A person stands in the diagonal produced of the square base of a church tower, at a distance 2a from it, and observes the angle of elevation of each of the two outer corners of the top of the tower to be 30°, while that of the nearest corner is 45°. Find the breadth of the tower.


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Question Number 84651 by ajfour last updated on 14/Mar/20

$A p e r s o n s t a n d s i n t h e d i a g o n a l$ $p r o d u c e d o f t h e s q u a r e b a s e o f a$ $c h u r c h t o w e r, a t a d i s t a n c e 2 a$ $f r o m i t, a n d o b s e r v e s t h e a n g l e$ $o f e l e v a t i o n o f e a c h o f t h e t w o$ $o u t e r c o r n e r s o f t h e t o p o f t h e$ $t o w e r t o b e 30 °, w h i l e t h a t o f t h e$ $n e a r e s t c o r n e r i s 45 ° . F i n d t h e$ $b r e a d t h o f t h e t o w e r .$
	Answered by mr W last updated on 14/Mar/20

	$b = b r e a d t h o f t o w e r$ $h = h e i g h t o f t o w e r$ $\tan 45 ° = \frac{h}{2 a} \Rightarrow h = 2 a$ $h = \sqrt{{(\frac{b}{\sqrt{2}})}^{2} + {(\frac{b}{\sqrt{2}} + 2 a)}^{2}} \tan 30 °$ $2 a = \sqrt{{(\frac{b}{\sqrt{2}})}^{2} + {(\frac{b}{\sqrt{2}} + 2 a)}^{2}} \times \frac{1}{\sqrt{3}}$ $12 a^{2} = 2 {(\frac{b}{\sqrt{2}})}^{2} + 4 a (\frac{b}{\sqrt{2}}) + 4 a^{2}$ ${(\frac{b}{\sqrt{2}})}^{2} + 2 a (\frac{b}{\sqrt{2}}) - 4 a^{2} = 0$ ${(\frac{b}{\sqrt{2}} + a)}^{2} = 5 a^{2}$ $\frac{b}{\sqrt{2}} + a = \sqrt{5} a$ $\Rightarrow b = (\sqrt{10} - \sqrt{2}) a \approx 1.748 a$
	Commented by ajfour last updated on 14/Mar/20

	$c o r r e c t s i r, a n d w e l l p r e s e n t e d,$ $t h a n k s .$
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