Geometry – Page 172

Question-105769

Tinku Tara June 4, 2023 Geometry 0 Comments

Question Number 105769 by ajfour last updated on 31/Jul/20 Commented by ajfour last updated on 31/Jul/20 $A C D B i s a s t r a i g h t l i n e . T h e$ $t h r e e c o l o u r e d r e g i o n s h a v e e q u a l$ $a r e a s . G i v e n O A = p, O B = q .$ $F i n d c o o r d i n a t e s o f A a n d B .$ …

Question-40213

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Question Number 40213 by ajfour last updated on 17/Jul/18 Terms of Service Privacy Policy Contact: info@tinkutara.com

Question-171255

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Question Number 171255 by mnjuly1970 last updated on 11/Jun/22 Commented by mr W last updated on 11/Jun/22 $S_{Δ A P B} = \frac{A B \times A P \cos θ}{2}$ $= \frac{A D \times A P \cos θ}{4}$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{AP}^{\mathrm{2}} }{\mathrm{4}} \…

Question-40170

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Question Number 40170 by ajfour last updated on 16/Jul/18 Commented by ajfour last updated on 16/Jul/18 Commented by ajfour last updated on 16/Jul/18 $${p}=\frac{\mathrm{1}}{\mathrm{2sin}\:\alpha}\:\:\:\:\:\:{q}=\frac{\mathrm{1}}{\mathrm{2sin}\:\beta} \…

Question-105630

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Question Number 105630 by I want to learn more last updated on 30/Jul/20 Commented by I want to learn more last updated on 30/Jul/20 $$\mathrm{Thanks}\:\mathrm{sir}…

Question-171132

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Question Number 171132 by cortano1 last updated on 08/Jun/22 Answered by kapoorshah last updated on 08/Jun/22 $P A = (- 3, - 4)$ $P T = (- 1, - 4)$ $\cos θ = \frac{P A . P T}{∣ P A ∣ . ∣ P T ∣}$ $= \frac{3 + 16}{5 . \sqrt{17}}$ $$\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{19}}{\mathrm{5}\sqrt{\mathrm{17}}}…

Question-40063

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Question Number 40063 by ajfour last updated on 15/Jul/18 Answered by ajfour last updated on 16/Jul/18 $Q \equiv (a, \sqrt{R^{2} - a^{2}})$ $T L = Q L$ $$\Rightarrow\:\:{T}\left(−{a},\:\sqrt{{R}^{\mathrm{2}} −{a}^{\mathrm{2}} }\:\right)…

Question-40031

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Question Number 40031 by ajfour last updated on 15/Jul/18 Commented by ajfour last updated on 16/Jul/18 $C E a n d D F ⊥ A F . F i n d θ .$ Answered by ajfour last updated on…

Question-40023

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Question Number 40023 by ajfour last updated on 15/Jul/18 Answered by MrW3 last updated on 15/Jul/18 $A B = 2 R \cos θ$ $A D = \frac{R}{\cos θ}$ $B D = A B - A D = R (2 \cos θ - \frac{1}{\cos θ}) = \frac{R \cos 2 θ}{\cos θ}$ $L e t E = m i d p o i n t o f B C .$ $${BE}={AB}\:\mathrm{sin}\:\theta={R}\:\mathrm{sin}\:\mathrm{2}\theta…

Question-39993

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Question Number 39993 by ajfour last updated on 14/Jul/18 Answered by tanmay.chaudhury50@gmail.com last updated on 14/Jul/18 $l e t ∠ A O T = θ$ $O T ∥ t o B C s o \frac{A O}{A C} = \frac{A T}{A B} = \frac{O T}{B C}$ $p o i n t T (c o s θ, s i n θ) a n d p o i n t C (- 1, 0)$ $e q n o f O T i s y = x t a n θ$ $${eqn}\:{ofCB}\:{is}\:{y}=\left({x}+\mathrm{1}\right){tan}\theta…

Category: Geometry