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Sia ? 6 years, 5 months ago
According to question, series is 105, 110, 115, 120, ………, 995
Here a = 105, d = 110 - 105 = 5 and an = 995
{tex}\therefore{/tex} an = a + (n - 1)d
{tex}\Rightarrow 995 = 105 + ( n - 1 ) \times 5{/tex}
{tex}\Rightarrow 995 - 105 = ( n - 1 ) \times 5{/tex}
{tex}\Rightarrow \frac { 890 } { 5 } = ( n - 1 ){/tex}
{tex}\Rightarrow{/tex} n = 178 + 1 = 179
Now, {tex}\mathrm { S } _ { n } = \frac { n } { 2 } ( a + l ){/tex}
{tex}\Rightarrow S _ { 199 } = \frac { 179 } { 2 } ( 105 + 995 ){/tex}
{tex}\Rightarrow S _ { 179 } = \frac { 179 } { 2 } \times 1100=98450{/tex}
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Sia ? 6 years, 5 months ago
tan 20° tan 40° tan 60° tan 80°
= (tan 20° tan 40° tan 80°) {tex}\sqrt {3}{/tex} {tex}\left[\because \tan 60^{\circ}=\sqrt{3}\right]{/tex}
{tex}=\left(\frac{\sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ}}{\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ}}\right) \sqrt{3}{/tex}
{tex}=\frac{\left(2 \sin 20^{\circ} \sin 40^{\circ}\right) \sin 80^{\circ} \times \sqrt{3}}{\left(2 \cos 20^{\circ} \cos 40^{\circ}\right) \cos 80^{\circ}}{/tex}
Applying
{tex}\Rightarrow{/tex} 2 sin A sin B = cos (A - B) - cos (A + B)
2 cos A cos B - cos (A + B) + cos (A - B)
{tex}\frac{\left(\cos \left(40^{\circ}-20^{\circ}\right)-\cos \left(40^{\circ}+20^{\circ}\right)\right) \sin 80^{\circ} \times \sqrt{3}}{\left(\cos \left(20^{\circ}+40^{\circ}\right)+\cos \left(40^{\circ}-20^{\circ}\right)\right) \cos 80^{\circ}}{/tex}
{tex}=\frac{\left(\cos 20^{\circ}-\cos 60^{\circ}\right) \sin 80^{\circ} \times \sqrt{3}}{\left(\cos 60^{\circ}+\cos 20^{\circ}\right) \cos 80^{\circ}}{/tex}
{tex}=\frac{\left(\cos 20^{\circ}-\frac{1}{2}\right) \sin 80^{\circ} \times \sqrt{3}}{\left(\frac{1}{2}+\cos 20^{\circ}\right) \cos 80^{\circ}}{/tex}
{tex}=\frac{\left(2 \sin 80^{\circ} \cos 20^{\circ}-\sin 80^{\circ}\right) \sqrt{3}}{\cos 80^{\circ}+2 \cos 20^{\circ} \cos 80^{\circ}}{/tex}
{tex}\Rightarrow{/tex} 2 sin A cos B - sin (A + B) + sin (A - B)
{tex}=\frac{\left(\sin \left(80^{\circ}+20^{\circ}\right)+\sin \left(80^{\circ}-20^{\circ} \right)-\sin \theta 0^{\circ}\right) \sqrt{3}}{\cos 80^{\circ}+\left(\cos \left(20^{\circ}+80^{\circ}\right)+\cos \left(80^{\circ}-20^{\circ}\right)\right)}{/tex}
{tex}=\frac{\left(\sin 100^{\circ}+\sin 60^{\circ}-\sin 80^{\circ}\right) \sqrt{3}}{\cos 80^{\circ}+\cos 100^{\circ}+\cos 60^{\circ}}{/tex}
{tex}=\frac{\left(\sin \left(180^{\circ}-80^{\circ}\right)+\frac{\sqrt{3}}{2}-\sin 80^{\circ}\right) \sqrt{3}}{\cos 80^{\circ}+\cos \left(180^{\circ}-80^{\circ}\right)+\cos 60^{\circ}}{/tex}
{tex}=\frac{\left(\sin 80^{\circ}+\frac{\sqrt{3}}{2}-\sin 80^{\circ}\right) \sqrt{3}}{\cos 80^{\circ}-\cos 80^{\circ}+\cos 60^{\circ}}{/tex}
{tex}=\frac{\frac{3}{2}}{\frac{1}{2}}{/tex} = 3 = RHS
Yogiraj Kaushik 6 years, 5 months ago
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Sia ? 6 years, 6 months ago
Since AD is the bisector of {tex}\angle B A C{/tex}.
{tex}\Rightarrow \quad \frac { B D } { D C } = \frac { A B } { A C }{/tex}
Now, {tex}A B = \sqrt { ( 5 - 3 ) ^ { 2 } + ( 3 - 2 ) ^ { 2 } + ( 2 - 0 ) ^ { 2 } }{/tex}
[{tex}\because{/tex} distance {tex}= \sqrt { \left( x _ { 2 } - x _ { 1 } \right) ^ { 2 } + \left( y _ { 2 } - y _ { 1 } \right) ^ { 2 } + \left( z _ { 2 } - z _ { 1 } \right) ^ { 2 } } ]{/tex}
{tex}= \sqrt { 2 ^ { 2 } + 1 ^ { 2 } + 2 ^ { 2 } } = \sqrt { 4 + 1 + 4 } = \sqrt { 9 } = 3{/tex}
and {tex}A C = \sqrt { ( - 9 - 3 ) ^ { 2 } + ( 6 - 2 ) ^ { 2 } + ( - 3 - 0 ) ^ { 2 } }{/tex}
{tex}= \sqrt { ( - 12 ) ^ { 2 } + ( 4 ) ^ { 2 } + ( - 3 ) ^ { 2 } }{/tex}
{tex}= \sqrt { 144 + 16 + 9 } = \sqrt { 169 } = 13{/tex}
{tex}\Rightarrow \quad \frac { B D } { D C } = \frac { 3 } { 13 }{/tex}
Hence, D divides BC in the ratio 3:13. Then, coordinates of D are
{tex}\left[ \frac { 3 ( - 9 ) + 13 ( 5 ) } { 3 + 13 } , \frac { 3 ( 6 ) + 13 ( 3 ) } { 3 + 13 } , \frac { 3 ( - 3 ) + 13 ( 2 ) } { 3 + 13 } \right]{/tex} [using internal division formula]
{tex}= \left( \frac { - 27 + 65 } { 16 } , \frac { 18 + 39 } { 16 } , \frac { - 9 + 26 } { 16 } \right){/tex}
{tex}= \left( \frac { 38 } { 16 } , \frac { 57 } { 16 } , \frac { 17 } { 16 } \right) = \left( \frac { 19 } { 8 } , \frac { 57 } { 16 } , \frac { 17 } { 16 } \right){/tex}
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Paras Chauhan 6 years, 5 months ago
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