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Question Number 218475 by lmcp1203 last updated on 10/Apr/25

	Answered by Nicholas666 last updated on 10/Apr/25

	$∙ ∠ C B D + ∠ B D C + ∠ B C D = 180 °$ $∙ x + 140 ° + 30 ° = 180 °$ $∙ x + 170 ° = 180 °$ $∙ x = 180 ° - 170 °$ $∙ x = 10 °$
	Commented by mr W last updated on 10/Apr/25

	$h o w c a n y o u s a y ∠ B D C = 140 ° ?$ $b e s i d e s y o u h a v e i g n o r e d t h e g i v e n$ $c o n d i t i o n A C = B D .$
	Commented by Nicholas666 last updated on 10/Apr/25

	$y o u h a v e t o l o o k a t t h e q u e s t i o n c a r e f u l l y, i t i s v e r y o b v i o s$
	Commented by mr W last updated on 10/Apr/25

	$i t h i n k y o u {d i d n}^{'} t l o o k c a r e f u l l y .$ $w h e n y o u t o o k ∠ B D C = 140 °, y o u$ $a s s u m e d ∠ D B A = 40 ° .$ $o r h o w i s i t o b v i o u s t h a t ∠ B D C = 140 ° ?$
	Commented by Nicholas666 last updated on 10/Apr/25

	$I {d o n}^{'} t r e a l l y l i k e t y p i n g a t l e n g h t, b u t o k a y, I^{'} l l e x p l a i n t o y o u h o w I g o t 10 °$ $p a y a t t e n t i o n t o t h e a n g l e a t A i s 100 °$ $s i n c e l i n e s A D a n d B D h a v e t h e s a m e s i g n$ $(j a g g e d l i n e s), t h i s s h o w t h a t t r i a n g l e A B D$ $i s a n i s o s c e l e s t r i a n g l e A D = B D$ $i n a n i s o s c e l e s t r i a n g l e,$ $t h e a n g l e s o p p o s i t e t h e s i d e s o f t h e s a m e l e n g h t a r e e q u a l .$ $s o, a n g l e A B D = A D B$ $c a l c u l a t e t h e a n g l e A B D a n d A D B$ $t h e s u m o f t h e a n g l e s i n a t r i a n g l e i x 180 °$ $i n a t r i a n g l e; A B D : ∠ B A D + ∠ A B D + ∠ A D B = 180 °$ $100 ° + ∠ A B D + ∠ A B D = 180 ° (b e c a u s e ∠ A B D = ∠ A D B)$ $2 \times ∠ A B D = 180 ° - 100 °$ $2 \times ∠ A B D = 80 °$ $∠ A B D = 40 °$ $s o, ∠ A D B + ∠ B D C = 180 °$ $40 ° + ∠ B D C = 180 ° - 40 °$ $∠ B D C = 140 °$ $s o, ∠ C B D + ∠ B D C + ∠ B C D$ $x + 140 ° + 30 ° = 180 °$ $x + 170 ° = 180 °$ $x = 180 ° - 170 °$ $x = 10 °$ $t h e v a l u e o f t h e a n g l e i s x = 10 °$
	Commented by mr W last updated on 11/Apr/25

	$s o r r y, n o t A D a n d B D h a v e t h e$ $s a m e s i g n, b u t A C a n d B D h a v e$ $t h e s a m e s i g n . t h a t m e a n s A C = B D,$ $n o t A D = B D a s y o u s a i d . b e s i d e s,$ $e v e n i f {y o u}^{'} r e r i g h t w i t h A D = B D,$ $t h e n y o u s h o u l d n o t g e t$ $∠ A B D = ∠ A D B, b u t g e t$ $∠ A B D = ∠ D A B = 100 ° . {t h a t}^{'} s$ $d e f i n i t i v e l y n o t r i g h t .$ $p l e a s e r e c h e c k s i r .$
	Answered by lmcp1203 last updated on 10/Apr/25

	$t h a n k s$
	Answered by vnm last updated on 10/Apr/25

	$A C = B D = a$ $\frac{A B}{\sin 30 °} = \frac{A C}{\sin ∠ A B C}$ $A B = \frac{a}{2 \sin 50 °}$ $E is the base of the perpendicular$ $drawn from B to the line AC$ $B E = A B \sin 80 ° = \frac{a \sin 80 °}{2 \sin 50 °} = B D \sin ∠ B D A = a \sin ∠ B D A$ $\sin ∠ B D A = \frac{\sin 80 °}{2 \sin 50 °} = \sin ∠ B D C$ $∠ B D C = 180 ° - \sin^{- 1} (\sin ∠ B D A)$ $x = ∠ D B C = 180 ° - (∠ D C B + ∠ B D C) =$ $180 ° - (30 ° + 180 ° - \sin^{- 1} (\frac{\sin 80 °}{2 \sin 50 °})) =$ $\sin^{- 1} (\frac{\sin 80 °}{2 \sin 50 °}) - 30 °$ $\frac{\sin 80 °}{2 \sin 50 °} = \frac{2 \sin 40 ° \cos 40 °}{2 \sin (90 ° - 40 °)} = \sin 40 °$ $x = \sin^{- 1} (\sin 40 °) - 30 ° = 10 °$
	Commented by mr W last updated on 11/Apr/25
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	Answered by lmcp1203 last updated on 10/Apr/25

	$t h a n k [y o u$
	Answered by Spillover last updated on 11/Apr/25

	Commented by lmcp1203 last updated on 16/Apr/25

	$t h a n k y o u$
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