Maharashtra Board Class 10 Maths Chapter 6 Trigonometry Set 6.2 Solutions

Question 1. A person is standing at a distance of 80 m from a church looking at its top. The angle of elevation is of 45°. Find the height of the church.
Solution:
Let AB represent the height of the church and point C represent the position of the person.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज ABC दिखाया गया है, जहाँ B पर समकोण है। कोण C 45° है और आधार BC 80 मीटर है। AB चर्च की ऊँचाई को दर्शाता है और C प्रेक्षक की स्थिति को।
BC = 80 m
Angle of elevation = ∠ACB = 45°
In right angled ∆ABC,
tan 45° = \( \frac{AB}{BC} \) ... [By definition]

\( \implies \) 1 = \( \frac{AB}{80} \)

\( \implies \) AB = 80m

\( \implies \) The height of the church is 80 m.
Answer: The height of the church is 80 m.
In simple words: Using the tangent function for the given angle of elevation and distance, we can calculate the vertical height of the church. Since tan 45° is 1, the height equals the base distance.

🎯 Exam Tip: Always draw a clear diagram to represent the given information in word problems involving trigonometry. This helps in correctly identifying the right-angled triangle and applying the appropriate trigonometric ratios.

Question 2. From the top of a lighthouse, an observer looking at a ship makes angle of depression of 60°. If the height of the lighthouse is 90 metre, then find how far the ship is from the lighthouse. ( √3 = 1.73)
Solution:
Let AB represent the height of lighthouse and point C represent the position of the ship.
AB = 90 m
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक प्रकाशस्तंभ AB को दर्शाता है जिसकी ऊँचाई 90 मीटर है। बिंदु A से, एक जहाज C को देखने पर 60° का अवनमन कोण बनता है। चित्र में क्षैतिज AP और रेखा AC के बीच 60° का कोण दिखाया गया है, जो BC के समांतर है।
Angle of depression = ∠PAC = 60°
Now, ray AP || seg BC

\( \implies \) ∠ACB = ∠PAC ... [Alternate angles]

\( \implies \) ∠ACB = 60°
In right angled ∆ABC,
tan 60° = \( \frac{AB}{BC} \) [By definition]

\( \implies \) \( \sqrt{3} \) = \( \frac{90}{BC} \)

\( \implies \) BC = \( \frac{90}{\sqrt{3}} \)
= \( \frac{90}{\sqrt{3}} \) × \( \frac{\sqrt{3}}{\sqrt{3}} \) [On rationalising the denominator]
= \( \frac{90\sqrt{3}}{3} \)
= 30\( \sqrt{3} \) = 30 × 1.73 = 51.90 m

\( \implies \) The ship is 51.90 m away from the lighthouse.
Answer: The ship is 51.90 m away from the lighthouse.
In simple words: The angle of depression from the lighthouse top to the ship is equal to the angle of elevation from the ship to the lighthouse. Using tan 60° and the lighthouse height, we can find the distance of the ship from the base of the lighthouse.

🎯 Exam Tip: Remember that the angle of depression from an observer's eye to an object is equal to the angle of elevation from the object to the observer's eye, due to alternate interior angles with a horizontal line.

Question 3. Two buildings are facing each other on a road of width 12 metre. From the top of the first building, which is 10 metre high, the angle of elevation of the top of the second is found to be 60°. What is the height of the second building?
Solution:
Let AB and CD represent the heights of the two buildings, and BD represent the width of the road.
ℹ️ चित्र व्याख्या (Diagram Explanation): चित्र में दो इमारतें AB (ऊँचाई 10 मीटर) और CD को दिखाया गया है, जिनके बीच की सड़क BD की चौड़ाई 12 मीटर है। पहली इमारत के शीर्ष A से दूसरी इमारत के शीर्ष C तक का उन्नयन कोण 60° है। AM को CD पर लंबवत खींचा गया है, जिससे ABDM एक आयत बनता है।
Draw seg AM ⊥ seg CD.
Angle of elevation = ∠CAM = 60°
AB = 10 m
BD= 12 m
In JABDM,
∠B = ∠D = 90°

\( \implies \) ∠A = 90° ... [Remaining angle of JABDM]

\( \implies \) JABDM is a rectangle .... [Each angle is 90°]

\( \implies \) AM = BD = 12 m opposite sides
DM = AB = 10 m of a rectangle
In right angled AAMC,
tan 60° = \( \frac{CM}{AM} \) ...[By definition]

\( \implies \) \( \sqrt{3} \) = \( \frac{CM}{12} \)

\( \implies \) CM = 12\( \sqrt{3} \) m
Now, CD = DM + CM ... [C - M - D]

\( \implies \) CD = (10 + 12\( \sqrt{3} \))m
= 10 + 12 × 1.73
= 10 + 20.76 = 30.76

\( \implies \) The height of the second building is 30.76 m.
Answer: The height of the second building is 30.76 m.
In simple words: We create a right-angled triangle by drawing a horizontal line from the top of the first building. This allows us to find the vertical height difference using the angle of elevation and the road width, which is then added to the first building's height.

🎯 Exam Tip: When dealing with two objects of different heights, construct a rectangle and a right-angled triangle. The height difference will be the side of the triangle, and the base of the triangle will be equal to the distance between the objects.

Question 4. Two poles of heights 18 metre and 7 metre are erected on a ground. The length of the wire fastened at their tops is 22 metre. Find the angle made by the wire with the horizontal.
Solution:
Let AB and CD represent the heights of two poles, and AC represent the length of the wire.
ℹ️ चित्र व्याख्या (Diagram Explanation): चित्र में दो खंभे AB (7 मीटर) और CD (18 मीटर) दिखाए गए हैं। इनके शीर्षों को जोड़ने वाले तार AC की लंबाई 22 मीटर है। A से CD पर लंब AM खींचा गया है, जिससे ABDM एक आयत बनता है। तार AC क्षैतिज रेखा AM के साथ θ कोण बनाता है।
Draw seg AM ⊥ seg CD.
Angle of elevation = ∠CAM = θ
AB = 7 m
CD = 18 m
AC = 22 m
In JABDM,
∠B = ∠D = 90°

\( \implies \) ∠A = 90° ... [Remaining angle of JABDM]

\( \implies \) ABDM is a rectangle. ... [Each angle is 90°]

\( \implies \) DM = AB = 7 m [Opposite sides of a rectangle]
Now, CD = CM + DM ... [C - M - D]

\( \implies \) 18 = CM + 7

\( \implies \) CM = 18-7 = 11 m
In right angled ∆AMC,
sin θ = \( \frac{CM}{AC} \) ..... [By definition]

\( \implies \) sin θ = \( \frac{11}{22} \) = \( \frac{1}{2} \)
But, sin 30° = \( \frac{1}{2} \)

\( \implies \) θ = 30°

\( \implies \) The angle made by the wire with the horizontal is 30°.
Answer: The angle made by the wire with the horizontal is 30°.
In simple words: By creating a horizontal line from the top of the shorter pole to the taller pole, we form a right-angled triangle. The vertical side of this triangle is the difference in pole heights, and the hypotenuse is the wire length. Using the sine ratio, we can find the angle.

🎯 Exam Tip: When finding an unknown angle, calculate the trigonometric ratio (sin, cos, or tan) using the known sides, and then use the inverse trigonometric function or a standard trigonometric table to determine the angle.

Question 5. A storm broke a tree and the treetop rested 20 m from the base of the tree, making an angle of 60° with the horizontal. Find the height of the tree.
Solution:
Let AB represent the height of the tree.
Suppose the tree broke at point C and its top touches the ground at D.
AC is the broken part of the tree which takes position CD such that ∠CDB = 60°

\( \implies \) AC = CD ...(i)
BD = 20 m
In right angled ΔCBD,
ℹ️ चित्र व्याख्या (Diagram Explanation): चित्र में एक पेड़ AB दर्शाया गया है जो बिंदु C पर टूट गया। पेड़ का टूटा हुआ ऊपरी भाग AC, जमीन पर बिंदु D पर टिका है। आधार से D की दूरी BD 20 मीटर है, और टूटे हुए भाग (CD) द्वारा जमीन के साथ बनाया गया कोण 60° है।
tan 60° = \( \frac{BC}{BD} \) [By definition]

\( \implies \) \( \sqrt{3} \) = \( \frac{BC}{20} \)

\( \implies \) BC = 20\( \sqrt{3} \) m
Also, cos 60° = \( \frac{BC}{CD} \) [By definition]

\( \implies \) \( \frac{1}{2} \) = \( \frac{20}{CD} \)

\( \implies \) CD = 20 × 2 = 40 m

\( \implies \) AC = 40 m ... [From(i)]
Now, AB = AC + BC ....[A - C - B]
= 40 + 20\( \sqrt{3} \)
= 40 + 20 × 1.73
= 40 + 34.6
= 74.6

\( \implies \) The height of the tree is 74.6 m.
Answer: The height of the tree is 74.6 m.
In simple words: When a tree breaks, its original height is the sum of the unbroken part and the broken part. We use tangent to find the unbroken part (BC) and cosine to find the broken part (CD = AC), then add them up.

🎯 Exam Tip: For problems involving broken objects, remember that the broken part becomes the hypotenuse of the right-angled triangle, and its length remains the same as its original length before breaking.

Question 6. A kite is flying at a height of 60 m above the ground. The string attached to the kite is tied at the ground. It makes an angle of 60° with the ground. Assuming that the string is straight, find the length of the string. (√3 = 1.73)
Solution:
Let AB represent the height at which kite is flying and point C represent the point where the string is tied at the ground.
∠ACB is the angle made by the string with the ground.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक समकोण त्रिभुज ABC दर्शाया गया है, जहाँ B पर समकोण है। AB पतंग की जमीन से ऊँचाई है, जो 60 मीटर है। C वह बिंदु है जहाँ डोरी जमीन पर बंधी है। डोरी AC द्वारा जमीन के साथ बनाया गया कोण (∠ACB) 60° है।
∠ACB = 60°
AB = 60 m
In right angled ∆ABC,
sin 60° = \( \frac{AB}{AC} \) ... [By definition]

\( \implies \) \( \frac{\sqrt{3}}{2} \) = \( \frac{60}{AC} \)

\( \implies \) AC = \( \frac{60 \times 2}{\sqrt{3}} \)
= \( \frac{120}{\sqrt{3}} \)
= \( \frac{120}{\sqrt{3}} \) × \( \frac{\sqrt{3}}{\sqrt{3}} \) [On rationalising the denominator]
= \( \frac{120\sqrt{3}}{3} \)

\( \implies \) AC = 40\( \sqrt{3} \) = 40 × 1.73 = 69.20 m

\( \implies \) The length of the string is 69.20 m.
Answer: The length of the string is 69.20 m.
In simple words: Given the height of the kite and the angle the string makes with the ground, we can use the sine ratio to find the length of the string, as it represents the hypotenuse of the right-angled triangle formed.

🎯 Exam Tip: When the height and angle are known, and the hypotenuse (e.g., string length or ladder length) is required, the sine ratio (opposite/hypotenuse) is usually the most direct method.