Saved Bookmarks
| 1. |
The shadow of a vertical tower on level ground increasesby 10 metres, when the altitude of the sun changes fromangle of elevation 45° to 30°. Find the height of thetower, correct to one place of decimal. (Take 3 1.73) |
|
Answer» As the shadow increase by 10 mlet the total distance when the sun elevation was 45 is xwhereas when the sun elevation was 30 the distance increased by 10 m so x + 10 let h be the height of towerwhen the sun elevationtan 45 = h/x1 = h/xx = h .............(i)when the sun elevation was 30tan 30 = h / (x +10)1/√3 = h / (x + 10)(x + 10) /√3 = h..................(ii) from (i) and (ii) we getx =(x + 10) /√3√3x = x + 10√3 x - x = 10x(√3 - 1) = 10x = 10 / (√3 - 1)x = 10 / (√3 - 1) * (√3 +1)/ (√3 +1) ................ rationalising the denominatorx = 10 ( √3 +1)/ (√3)² - (1)²x = 10 ( √3 +1) /3 -1x = 10 ( √3 +1) /2x = 5 ( √3 +1)from (i) we getx = h5 ( √3 +1) m = height of tower |
|