{s i n (23 + 48)} = 3.83$ $i n Δ B D E : {D E}^{2} = {B D}^{2} + {B E}^{2} - 2 B D . B E . c o s (\frac{B}{3}) =$ $= 3 . 97^{2} + 3 . 83^{2} - 2 \times 3.97 \times 3.83 \times c o s (24)$ $\Rightarrow D E = 1.6274 ◼$
	Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 07/Apr/17

	$i f a n y o n e h a v e a n o t h e r s o l o t i o n b y u s i n g$ $g e o m e t r i c o r v e c t o r m e t h o d s, p l e a s e$ $p o s t i t h e r e .$
	Commented by mrW1 last updated on 05/Apr/17

	$I u s e d a n o t h e r w a y a n d g o t a d i f f e r e n t$ $r e s u l t, s e e b e l o w .$
	Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 05/Apr/17

	$t h a n k y o u v e r y m u c h d e a r m r W 1 .$ $y o u a r e r i g h t . t h i s i s m y f a u l t . a n s w e r$ $i s f i x e d . y o u r m e t h o d i s a b i t l o n g .$ $b u t i t i s s o p o w e r f o l a n d a m a z i n g .$ $t h a n k s a l o t .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum