Question

a girl stand on tree building of 60mts height and observe bottoms of two trees which are either side of building at the angles of depression 45° and 30° . find the distance between 2 trees​

Answer 1

Refer to the attachment

hope it helps

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{Distance\:between\:two\:trees}}\approx43 m}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline \bold{given : }} \\ : \implies \text{height \: of \: building = 60 \: m} \\ \\ : \implies \text{angle \: of \: depression \: = 45 \degree \: and \: 30 \degree} \\ \\ \red{ \underline \bold{to \: find : }} \\ \implies \text{distance \: between \: two \: trees = ?}$

• According to given question :

$\text{In \: right }\triangle \text{ABC} \\ : \implies tan \theta = \frac{p}{b} \\ \\ : \implies tan \: 45 \degree = \frac{60}{b} \\ \\ : \implies 1 = \frac{60}{BC} \\ \\ : \implies \text{BC = 60 \: m} \\ \\ \text{In \: right }\triangle \text{ABD} \\ : \implies tan \: \theta = \frac{p}{b} \\ \\ : \implies tan \: 30 \degree = \frac{60}{bd} \\ \\ : \implies \frac{1}{ \sqrt{3} } = \frac{60}{BD} \\ \\ : \implies \text{BD = 60} \sqrt{3} \: m \\ \\ \bold{For \: distance \: between \: two\: trees}\\ : \implies Distance = BD- BC\\ \\ : \implies Distance = 60 \sqrt{3}-60 \\ \\ : \implies Distance = 60 ( \sqrt{3}-1 ) \\ \\ : \implies Distance = 60 \times 0.73 \\ \\ \green{ : \implies \text{Distance}\approx 43 m}$