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Preeti Dabral 1 year, 11 months ago
{tex}\begin{aligned} & \text { LHS }=\frac{\sin A-\sin B}{\cos A+\cos B}+\frac{\cos A-\cos B}{\sin A+\sin B} \\ & =\frac{\sin ^2 A-\sin ^2 B+\cos ^2 A-\cos ^2 B}{(\cos A+\cos B)(\sin A+\sin B)} \\ & =\frac{\left(\sin ^2 A+\cos ^2 A\right)-\left(\sin ^2 B+\cos ^2 B\right)}{(\cos A+\cos B)(\sin A+\sin B)} \\ & =(\cos A+\cos B)(\sin A+\sin B) \\ & =0=\text { RHS } \end{aligned}{/tex}
Posted by Bunny . 1 year, 11 months ago
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Preeti Dabral 1 year, 11 months ago
Let (-4, 6) divide AB internally in the ratio k:1. Using the section formula, we get
{tex}( - 4,6 ) = \left( \frac { 3 k - 6 } { k + 1 } , \frac { - 8 k + 10 } { k + 1 } \right){/tex}
So, {tex}- 4 = \frac { 3 k - 6 } { k + 1 }{/tex}
i.e., -4k - 4 = 3k - 6
i.e., 7k = 2
i.e., k:1 = 2:7
The same can be checked for the y-coordinate also.
Therefore, the ratio in which the point (-4,6) divides the line segment AB is 2: 7.
Posted by Aniket Topwal 1 year, 11 months ago
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Preeti Dabral 1 year, 11 months ago
Let A(a, a), B(-a, -a) C{tex}\left( { - \sqrt 3 a,,\sqrt 3 a} \right){/tex}
{tex}AB = \sqrt {{{( - a - a)}^2} + {{( - a - a)}^2}} = \sqrt {8{a^2}} = 2\sqrt 2 a{/tex}
{tex}BC = \sqrt {{{( - \sqrt 3 a + a)}^2} + {{(\sqrt 3 a + a)}^2}} = 2\sqrt 2 a{/tex}
{tex}AC = \sqrt {{{( - \sqrt 3 a - a)}^2} + {{(\sqrt 3 a - a)}^2}} = 2\sqrt 2 a{/tex}
{tex}\therefore {/tex} AB = BC = AC = {tex}2\sqrt 2 a{/tex}
{tex}ar\Delta ABC = \frac{{\sqrt 3 }}{4} \times {(side)^2}{/tex}
{tex} = \frac{{\sqrt 3 }}{4} \times \left( {2\sqrt 2 a} \right)^2{/tex}
{tex} = 2\sqrt 3 {a^2}{/tex} sq. units
Posted by Ritu Kumari 1 year, 11 months ago
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Posted by Vis H 1 year, 11 months ago
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Preeti Dabral 1 year, 11 months ago
Given: A circle with centre O. A tangent CD at C.
Diameter AB is produced to D.
BC and AC chords are joined, ∠BAC = 30°
To prove: BC = BD
Proof: DC is tangent at C and, CB is chord at C.
Therefore, ∠DCB = ∠BAC [∠s in alternate segment of a circle]
⇒ ∠DCB = 30° …(i) [∵ ∠BAC = 30° (Given)]
AOB is diameter. [Given]
Therefore, ∠BCA = 90° [Angle in s semi circle]
Therefore, ∠ABC = 180° - 90° - 30° = 60°
In ΔBDC,
Exterior ∠B = ∠D + ∠BCD
⇒ 60° = ∠D + 30°
⇒ ∠D = 30° …(ii)
Therefore, ∠DCB = ∠D = 30° [From (i), (ii)]
⇒ BD = BC [∵ Sides opposite to equal angles are equal in a triangle]
Hence, proved.
Posted by Aishiki Jana 1 year, 11 months ago
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Pranjal Gupta 1 year, 11 months ago
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Ankit Kumar 1 year, 11 months ago
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