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Based on the following informationA man standing on a boat south of a light house observes his shadow to be 24 meters long, as measured at the sea level. On sailing 300 meters eastwards, he finds his shadow as 30 meters long, measured in a similar manner. The height of the man is 6 meters above sea level.The height of the light house above the sea level is:
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By: anil on 05 May 2019 02.37 am
KL = lighthouse BA = initial position man of man and BC = shadow After moving 300 m east, DE = new position of man and EF = shadow Given : AB = DE = 6 m BC = 24 m and EF = 30 m and BE = 300 m $$ riangle$$ LBE is right angled triangle (sea level). To find : KL = ? Solution : In $$ riangle$$ KLF and $$ riangle$$ DEF => $$angle KLF = angle DEF = 90$$ $$angle KFL = angle DFE$$ (common angle) => $$ riangle KLF sim riangle DEF$$ => $$frac{KL}{DE} = frac{LF}{EF}$$ -----------Eqn(I) Similarly, $$ riangle KLC sim riangle ABC$$ => $$frac{KL}{AB} = frac{LC}{BC}$$ ----------Eqn(II) From eqn (I) and (II), and using AB = DE => $$frac{LC}{BC} = frac{LF}{EF}$$ => $$frac{LC}{24} = frac{LF}{30}$$ => $$frac{LC}{LF} = frac{24}{30} = frac{4}{5}$$ If, LC is 4 part $$equiv$$ LF is 5 part => $$LB = 4x$$ and $$LE = 5x$$ $$ecause$$ $$ riangle$$ LBE is right angled triangle => $$(LE)^2 - (LB)^2 = (BE)^2$$ => $$25^2 - 16X^2 = 90000$$ => $$x^2 = frac{90000}{9} = 10000$$ => $$x = sqrt{10000} = 100$$ => $$LB = 400$$ and $$LE = 500$$
=> $$LC = LB + BC = 400 + 24 = 424$$ Now, using Eqn (II), we get : => $$KL = frac{LC}{BC} imes AB$$ = $$frac{424}{24} imes 6 = frac{424}{4}$$ = $$106 m$$
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=> $$LC = LB + BC = 400 + 24 = 424$$ Now, using Eqn (II), we get : => $$KL = frac{LC}{BC} imes AB$$ = $$frac{424}{24} imes 6 = frac{424}{4}$$ = $$106 m$$