Class 10 Mathematics · Chapter 9 NotesSome Applications of Trigonometry

When you look at an object, the straight line from your eye to that point on the object is called the line of sight. If the object is above your horizontal level — you raise your head to see it — the angle between the line of sight and the horizontal is the angle of elevation. If the object is below your horizontal level — you lower your head to see it — the angle between the line of sight and the horizontal is the angle of depression. In diagrams, the horizontal is drawn through the eye of the observer, so the angle of elevation opens upward and the angle of depression opens downward.

• Line of sightthe line from the observer's eye to the point being viewed.
• Angle of elevationformed with the horizontal when the point is above the horizontal level.
• Angle of depressionformed with the horizontal when the point is below the horizontal level.

Most problems in this chapter are solved by drawing a right-angled triangle. The vertical object (tower, pole, building, tree) forms one perpendicular side, the horizontal ground distance forms the base, and the line of sight forms the hypotenuse. The angle of elevation or depression sits at the observer's position. Once the triangle is drawn, label the known values and the unknown quantity. If the observer has a height, the horizontal line through the eyes is parallel to the ground, so the observer's height is added separately to the calculated vertical length.

• Vertical object → perpendicular side of the triangle.
• Horizontal distance on the ground → base of the triangle.
• Line of sight → hypotenuse.
• Observer's height is added to the calculated height when the eyes are above the ground.

After drawing the triangle, decide which ratio connects the two known quantities with the unknown one. If the perpendicular and base are involved, use tan θ = perpendicular ÷ base. If the perpendicular and hypotenuse are involved, use sin θ = perpendicular ÷ hypotenuse. If the base and hypotenuse are involved, use cos θ = base ÷ hypotenuse. Substitute the known values, use standard values such as tan 45° = 1, tan 30° = 1/√3, tan 60° = √3, sin 30° = 1/2 and sin 60° = √3/2, and solve the equation for the unknown.

• tan θ = perpendicular ÷ base
• sin θ = perpendicular ÷ hypotenuse
• cos θ = base ÷ hypotenuse
• Standard valuestan 45° = 1, tan 30° = 1/√3, tan 60° = √3, sin 30° = 1/2, sin 60° = √3/2

The simplest problems involve one right-angled triangle. For example, if a tower stands on level ground and a point 15 m from its foot has an angle of elevation of 60°, then tan 60° = height ÷ 15, so height = 15√3 m. If a ladder makes 60° with the horizontal and must reach a point 3.7 m above the ground, then sin 60° = 3.7 ÷ ladder length, giving the ladder length, and cot 60° = distance from the pole ÷ 3.7 gives the distance from the foot of the pole. In every case, write the ratio, substitute, and solve.

• One triangle → one equation → one unknown.
• Use tan when height and ground distance are involved.
• Use sin or cos when the ladder or rope (hypotenuse) is involved.

Many real situations need two right-angled triangles that share a common side. A building with a flagstaff on top, a tower with a statue on it, or a point from which two different angles of elevation are measured all produce two triangles. Write a trigonometric equation for each triangle, then link them using the common side or a known total length. For instance, if the angle of elevation of the top of a 10 m building is 30° and of the top of the flagstaff is 45° from the same point, the first triangle gives the ground distance and the second gives the total height, from which the flagstaff length is found.

• Draw both triangles separately and label the common side.
• Write one equation per triangle.
• Solve the simpler equation first, then substitute into the other.

When the observer is at the top of a building or tower and looks down, the angle of depression is measured from the horizontal through the observer's eye. Because the horizontal line through the eye is parallel to the ground, the angle of depression equals the corresponding angle of elevation at the object, by alternate angles. This lets you place the given angle inside the right-angled triangle at the ground level and use the usual ratios. For example, from the top of a multi-storeyed building, the angles of depression of the top and bottom of an 8 m building are 30° and 45°; these become angles of elevation 30° and 45° in the triangles used for calculation.

• Angle of depression = corresponding angle of elevation (alternate angles).
• Draw the horizontal through the observer's eye.
• Transfer the angle to the ground-level triangle before applying ratios.

The same method answers a wide range of practical questions: the height of a tower from its shadow, the width of a river from a bridge, the distance between two ships from a lighthouse, the distance travelled by a balloon as its angle of elevation changes, and the time a car takes to reach a tower as the angle of depression changes. In each case, draw the diagram, mark the right angles, write the correct trigonometric equation, and solve. When a quantity changes over time, use the two positions to form two triangles and subtract the distances to find the change.

• Shadow problemsthe shadow length is the base of the triangle.
• River problemsthe width is the sum of two base lengths.
• Moving-object problemscompare two positions using two triangles.