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Question Number 200533 by ajfour last updated on 19/Nov/23

Commented by mr W last updated on 20/Nov/23

$A B ⊥ A D i s g i v e n ?$ $o t h e r w i s e t h e r e i s n o u n i q u e s o l u t i o n .$
Commented by mr W last updated on 20/Nov/23

Commented by ajfour last updated on 20/Nov/23

$p + q = 13 w o n t d o s i r, o b v i o u s!$
Commented by ajfour last updated on 20/Nov/23

$y e s s i r g i v e n A B ⊥ A D . f o r g o t t o$ $m a r k s o .$
Commented by a.lgnaoui last updated on 20/Nov/23

$look at the exact answer ↓$
	Answered by a.lgnaoui last updated on 20/Nov/23
	$NB=15−7=8 DE=MD=12−7=5 CE=p−5 CN=q−8 q^2 =OB^2 +OC^2 −2OB×OCcos 𝛌 (1) 𝛌=∡(BOC)=∡BON+∡NOC=𝛂+𝛃 tan (∡BON)=(8/7) tan(∡ NOC)=((q−8)/7) ⇒(1/(cos^2 α))=1+((64)/(49))=((113)/(49)) cos 𝛂=((7(√(113)))/(113)) (1/(cos^2 𝛃))=1+(((q−8)^2 )/7^2 )=((49+(q−8)^2 )/(49)) cos β=(7/( (√(49+(q−8)^2 )))) CE=CN ⇒p−5=q−8 ⇒ q−p=3 (2) (1) q^2 =[(49+64)]+[(49+(q−8)^2 ] 2[((√(113)) ×(√(49+q−8)^2 )) ]cos 𝛌 cos 𝛌=cos α𝛂cos 𝛃−sin 𝛂αsin 𝛃 { ((sin 𝛂=((8(√(113)))/( 113)))),((sin 𝛃 =((q−8)/( (√(49+(q−8)^2 )))))) :} ⇒q^2 =(113+49)+(q−8)^2 −2×(√(113))(((49−8(q−8))/( 1]))) q^2 =162+(q−8)^2 )−2(√(113)) [49−8(q−8)] q^2 =(q−8)^2 −16(√(113 )) (q−8) +98(√(113)) −162 posons x=q−8 (x+8)^2 −x^2 −16x(√(113)) +879,754=0 16x−16x(√(113)) +879,754=0 16x((√(113)) −1)=879,754 x=((879,754)/( 154))=5,712 ⇒q=8+5,712=13,712 (2)⇒p= q−3 ⇒ p=10,712 alors p+q=24,424$
	$NB = 15 - 7 = 8 DE = MD = 12 - 7 = 5$ $CE = p - 5 CN = q - 8$ $q^{2} = {OB}^{2} + {OC}^{2} - 2 OB \times OC \cos λ (1)$ $λ = ∡ (BOC) = ∡ BON + ∡ NOC = α + β$ $\tan (∡ BON) = \frac{8}{7} \tan (∡ NOC) = \frac{q - 8}{7}$ $\Rightarrow \frac{1}{\cos^{2} α} = 1 + \frac{64}{49} = \frac{113}{49} \cos α = \frac{7 \sqrt{113}}{113}$ $\frac{1}{\cos^{2} β} = 1 + \frac{{(q - 8)}^{2}}{7^{2}} = \frac{49 + {(q - 8)}^{2}}{49}$ $\cos β = \frac{7}{\sqrt{49 + {(q - 8)}^{2}}}$ $CE = CN$ $\Rightarrow p - 5 = q - 8 \Rightarrow q - p = 3 (2)$ $(1) q^{2} = [(49 + 64)] + [(49 + {(q - 8)}^{2}]$ $2 [(\sqrt{113} \times \sqrt{{49 + q - 8)}^{2}}] \cos λ$ $\cos λ = \cos α α \cos β - \sin α α \sin β$ ${\begin{cases} \sin α = \frac{8 \sqrt{113}}{113} \\ \sin β = \frac{q - 8}{\sqrt{49 + {(q - 8)}^{2}}} \end{cases}$ $\Rightarrow q^{2} = (113 + 49) + {(q - 8)}^{2}$ $- 2 \times \sqrt{113} (\frac{49 - 8 (q - 8)}{1]})$ $q^{2} = 162 + {(q - 8)}^{2}) - 2 \sqrt{113} [49 - 8 (q - 8)]$ $q^{2} = {(q - 8)}^{2} - 16 \sqrt{113} (q - 8)$ $+ 98 \sqrt{113} - 162$ $posons x = q - 8$ ${(x + 8)}^{2} - x^{2} - 16 x \sqrt{113} + 879, 754 = 0$ $16 x - 16 x \sqrt{113} + 879, 754 = 0$ $16 x (\sqrt{113} - 1) = 879, 754$ $x = \frac{879, 754}{154} = 5, 712$ $\Rightarrow q = 8 + 5, 712 = 13, 712$ $(2) \Rightarrow p = q - 3$ $\Rightarrow p = 10, 712$ $alors p + q = 24, 424$
	Commented by a.lgnaoui last updated on 20/Nov/23

	Commented by ajfour last updated on 20/Nov/23
	I ll check your solution in detail, there must be some mistake, perhaps. huge values!
	Answered by mr W last updated on 20/Nov/23

	Commented by mr W last updated on 20/Nov/23

	$7 (\tan α + \tan β) = 15$ $7 (\tan α + \tan γ) = 12$ $7 (\tan γ + \tan δ) = p$ $7 (\tan β + \tan δ) = q$ $p + q = 7 (\tan β + \tan γ + 2 \tan δ)$ $δ = π - (α + β + γ)$ $\tan δ = - \frac{\tan (α + β) + \tan γ}{1 - \tan (α + β) \tan γ}$ $= - \frac{\frac{\tan α + \tan β}{1 - \tan α \tan β} + \tan γ}{1 - \frac{\tan α + \tan β}{1 - \tan α \tan β} \times \tan γ}$ $= \frac{\tan α + \tan β + \tan γ - \tan α \tan β \tan γ}{\tan α \tan β + \tan β \tan γ + \tan γ \tan α - 1}$ $g i v e n : α = 45 °, \tan α = 1$ $\tan β = \frac{15}{7} - \tan α = \frac{8}{7}$ $\tan γ = \frac{12}{7} - \tan α = \frac{5}{7}$ $\tan δ = \frac{1 + \frac{8}{7} + \frac{5}{7} - 1 \times \frac{8}{7} \times \frac{5}{7}}{1 \times \frac{8}{7} + \frac{8}{7} \times \frac{5}{7} + \frac{5}{7} \times 1 - 1} = \frac{50}{41}$ $p = 7 \times (\frac{5}{7} + \frac{50}{41}) = \frac{555}{41}$ $q = 7 \times (\frac{8}{7} + \frac{50}{41}) = \frac{678}{41}$ $p + q = \frac{1233}{41} \approx 30.07$
	Commented by mr W last updated on 20/Nov/23

	Commented by ajfour last updated on 20/Nov/23

	$T h a n k s S i r, u t t e r g o o d p r e s e n t e d!$
	Commented by justenspi last updated on 21/Nov/23

	$Sir how can I contact you, I am available$ $through any social media app . Thanks in advance .$
	Commented by mr W last updated on 21/Nov/23

	$b u t i^{'} m a v a i l a b l e o n l y i n t h i s f o r u m .$ $w h e n y o u t h i n k i c a n h e l p y o u, t h e n$ $p o s t y o u r q u e s t i o n s h e r e . i^{'} l l t r y m y$ $b e s t .$
	Commented by justenspi last updated on 21/Nov/23

	$T h a n k s s i r r e a l l y a p p r e c i a t e u r t i m e$ $I a m i n m i d d l e s c h o o l, I w a n t e d t o$ $a s k y o u h o w t o g e t b e t t e r a t t r i g o n o m e t r y$ $a n d g e o m e t r y i n t h a t m a n n e r, a r e t h e r e a n y$ $b o o k s o r p r o b l e m b o o k s I s h o u l d d o!$
	Commented by mr W last updated on 21/Nov/23

	$i^{'} v e l e f t t h e s c h o o l f o r a l o n g t i m e,$ $i r e a l l y {d o n}^{'} t k n o w w h a t b o o k s i c a n$ $r e c o m m e n d .$
	Commented by justenspi last updated on 21/Nov/23

	$O k a y s i r, t h a n k s$ 😊
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