Question

Q.4] The angle of depression of two ships from the top of a lighthouse are 45 degree and 30 degree. If the ships are 120 metre apart, find the height of the lighthouse.प्र.4] दीपगृहाच्या टोकापासून दोन जहाजांचा होणारा अवनत कोन अनुक्रमे 45 ◦ आणि 30 ◦ आहे ,जर दोन जहाजांमधील अंतर 120 मी असेल तर दीपगृहाची उंची काढा . (√3 = 1.73) *

a) 40m
b) 44 m
c) 50 m
D) 60 m​

Answer 1

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Answer 2

Given:

• Angle of depression of two ships = 45° and 30°
• ships are 120 metre apart

Find:

• Height of the lighthouse

Solution:

$\sf In \: \triangle ABC,$

$\sf \tan {45}^{ \circ} = \frac{AB}{BC}$

$\sf \implies l = \frac{h}{BC}$

$\sf \implies BC = h....(i)$

$\sf In \: \triangle ABD,$

$\sf \tan {30}^{ \circ} = \frac{AB}{BD}$

$\sf \implies \frac{l}{ \sqrt{3} } = \frac{h}{120 + BC }$

$\sf \implies h \sqrt{3} = 120 + BC$

$\sf \implies h = \frac{120 + BC}{ \sqrt{3} }$

$\sf \implies h = \frac{120 +h}{ \sqrt{3} } \qquad [from \: eq(i)]$

$\sf \implies h \sqrt{3} = 120 +h$

$\sf \implies h \sqrt{3} - h = 120$

$\sf \implies h( \sqrt{3} - 1) = 120$

$\sf \implies h( 1.732- 1) = 120$

$\sf \implies 0.732h = 120$

$\sf \implies h = \frac{120}{0.732} = 163.93m$

$\sf \hookrightarrow h = 163.9m$

$\rule{221}{4}$

Hence, the height of the lighthouse will be 163.9m