Question

Determine the height of a mountain if the elevation of its top at an unknown distance from the base is 30° and at a distance 10 km further off from the mountain, along the same line, the angle of elevation is 15°. (use tan 15° = 0.27)​

Answer 1

Answer:

Step-by-step explanation:

Mountain of height h = AB.

Point at a distance of x  = C .

Angle of elevation of the top at C =  30  ∘

Point at a distance of 10 km from C = D

Such that the angle of elevation at D is of15  ∘

 

In △ CAB , we have  

tan30  ∘  =  $\frac{AC}{AB}$

 

⇒ $\frac{1}{\sqrt{3} }$ = $\frac{h}{x}$

⇒x=  $\sqrt{3x}$

​  

 In △ DAB  we have  

tan15  ∘  =  $\frac{AD}{AB}$

​    

⇒0.27=  $\frac{h}{x+10}$

 

⇒ (0.27) (x + 10) = h  

substituting x = $\sqrt{3h}$ obtained from equation (i) in equation (ii)  we get  

0.27 (  

+ 10)  = h  

⇒0.27×10=h−0.27×  $\sqrt{3h}$

​  

 ⇒ h (1 - 0.27  ×  3  )  = 2.7

⇒ h (1 - 0.46 )  = 2 . 7  

⇒h=  0.54

/2.7

​    = 5  

Hence  , the height  of the mountains is 5 km

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Answer 2

⋆ Reference of image is shown in diagram

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⠀

Here,

• Let AB be the mountain of height h km
• Let C be a point at a distance of x km
• Let D be a point at s distance of 10 km
• Angle of elevation of the top at C is 30°
• Angle of elevation at D is of 15°

⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

${\underline{\sf{\bigstar\;In\;\triangle\;CAB,}}}$

$\qquad\qquad\sf tan\;30^\circ = \dfrac{AB}{AC}$

$\qquad\quad:\implies\sf \dfrac{1}{ \sqrt{3}} = \dfrac{h}{x}$

$\qquad\quad:\implies\bf x = \sqrt{3} h\qquad\qquad\bigg\lgroup\bf eq.\;(1) \bigg\rgroup$

${\underline{\sf{\bigstar\;Now,\;In\;\triangle\;CAB,}}}$

$\qquad\qquad\sf tan\;15^\circ = \dfrac{AB}{AD}$

$\quad\quad:\implies\sf 0.27 = \dfrac{h}{x + 10}$

$\quad:\implies\bf (0.27)(x + 10) = h\qquad\qquad\bigg\lgroup\bf eq.\;(2) \bigg\rgroup$

⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

⠀⠀☯ Substituting x = $\bf \sqrt{3}h$ obtained from eq. (1) in eq. (2), we get

$\qquad\qquad\sf 0.27( \sqrt{3}h + 10) = h$

$:\implies\sf 0.27 \times 10 = h - 0.27 \times \sqrt{3} h$

$\quad:\implies\sf h(1 - 0.27 \times \sqrt{3}) = 2.7$

$\qquad \: :\implies\sf h(1 - 0.46) = 2.7$

$\qquad \: \quad:\implies\sf 0.54 h = 2.7$

$\qquad\quad \: \: :\implies\sf h = \cancel{ \dfrac{2.7}{0.54}}$

$\qquad\qquad:\implies{\boxed{\frak{\pink{h = 5}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;the\;height\;of\;the\; Mountain\;is\; \bf{5\;km}.}}}$