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At the foot of a mountain, the elevation of its summit is \( \Large 45 ^{\circ} \). After ascending 2km towards the mountain upon an incline of \( \Large 30 ^{\circ} \), the elevation changes to \( \Large 60 ^{\circ} \). The height of the mountain is

A) \( \Large \left(\sqrt{3} - 1\right) km \)
B) \( \Large \left(\sqrt{3} + 1\right) km \)
C) \( \Large \left(\sqrt{3} + 2\right) km \)
D) \( \Large \left(\sqrt{3} - 2\right) km \)

Correct Answer:


B) \( \Large \left(\sqrt{3} + 1\right) km \)

Description for Correct answer:

Let AB = h km

In \( \Large \triangle OAB, \tan 45 ^{\circ} = \frac{AB}{OB} \)

=> \( \Large OB = h km \)

In \( \Large \triangle OLM, \)

=> \( \Large \frac{OM}{OL}= \cos 30 ^{\circ} \)

\( \Large OM = 2 \cos 30 ^{\circ} \left[ OL=2 km \right] = \sqrt{3} km \)

\( \Large \therefore LN = BM = OB - OM = \left(h-\sqrt{3}\right) km \)

In \( \Large \triangle OLM, \sin 30 ^{\circ} =\frac{LM}{OL} \)

\( \Large LM = 2 \sin 30 ^{\circ} = 1 km \)

\( \Large \therefore BN = LM = 1 km \)

In \( \Large \triangle ALN, \tan 60 ^{\circ} = \frac{AN}{LN} \)

=> \( \Large \sqrt{3}=\frac{AB - BN}{LN} \)

=> \( \Large \sqrt{3}=\frac{h-1}{h-\sqrt{3}} => \sqrt{3}h-3 = h - 1 \)

=> \( \Large h = \frac{2}{\sqrt{3}-1} \)

Therefore, \( \Large h=\frac{2}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}= \left(\sqrt{3}+1\right) km \)

Part of solved Height and Distance questions and answers : >> Elementary Mathematics >> Height and Distance

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Similar Questions

1). A man standing in one corner of a square football field observes that the angle subtended by a pole in the corner just diagonally opposite to this corner is \( \Large 60 ^{\circ} \). When he retires 80 m from the corner, along the same straight line, he finds the angle to be \( \Large 30 ^{\circ} \). The length of the field is

A). 20 m

B). \( \Large 40\sqrt{2} \) m

C). 40 m

D). \( \Large 20\sqrt{2} \) m

2). A spherical balloon of radius r subtends angle \( \Large 60 ^{\circ} \) at the eye of an observer. If the angle of elevation of its centre is \( \Large 60 ^{\circ} \) and h is the height of the centre of the balloon, then which one of the following is correct?

A). h=r

B). \( \Large h = \sqrt{2}r \)

C). \( \Large h = \sqrt{3}r \)

D). h=2r

3). If the roots of equations \( \Large ax^{2}+bx+c=0 \) be \( \Large \ \alpha \ and\ \beta \) then the roots of equations \( \Large cx^{2}+bx+a=0 \) are:

A). \( \Large - \alpha ,\ - \beta \)

B). \( \Large \alpha ,\ \frac{1}{ \beta } \)

C). \( \Large \frac{1}{ \alpha },\ \frac{1}{ \beta } \)

D). none of these

4). Both the roots of the given equation. \( \Large \left(x-a\right) \left(x-b\right)+ \left(x-b\right) \left(x-c\right)+ \left(x-c\right) \left(x-a\right)=0 \) are always

A). positive

B). negative

C). real

D). imaginary

5). If the difference of roots of the equation \( \Large x^{2}-bx+c=0 \) be 1, then;

A). \( \Large b^{2}-4c-1=0 \)

B). \( \Large b^{2}-4c=0 \)

C). \( \Large b^{2}-4c+1=0 \)

D). \( \Large b^{2}+4c-1=0 \)

6). If \( \Large 2+i\sqrt{3} \) is a root of the equation \( \Large x^{2}+px+q=0 \) where p and q are real, then \( \Large p, q \) is equal to:

A). (-4, 7)

B). (4,-7)

C). (4, 7)

D). (-4, -7)

7). If \( \Large x=\sqrt{1+\sqrt{1+\sqrt{1}+.....}} \), the x is equal to

A). \( \Large \frac{1+\sqrt{5}}{2} \)

B). \( \Large \frac{1-\sqrt{5}}{2} \)

C). \( \Large \frac{1\pm \sqrt{5}}{2} \)

D). none of these .

8). If \( \Large \alpha , \beta \) are the roots of equation \( \Large ax^{2}+bx+c=0 \), then \( \Large \frac{ \alpha }{ \alpha \beta +b}+\frac{ \beta }{ \alpha x+b} \) is equal to

A). 2/a

B). 2/b

C). 2/c

D). -2/a

9). If the roots of the equation \( \Large \frac{ \alpha }{x- \alpha }+\frac{ \beta }{x- \beta }=1 \) be equal in magnitude but opposite in sign, then \( \Large \alpha + \beta \) is equal to:

A). 0

B). 1

C). 2

D). none of these

10). If the roots of the equation \( \Large px^{2}+2qx+r=0 \) and \( \Large qx^{2}-2\sqrt{pr}x+q=0 \) be real, then:

A). \( \Large p=q \)

B). \( \Large q^{2}=pr \)

C). \( \Large p^{2}=qr \)

D). \( \Large r^{2}=pq \)