Trigonometry Chapter 4 Application Of Trigonometric Ratios Height And Distance
⇔ The angle of Elevation: When an observer stands on a base, the point being viewed is above the horizontal level, the angle formed by the line of sight, of an observer with a horizontal line, is called the angle of elevation.
⇒ ∠AOP is the angle of elevation.
⇔ The angle of Depression: When the observer is looking down at an object, then the angle so formed by his line of sight with the horizontal line is called the angle of depression.
⇒ ∠BPO is the angle of depression.
⇒ As BP || OA
∴ Angle of elevation = Angle of depression
Read and Learn More WBBSE Solutions for Class 10 Maths
Trigonometry Chapter 4 Application Of Trigonometric Ratios Height And Distance True Or False
Example 1. In ΔABC, ∠B = 90°, if AB = BC then ∠C = 60°.
Solution: In ΔABC, ∠B = 90°
⇒ tan C = \(\frac{AB}{BC}\)
= \(\frac{AB}{AB}\) [as AB = BC]
= 1
⇒ tan C = tan 45° ⇒ ∠C = 45°
∴ The statement is false.
Example 2. PQ is the height of a building, QR is the base, and the angle of depression from a point R is ∠SPR; So, ∠SPR = ∠PRQ.
Solution: Clearly, the statement is true.
Class 10 Maths Trigonometry Chapter 4 Solutions
Example 3. If the angle of depression of a point is 30°, then the angle of elevation of that point is 30°
Solution: AB || DC and AC is intersection
∴ ∠ACB = ∠BAC = 30°
∴ The statement is true.
Example 4. If the length of the shadow on the ground of a post is \(\frac{1}{\sqrt{3}}\) times its height, then the angle of elevation is 60°
Solution: AB is the height and BC is the shadow of the post.
∠ABC = 90°, Let angle of elevation is θ.
∴ ∠ACB = θ and BC = \(\frac{A B}{\sqrt{3}}\)
tan θ = \(\frac{AB}{BC}\) = √3
⇒ tan θ = tan 60°
⇒ θ = 60°
∴ The statement is true
Trigonometry Chapter 4 Application Of Trigonometric Ratios Height And Distance Fill In The Blanks
Example 1. If the sun’s angle of elevation increases from 30° to 60°, the length of the shadow of a post _______ (decreases / increases).
Solution: When the angle of elevation is 30° then the length of the shadow of the post-AB is BD; If the angle of elevation is 60°, then the length of the shadow of AB is BC; As BD > BC.
∴ The length of the shadow of a post decreases.
Example 2. If the angle of elevation of sun is 45°, the length of shadow and length of post are _________
Solution: Let the length of the post is AB and the length of its shadow is BC
In ΔABC, \(\frac{AB}{BC}\) = tan 45°
\(\frac{AB}{BC}\) = 1
AB = BC
∴ Length of the shadow and the length of the post are equal.
Trigonometric Equations Class 10 Solutions
Example 3. If the angle of elevation of the sun is ________ 45°, the length of the shadow of the tower will be less than the height of tower.
Solution: Greater
[BD > BC]
Example 4. If the angle of elevation of sun decreases from 65° to 50°, the length of the shadow of a tower _______
Solution: increases.
Trigonometry Chapter 4 Application Of Trigonometric Ratios Height And Distance Short Answer Type Questions
Example 1. If the angle of elevation of a kite is 60° and the length of the thread is 20√3 metres, then calculate the height of the kite above the ground.
Solution: Let the height of the kite above the ground is AB and the length of the thread is AC.
The angle of elevation of the kite is ∠ACB
So, ∠ACB = 60° and AC = 20√3 metres.
In ΔABC, \(\frac{AB}{AC}\) = sin 60°
⇒ \(\frac{A B}{20 \sqrt{3}}m=\frac{\sqrt{3}}{2}\)
⇒ AB = \(\frac{60}{2}\) metre = 30 metre.
∴ The height of the kite is 30 metres.
Example 2. AC is the hypotenuse witha length of 100 metres of a right-angled triangle ABC and if AB = 50√3 metre, then find the value of ∠C.
Solution: In ΔABC, ∠ABC = 90°
∴ sin C = \(\frac{AB}{AC}\)
sin C = \(\frac{50 \sqrt{3}}{100}=\frac{\sqrt{3}}{2}\)
⇒ sin C = sin 60°
⇒ ∠C = 60°
Class 10 Trigonometry Chapter 4 Solved Examples
Example 3. A tree breaks due to storm and its top touches the ground in such a manner that the distance from the top of the tree to the base of the tree and present height are equal. Calculate how much angle is made by the top of the tree with the base.
Solution: Let the present height of the tree is AB and the distance from the top of the tree to the base of the tree is BC where AB = AC
⇒ tan ∠ACB = \(\frac{AB}{BC}\) = \(\frac{AB}{AB}\) = 1
⇒ tan ∠ACB = tan 45°
⇒ ∠ACB = 45°
∴ The angle is 45°
Example 4. In the right-angled triangle, ABC, ∠B = 90°, D is such a point on AB that AB: BC: BD = √3: 1: 1. Find the value of ∠ACD.
Solution: AB : BC : BD = √3 : 1 : 1
Let AB = √3 x unit, BC = x unit and BD = x unit where x is common multiple and x > 0.
In ΔDBC, ∠DBC = 90°
tan ∠BDC = \(\frac{BC}{BD}\) = \(\frac{x}{x}\) = 1
⇒ tan ∠BDC = tan 45° ⇒ ∠BDC = 45°
In ΔABC, tan ∠BAC = \(\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{x}{\sqrt{3} x}=\frac{1}{\sqrt{3}}\)
⇒ tan ∠BAC = tan 30°
⇒ ∠BAC = 30° i.e. ∠DAC = 30°
∴ ∠ACD = ∠BDC – ∠DAC = 45° – 30° = 15°
∴ The value of ∠ACD is 15°.
Example 5. If the ratio between length of shadow of a tower and height of tower is 3: 1, find the angle of elevation of the sun.
Solution: Let length of shadow of a tower is BC and height of tower is AB.
Where = \(\frac{\mathrm{BC}}{\mathrm{AP}}=\frac{\sqrt{3}}{1}\)
The angle of elevation of tower is ∠ACB
∴ In ΔABC, ∠ABC = 90°
∴ tan ACB = \(\frac{A B}{B C}=\frac{1}{\sqrt{3}}\)
⇒ tan ∠ACB = tan 30°
⇒ ∠ACB = 30°
∴ The angle of elevation is 30°.
Wbbse Class 10 Trigonometry Notes
Example 6. The angles of elevation of the top of a tower from two points at a distance 4 metres and 9 metres from the base and in the same straight line with it are complementary, hind the height of the tower.
Solution: Let the height of the tower AB is h metre, [h > 0]
Let C and D be two points at distances 4 metres and 9 metres respectively from the base of the tower.
BC = 4 metre and BD = 9 metre.
Let ∠ADB = θ and ∠ACB = 90° – θ
In ΔADB, ∠B = 90° tanθ = \(\frac{AB}{BD}\)
⇒ tanθ = \(\frac{h}{9}\)……(1)
In ΔABC, tan (90°- θ) = \(\frac{AB}{BC}\)
⇒ cotθ = \(\frac{h}{4}\)……..(2)
From (1) and (2), we have,
\(\frac{h}{9}\) x \(\frac{h}{4}\) = tanθ.cotθ
⇒ \(\frac{h^2}{36}=1\) ⇒ h2 = 36 ⇒ h = √36 = 6
∴ The length of the tower is 6 metre.
Example 7. If AB = 5 m and BC = 5√3 m, then find the value of θ.
Solution: tanθ = \(\frac{A B}{B C}=\frac{5}{5 \sqrt{3}}=\frac{1}{\sqrt{3}}\)
⇒ tanθ = tan 30°
⇒ θ = 30°
Solving Trigonometric Equations Class 10
Example 8. On the same side of a house, two objects are located. When observed from the top of the house, their angles of depression are 45° and 60°. If the height of the house is 50 m find the distance between the objects.
Solution: Let, the height of the house = AB = 50m.
Distance between the two objects = CD = x m. [x>0]
BC = y m [y>0]
The angle of depression from the point A of AB at the point C is 60° and angle of depression at the point D is 45°.
∴ ∠CAE = 60° and ∠DAE = 45°
∴ ∠ACB = ∠CAE = 60° [As AE || BD]
and ∠ADB = ∠DAE = 45°
In ΔABC, ∠ABC = 90°
∴ \(\frac{AB}{BC}\) = tan 60°
⇒ \(\frac{50}{y}\) = √3
⇒ y = \(\frac{50}{\sqrt{3}}\)
In ΔABD, \(\frac{AB}{BD}\) = tan 45°
\(\frac{50}{y+x}\) = 1
Class 10 Maths Trigonometry Important Questions
⇒ x + y = 50
x = 50 – y = 50 – \(\frac{50}{\sqrt{3}}\) = 50 – \(\frac{50 \sqrt{3}}{3}\)
= 50 – \(\frac{50 \times 1 \cdot 732}{3}\) = 50 – 28.87 = 21.13 (approx)
∴ The distance between the objects is 21.13 m (approx)