Symmetric Angles

Author name: anushka kumari

The correct answer is: $ \displaystyle x$ = 1, 2, –3 Explanation: We are asked to solve the equation x³ – 7x + 6 = 0, which is a cubic equation. We will find the roots using the trial-and-error method followed by factorization. 1. Check for a Root via Substitution: We can start by trying […]

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If $ \displaystyle x$³ – 6$ \displaystyle x$² + 11$ \displaystyle x$ – 6 is factorized as ($ \displaystyle x-a$)($ \displaystyle x-b$) ($ \displaystyle x-c$), what are the values of $ \displaystyle a$, $ \displaystyle b$, and $ \displaystyle c$?

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± 4 Explanation: According to the question, the distance between the points (4, p) and (1, 0) = 5 i.e., $ \displaystyle \sqrt{{{{{(1-4)}}^{2}}+{{{(0-p)}}^{2}}}}=5$ [∵ distance between the points $ \displaystyle {{x}_{1}},{{y}_{1}}$ and $ \displaystyle {{x}_{2}},{{y}_{2}}$, d= $ \displaystyle \sqrt{{{{{\left( {{{x}_{2}}-{{x}_{1}}} \right)}}^{2}}+{{{\left( {{{y}_{2}}-{{y}_{1}}} \right)}}^{2}}}}$​] $ \displaystyle \begin{array}{l}\Rightarrow \sqrt{{{{{(-3)}}^{2}}+{{p}^{2}}}}=5\\\Rightarrow \sqrt{{9+{{p}^{2}}}}=5\end{array}$​ On squaring both the sides, we get

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sec²θ(1 + sinθ)(1 − sinθ) = sec²θ(1 − sin²θ) = sec²θ × cos²θ = $ \displaystyle \left( {\dfrac{1}{{{{{\cos }}^{2}}\theta }}} \right)\times {{\cos }^{2}}\theta $ = 1

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The correct answer is: 2 Explanation: Given, ​DE∥AB ∴ $ \displaystyle \dfrac{{\text{CD}}}{{\text{AD}}}=\dfrac{{\text{CE}}}{{\text{BE}}}$​​ [by basic proportionality theorem] ⇒ $ \displaystyle \dfrac{{x+3}}{{3x+19}}=\dfrac{x}{{3x+4}}$ ⇒(x + 3)(3x + 4) = x(3x + 19) ⇒ 3$ \displaystyle {{x}^{2}}$ + 4x + 9x + 12 = 3$ \displaystyle {{x}^{2}}$ + 19x 19x − 13x = 12 6x = 12 x = $

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In △ADC and △BAC, ∠ADC = ∠BAC (Already given) ∠ACD = ∠BCA (Common angles) ∴ △ADC ∼ △BAC (AA similarity criterion) We know that corresponding sides of similar triangles are in proportion. $ \displaystyle \dfrac{{\text{CA}}}{{\text{CB}}}=\dfrac{{\text{CD}}}{{\text{CA}}}$ ​∴ $ \displaystyle {\Rightarrow \text{C}{{\text{A}}^{2}}=\text{CB}\text{.CD}\text{.}}$  Hence, proved.

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The correct answer is: Sublimation Explanation: Radius of bigger circle $ \displaystyle {{{r}_{1}}}$​ = OA = 7 Radius of smaller circle $ \displaystyle {{{r}_{2}}}$ = $ \displaystyle \dfrac{{{{r}_{1}}}}{2}$ = $ \displaystyle \dfrac{7}{2}$​ Area of the bigger circle $ \displaystyle {{{C}_{1}}}$​ = $ \displaystyle {\pi r_{1}^{2}}$​ ​= $ \displaystyle \dfrac{22}{7}$ ​× 7 × 7 = 154 $ \displaystyle \text{c}{{\text{m}}^{2}}$​

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Sublimation Explanation: Given, in △ABC, circle touch the triangle at point D, F and E respectively and let the lengths of the segment AF be x. So, BF = BD = 6 cm [Tangent drawn from same point are equal] CE = CD = 9 cm [Tangent drawn from same point are equal ] and AE =

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3293.5 $ \displaystyle {{\text{m}}^{2}}$ Explanation: Length of rectangular field = L = 70 m Breadth of rectangular field = B = 52 m So, Area of the field = L × B = 70 × 52 = 3640 Area of the field is 3640 $ \displaystyle {{\text{m}}^{2}}$ And, Area it can graze = Area of quadrant of

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The correct answer is: $ \displaystyle 3\sqrt{3}$ cm Explanation: Let P be an external point and a pair of tangents is drawn from point P and angle between these two tangents is 60°. Join OA and OP. Also, OP is a bisector line of ∠APC. ∴ ​∠APO = ∠CPO = 30°  Also,  OA⊥AP​ Tangent at any

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