Some Applications of Trigonometry — Karnataka (SSLC) Class 10 Mathematics Solutions (Free)
Free step-by-step Karnataka (SSLC) Class 10 Mathematics solutions for "Some Applications of Trigonometry" — important questions with detailed answers, download PDF for board exam preparation.
TL;DR: Free step-by-step Karnataka (SSLC) Class 10 Mathematics solutions for "Some Applications of Trigonometry" — important questions with detailed answers,…
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Q1: Define angle of elevation and angle of depression. Draw a diagram showing both concepts.
Angle of Elevation:
The angle of elevation is the angle between the horizontal line of sight and the line of sight upward to an object that is above the horizontal.
When an observer looks upward at an object, the angle formed below the object's line of sight from the horizontal is called the angle of elevation.
Angle of Depression:
The angle of depression is the angle between the horizontal line of sight and the line of sight downward to an object that is below the horizontal.
When an observe…
Q2: A ladder 10 m long leans against a wall. The ladder makes an angle of 60° with the ground. Find the height of the wall reached by the ladder and the distance of the foot of the ladder from the wall.
Given:
Length of ladder = 10 m (hypotenuse)
Angle with ground = 60°
Find:
1. Height of wall reached by ladder (opposite side)
2. Distance from wall to foot of ladder (adjacent side)
Diagram (text description):
Wall
|
| Height (h)
|/
Ladder (10 m)
/|
/ |
/ |
/60°|
Ground
Distance (d)
Step 1: Find the height (opposite side).
sin 60° = height / ladder
sin 60° = h / 10
√3/2 = h / 10
h = 10 × (√3/2)
h = 5√3 m
h ≈ 5 × 1.732 = 8.66 m
Step 2: Find the …
Q3: A man standing on a level ground observes the angle of elevation to the top of a building as 30°. He walks 50 m closer to the building and observes the angle of elevation as 60°. Find the height of the building.
Given:
Initial angle of elevation = 30°
Angle of elevation after walking 50 m closer = 60°
Distance walked = 50 m
Find: Height of building
Diagram (text description):
Building (height = h)
|
|
| 60°
|/
---*--- (closer position, 50m away)
/ |
30° | h
/ |
*-------+-------* (original position)
50m d
Total distance = d + 50 m
Step 1: From the original position (distance = d + 50):
tan 30° = …
Q4: An observer on a lighthouse 60 m high observes two ships at angles of depression 30° and 60°. The ships are on the same side of the lighthouse and are on the same horizontal line. Find the distance between the ships.
Given:
Height of lighthouse = 60 m
Angle of depression to ship 1 = 30°
Angle of depression to ship 2 = 60°
Both ships are on the same side and same horizontal line
Find: Distance between the ships
Diagram (text description):
Lighthouse
|\60 m
| \
| \ 30°
| \
| \_____ Ship 1 (distance d₁ from base)
|
| \ 60°
| \
|___\
Base| Ship 2 (distance d₂ from base)
Note: Angle of depression …
Q5: From a point on the ground 100 m from the foot of a building, the angle of elevation to the top is 45°. Find the height of the building. If the angle changes to 60° after moving 25 m closer, verify your answer.
Given:
Initial distance from building = 100 m
Initial angle of elevation = 45°
After moving 25 m closer, new angle = 60°
Find: Height of building
Part 1: Using initial observation
Step 1: Use angle of elevation formula.
tan 45° = height / distance
1 = h / 100
h = 100 m
Part 2: Verification with second observation
Step 2: New distance after moving 25 m closer.
New distance = 100 - 25 = 75 m
Step 3: Check with the second angle.
tan 60° = h / new distance
√3 = h / 75
h = 75√3 m
h ≈ 75 × 1.732…
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