Geometry – Page 144

Question-111267

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Question Number 111267 by I want to learn more last updated on 03/Sep/20 Answered by Her_Majesty last updated on 03/Sep/20 $t a n α - t a n β = \frac{s i n (α - β)}{c o s α c o s β}; c o s α c o s β = \frac{c o s (α - β) + c o s (α + β)}{2}$ $\frac{a}{b} = \frac{t a n 10 °}{t a n 50 ° - t a n 40 °} = \frac{\frac{s i n 10 °}{c o s 10 °}}{\frac{s i n 10 °}{c o s 50 ° c o s 40 °}} = \frac{c o s 50 ° c o s 40 °}{c o s 10 °} =$ $$=\frac{\frac{{cos}\mathrm{10}°+{cos}\mathrm{90}°}{\mathrm{2}}}{{cos}\mathrm{10}°}=\frac{\frac{{cos}\mathrm{10}°}{\mathrm{2}}}{{cos}\mathrm{10}°}=\frac{\mathrm{1}}{\mathrm{2}}…

Question-176776

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Question Number 176776 by cherokeesay last updated on 26/Sep/22 Answered by BaliramKumar last updated on 27/Sep/22 $\frac{3 - a}{a} = \frac{a}{4 - a}$ $a^{2} = a^{2} - 7 a + 12$ $a = \frac{12}{7}$ $$\frac{{a}^{\mathrm{2}}…

Question-45690

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Question Number 45690 by ajfour last updated on 15/Oct/18 Commented by ajfour last updated on 15/Oct/18 $R a d i u s o f c i r c l e i s u n i t y .$ $F i n d b i n t e r m s o f a .$ Answered by MrW3 last…

In-the-following-figure-if-S-1-S-2-S-3-6-pi-find-the-area-of-the-circular-crown-

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Question Number 176741 by HeferH last updated on 26/Sep/22 $I n t h e f o l l o w i n g f i g u r e, i f \frac{S_{1} + S_{2}}{\sqrt{S_{3}}} = \frac{6}{\sqrt{π}}$ $f i n d t h e a r e a o f t h e c i r c u l a r c r o w n$ Commented by HeferH last…

Question-111187

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Question Number 111187 by ajfour last updated on 02/Sep/20 Commented by ajfour last updated on 03/Sep/20 $F i n d s i d e s A B a n d B C o f r e c t a n g l e .$ Answered by ajfour last updated on…

Question-45630

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Question Number 45630 by ajfour last updated on 14/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 15/Oct/18 $\frac{1}{2} \times l \times x = \sqrt{s (s - a) (s - b) (s - c)}$ $s = \frac{R + x + x + r + l}{2} = x + \frac{R + r + l}{2}$ $a = R + x b = x + r c = l$ $\frac{1}{2} l x = \sqrt{(x + \frac{R + r + l}{2}) (\frac{r + l + R}{2} - R) (\frac{R + l + r}{2} - r) (x + \frac{R + r + l}{2} - l)}$ $$\frac{{l}^{\mathrm{2}}…

Question-111147

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Question Number 111147 by I want to learn more last updated on 02/Sep/20 Terms of Service Privacy Policy Contact: info@tinkutara.com

Question-111135

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Question Number 111135 by ajfour last updated on 02/Sep/20 Answered by mr W last updated on 02/Sep/20 $2 r + R - \sqrt{R^{2} - r^{2}} = 2 R$ $$\mathrm{2}{r}−{R}=\sqrt{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} } \…

Dans-la-figure-ci-joint-AB-A-B-AA-BB-CC-Determiner-le-rapport-aire-A-B-C-aire-ABC-

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Question Number 176531 by a.lgnaoui last updated on 20/Sep/22 $D a n s l a f i g u r e c i - j o i n t$ $A B ∣∣ A^{^{'}} B^{^{'}} {A A}^{^{'}} = {B B}^{^{'}} = {C C}^{^{'}}$ $D e t e r m i n e r l e r a p p o r t$ $\frac{a i r e Δ (A^{^{'}} B^{^{'}} C^{^{'}})}{a i r e Δ (A B C)} = ?$ …

Question-45448

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Question Number 45448 by MrW3 last updated on 13/Oct/18 Commented by MrW3 last updated on 13/Oct/18 $W h e n a n e l l i p s e i s r o t a t e d b y a n$ $a n g l e α, t h e f i v e p a r t i a l a r e a s a r e e q u a l .$ $F i n d t h e e q u a t i o n o f t h e e l l i p s e$ $f o r α = 90 ° a n d 45 ° r e s p e c t i v e l y .$ …

Category: Geometry