Example 9

The angle of elevation of the top of a pole is 15⁰ from a point on the ground. On walking 100 feet towards the pole the angle of elevation is found to be 30⁰ . Find the height of the pole.

Solution

Let PQ be the pole from point M when observed at P, ÐPMQ = 15⁰. In the direction MQ when walked 100 feet to point L, Ð PLQ = 30⁰ . Now ÐPLQ is the exterior angle of triangle PML \ Ð PML + Ð MPL = 30⁰ . But Ð PML = 15⁰

\ Ð MPL = 15⁰

i.e. Triangle is an isosceles triangle

ML = PL = 100 feet

Now from triangle PLQ, sin 300 =

\ PQ = 100 ´ sin 30⁰= 100 ´ 1/ 2 = 50

Hence PQ = 50 feet

\ Height of the pole is 50 feet.

Example 10

When seen from the top and bottom of a tower towards the top of a building, the angle of depression and the angle of elevation are 27 ⁰ and 63⁰ respectively. If the height of the building is 20 m. Find the height of the tower.

Solution

DC = 20 m, DE = 20 m. Let AE = x, BC = y

Now in right triangle BCD, tan 27⁰ =

Index

3. 1 Solving Right Triangles
3. 2 Law of Cosines
3. 3 Law of Sines
3. 4 The Ambiguous Case of Law of Sines
3. 5 Areas of Triangles
Supplementary Problems

Chapter 4