Question

From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45° respectively. Bridge is at a height of 2.5m from the banks. Take √3 = 1.732 

Diagram for above situation can be shown as:

P is a point on bridge and AB is the width of river. 

(a) Measure of DB is: 

(i). 3m 

(ii). 2m 

(iii).2.5 m 

(iv).None. 

(b) Length of AD is: 

(i). (√3 + 1) m 

(ii). 2.5 √3 m 

(iii).3.7 m 

(iv).None. 

(c) Apply Pythagoras theorem and find the distance PA: 

(i). 5 m 

(ii). 5√3 m 

(iii).4 m 

(iv).None. 

(d) Distance PB is:

(i). 2√3 m 

(ii). 4 m 

(iii).3.53 m 

(iv).None. 

(e) Width of the river, can be calculated as: 

(i). 6.83 m 

(ii). 7 m 

(iii).5 m 

(iv).None.

Answer 1

P is a point on bridge and AB is the width of the river. 

Given that the angle of depression of the banks on opposite sides of the river are 30° and 45°. 

∴ ∠PAB = 30° and ∠PBA = 45°. 

Given that height of the bridge from the banks is DP = 2.5 m. 

(a) In right angled triangle ∆BDP, tan 45° = \(\frac{DP}{DB}\)

⇒ DB = \(\frac{DP}{tan \,45°}\) = \(\frac{2.5}{1}\) . (∵ DP = 2.5 m and ∠PBD = ∠PBA = 45°.)

Hence, DB = 2.5 m. 

∴ Hence, option (iii) is correct. 

(b) In right angled triangle ∆ADP, tan 30° = \(\frac{DP}{AD}\)

⇒ AD = \(\frac{DP}{tan \,30°}\) = \(\frac{\frac{2.5}1}{\sqrt3}\) = 2.5√3 m.

(∵ DP = 2.5 m & ∠PAB = ∠PAD = 30° and tan 30° = \(\frac{1}{\sqrt3}\) ) 

∴ The length of AD is 2.5√3 m. 

Hence, option (ii) is correct. 

(c) In right angled triangle ∆ADP, 

PA² = AD² +PD² (By using Pythagoras theorem)

=(2.5√3)² + (2.5)² (∵ AD =2.5√3 m and PD = 2.5 m) 

= (2.5)² × (3 + 1) 

= 6.25 × 4 = 25.00. 

⇒ PA = \(\sqrt{25}\) = 5 m. 

Hence, distance PA is 5 m. 

Hence, option (i) is correct. 

(d) In right angled triangle ∆BDP, 

PB² = BD² + DP² = (2.5)² + (2.5)² =2 × (2.5)² = 6.25 × 2 = 12.50 

⇒ PB = \(\sqrt{12.5}\) = 3.53 m. 

Hence, distance PB is 3.53 m. 

Hence, option (iii) is correct. 

(e) The width of the river is AB = AD + BD 

= 2.5√3 + 2.5 (∵ BD = 2.5 m and AD = 2.5√3 m) 

= 2.5× 1.732 + 2.5 = 4.33 + 2.5 = 6.83 m. (√3 = 1.732)

Hence, the width of the river is 6.83 m. 

Hence, option (i) is correct.