Question

10. A radio mast PO.of height' metres, is standing vertically on the horizonte ground from
A, the angle of elevation of the top of the most is found to be 45. On moving som
slope of 15, the angle of elevation of P is found to be 75 from the horizontal through B is BC.
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Answer 1

Answer:

Step-by-step explanation:

The given information describes the measurement of the angle of elevation of a radio mast from two different points on the ground, Point A and Point B. The angle of elevation from Point A is 45 degrees and from Point B, it is 75 degrees. To determine the height of the radio mast, we need to use trigonometry. By drawing a right triangle with the radio mast as the hypotenuse and the ground as the base, we can calculate the height using the angle of elevation and the distance from Point A to Point B (denoted as BC). The formula for the height, h, of the radio mast can be determined using the tangent function:

h = BC * tan (angle of elevation)

By substituting the values of BC and the angles of elevation, we can calculate the height of the radio mast.

Answer 2

COMPLETE QUESTION

A radio mast PO.of height' metres, is standing vertically on the horizonte ground from

A, the angle of elevation of the top of the most is found to be 45. On moving som

slope of 15, the angle of elevation of P is found to be 75 from the horizontal through B is BC. Find AP and 'h'.

ANSWER

AP = 50√2cm and h = 50 cm

GIVEN

A radio mast PO.of height' metres, is standing vertically on the horizonte ground from

A, the angle of elevation of the top of the most is found to be 45. On moving slope of 15°, the angle of elevation of P is found to be 75 from the horizontal through B is BC

TO FIND

Find AP and 'h'.

SOLUTION

The above problem can be simply solved as follows;

ΔAQP and ΔPBC are isosceles triangles.

AP = AQ = h

∠QAP = 60°

From ΔPBC

∠B = 75°, ∠C = 90°

Let, ∠CPB = x

Therefore,

x + 90 + 75 = 180

x = 180 - 165

x = 15°

∠CPB = 15°

∠QPA = ∠CPB + ∠BPA

∠BPA = 60 - 15 = 45°

In ΔABP

∠APB = 180 - (45 + 45)

∠APB = 90°

Therefore,

ΔAPB is right angled triangle.

Cos45° = AB/AP

1/√2 = 50/AP

AP = 50√2

Applying pythagorus theorem in ΔAQP

AP² = AQ² + QP²

(50√2)² = h² + h²

2500× 2 = 2h²

h² = 2500

h = √2500 = 50 cm

Hence, AP = 50√2cm and h = 50 cm

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