Let PM be perpendicular from P on CB. Then, ∠CPM = 30° and ∠CPM = 60°.
Let CM = h. Then, CB = h + 100 and CB = h + 100.
In right 
CMP,
In right 
PMC'
From (i) and (ii), we get
Now, CB = CM + MB = h + 100 = 100 + 100 = 200
Hence, the height of the helicopter from the surface of the lake = 200 m
Let CD be the flagstaff whose height is 7 m, fixed on the tower BC of height h metres. From a point A on the ground the angles of elevation of top and bottom of the flagstaff are 45° and 30° respectively.
In right triangle ABC, we have
In right triangle ABD, we have
Comparing (i) and (ii), we get
Hence, the height of the tower = 9.56 m.
Let AB = x metres.
In right triangle ABD, we have
In right triangle ABC, we have
Comparing (i) and (ii), we get
et AB be the surface of the lake and 'P' be the position of the observer h metres above the lake. Let C be the cloud and C' be the reflection in the cloud. Then CB = C'B. It is also given that the angle of elevation of cloud from a point h m above a lake is α and angle of depression of its reflection be β. i.e., ∠CPQ = α and ∠QPC' = β. Let CQ = xm.
In right triangle PQC, we have
In right triangle PQC, we have
In right triangle PQC', we have
Comparing (ii) and (iii), we get
Comparing (i) and (iv), we get
Hence, the distance of the cloud from the point of observer is 
Let CD be the tree of height h m. Let B be the position of a man standing on the opposite bank of the river. After moving 40 m away from point B let new position of man be A i.e., AB = 40 m.
The angles of elevation of the top of the tree from point A and B are 30° and 60° respectively, i.e., ∠CAD = 30° and ∠CBD = 60°. Let BC = x m.
In right triangle BCD, we have
In right triangle ACD, we have
Comparing (i) and (ii), we get
Hence, the height of the tree is 34.64 metres. Now substituting the value of
Hence, the width of the river is 20 m.
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