The angle of elevation of the top P of a vertical tower P Q of height 10 from a point A on the horizontal ground is 45 ° . Let R be a point on A Q and from a point B , vertically above R , the angle of elevation of P is 60 ° . If ∠ B A Q = 30 ° , A B = d and the area of the trapezium P Q R B is α , then the ordered pair d , α is

Step-by-step solution

As per the given information, the figure would be Here, R A = d cos 30 ° = 3   d 2 Q R = A Q A - R A = 10 - 3   d 2 B R = d sin 30 ° = d 2 Now tan 60 ° = 10 - B R Q R = 10 - d 2 10 - 3   d 2 ⇒ 3 = 20 - d 20 - 3 d ⇒ 20 3 - 3 d = 20 - d ⇒ 2 d = 20 3 - 1 ⇒ d = 10 3 - 1 Now ar trap   P Q R B = 1 2 P Q + B R Q R = 1 2 10 + d 2 10 - 3 d 2 = 1 2 10 + 5 3 - 5 10 - 15 + 5 3 = 1 2 5 3 + 5 5 3 - 5 = 1 2 75 - 25 = 25 i.e. α = 25