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Sia ? 6 years, 4 months ago
Given that at the foot of a mountain the elevation of its summit is 45°; after ascending 1000 m towards the mountain up a slope of 30° inclination, the elevation is found to be 60°. We have to find the height of the mountain.
Let F be the foot and S be the summit of the mountain FOS. Then {tex}\angle O F S = 45 ^ { \circ }{/tex}and therefore, {tex}\angle O S F = 45 ^ { \circ }.{/tex}Consequently, OF = OS = h km (say). Let FP = 1000 m = 1 km be the slope so that {tex}\angle O F P = 30 ^ { \circ }.{/tex}Draw PM {tex}\perp {/tex}OF. join PS. It is given that {tex}\angle M P S = 60 ^ { \circ }.{/tex}
In {tex}\triangle F P L,{/tex}we have
{tex}\sin 30 ^ { \circ } = \frac { P L } { P F }{/tex}
{tex}\Rightarrow \quad P L = P F \sin 30 ^ { \circ } = \left( 1 + \frac { 1 } { 2 } \right) \mathrm { km } = \frac { 1 } { 2 } \mathrm { km }{/tex}
{tex}\therefore \quad O M = P L = \frac { 1 } { 2 } \mathrm { km }{/tex}
{tex}\Rightarrow \quad M S = O S - O M = \left( h - \frac { 1 } { 2 } \right) \mathrm { km }{/tex} ...(i)
Also, {tex}\cos 30 ^ { \circ } = \frac { F L } { P F }{/tex}
{tex}\Rightarrow \quad F L = P F \cos 30 ^ { \circ } = \left( 1 \times \frac { \sqrt { 3 } } { 2 } \right) \mathrm { km } = \frac { \sqrt { 3 } } { 2 } \mathrm { km }{/tex}
Now, h = OS = OF = OL + LF
{tex}\Rightarrow \quad h = O L + \frac { \sqrt { 3 } } { 2 }{/tex}
{tex}\Rightarrow \quad O L = \left( h - \frac { \sqrt { 3 } } { 2 } \right) \mathrm { km }{/tex}
{tex}\Rightarrow \quad P M = \left( h - \frac { \sqrt { 3 } } { 2 } \right) \mathrm { km }{/tex} ...(ii)
In {tex}\triangle S P M,{/tex} we have
{tex}\tan 60 ^ { \circ } = \frac { S M } { P M }{/tex}
{tex}\Rightarrow{/tex} SM = PM . tan60 °
{tex}\Rightarrow \quad \left( h - \frac { 1 } { 2 } \right) = \left( h - \frac { \sqrt { 3 } } { 2 } \right) \sqrt { 3 }{/tex}
{tex}\Rightarrow \quad h - \frac { 1 } { 2 } = h \sqrt { 3 } - \frac { 3 } { 2 }{/tex}
{tex}\Rightarrow \quad \sqrt { 3 } h - h = \frac { 3 } { 2 } - \frac { 1 } { 2 }{/tex}
{tex}\Rightarrow \quad h ( \sqrt { 3 } - 1 ) = 1{/tex}
{tex}\Rightarrow \quad h = \frac { 1 } { \sqrt { 3 } - 1 } = \frac { \sqrt { 3 } + 1 } { ( \sqrt { 3 } - 1 ) ( \sqrt { 3 } + 1 ) } = \frac { \sqrt { 3 } + 1 } { 2 } = \frac { 2.732 } { 2 } = 1.366 \mathrm { km }{/tex}
Hence, the height of the mountain is 1.366 km.
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=3/4 + 3/4 + 1
=3+3+4/4
=10/4
=5/2
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Sia ? 6 years, 4 months ago
In {tex}\triangle {/tex}AOB and {tex}\triangle {/tex}COD
{tex}\angle{/tex} AOB = {tex}\angle{/tex} COD [Vertically opposite angles]
{tex}\frac{{AO}}{{OC}} = \frac{{BO}}{{OD}} \Rightarrow \frac{{AO}}{{OB}} = \frac{{OC}}{{OD}}{/tex}[Given]
{tex}\therefore {/tex} {tex}\triangle {/tex}AOB {tex} \sim {/tex} {tex}\triangle {/tex}COD [By SAS similarity]
{tex}\therefore {/tex}{tex}\frac{{AO}}{{OC}} = \frac{{BO}}{{OD}} = \frac{{AB}}{{CD}}{/tex}
{tex}\frac{1}{2} = \frac{{AB}}{{DC}}\left[ {\frac{{AO}}{{OC}} = \frac{{BO}}{{OD}} = \frac{1}{2}} \right]{/tex}is given
{tex}\Rightarrow {/tex} {tex}\frac{1}{2} = \frac{5}{{DC}}{/tex}
{tex}\Rightarrow {/tex} DC = 10 cm
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