Question

A parachutist is descending vertically and makes angles of elevation of 45° and 60° at two observing points 100 m apart from each other on the left side of himself. Find the maximum height from which he falls and the distance of the point where he falls on the ground form the just observation point.

Answer 1

Answer:

The maximum height from which the parachutist falls is 236.6 m & the distance of point where parachutist falls on the ground from the observation point is 136.6 m .

Step-by-step explanation:

Let BC be the height of the parachutist (h).

Angle of elevation are ∠BAC = 45° & ∠BDC = 60°.  

Given : AD = 100 m

Let DC = x m

In right angle ∆BCD,

tan θ  = P/ B

tan 60° = BC/CD

√3 = h/x

x = h/√3 ……...…..(1)

In right angle triangle, ∆ACB

tan θ  = P/ B

tan 45° = BC/AC

1 = AB/(AD + CD)

1 = h/(x + 100)

h = x + 100

h = h/√3 + 100

[From eq 1]

h - h/√3 = 100

(h√3 - h)/√3 = 100

(h√3 - h) = 100√3

h(√3 - 1) = 100√3

h = (100√3) /(√3 - 1)

h = 100√3(√3 + 1)/[(√3 -1)x(√3 + 1)]

[By Rationalizing ]

h = (300 + 100√3)/(√3² - 1²)

[By using identity , (a + b) (a - b) = a² - b²]

h = 100(3 + √3)/(3 - 1)

h = 100(3 + √3)/2

h = 50(3 + √3).........(2)

h = 150 + 50√3

h = 150 + 50 × 1.732

[√3 = 1.732]

h = 150 + 86.6

h = 236.6 m

For Distance :  

x = h/√3

x = 50(3 + √3)/√3

x = 50 × √3(3 + √3) /√3 × √3

[By Rationalizing ]

x = 50(3√3 + 3)/3

x = 50 × 3(√3 + 1)/3

x = 50 (√3 + 1)

x = 50(1.732 + 1)

50 × 3(√3 + 1

x = 50 × 2.732

x = 136.6 m

Hence, the maximum height from which the parachutist falls is 236.6 m & the distance of point where parachutist falls on the ground from the observation point is 136.6 m .

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

$\huge\bigstar\mathfrak\purple{\underline{\underline{SOLUTION:}}}$

Let the height of the parachutist=h(m)

Let the distance of falling point from observation point =x(m)

In ∆ABC,

$tan45 \degree = \frac{ab}{bc} \\ \\ = > 1 = \frac{ab}{cd + bd} = > \frac{h}{100 + x} \\ \\ = > h = 100 + x ...............(1)$

In ∆ABD,

$tan60 \degree = \frac{ab}{bd} = > \frac{h}{x} \\ \\ = > \sqrt{3} = \frac{h}{x} \\ \\ = > h = \sqrt{3} x.............(2)$

From equation (1) & (2) we get,

=) 100+ x= √3x

=) 100= x(√3-1)

$= > x = \frac{100}{ \sqrt{3 - 1} } = > \frac{100( \sqrt{3} + 1) }{( \sqrt{3} - 1)( \sqrt{3} + 1) } \\ \\ = >x = \frac{100 \times \sqrt{3} + 1}{3 - 1} \\ \\ > x = \frac{100 \times 2.732}{2} \\ \\ = > x = 136.6m$

Now, using value of x in equation (1),

h= 100+ x

=) h= 100+ 136.6

=) h= 236.6m

Therefore, height of parachutist is 236.6m & distance of point where he falls is 136.6m.

hope it helps ☺️