Maharashtra Board Class 9 Maths Part 2 Geometry Chapter 8 Set 8 Trigonometry Solutions

Question 1. Choose the correct alternative answer for the following multiple choice questions.
 

(i) Which of the following statements is true?
(A) sin θ = cos (90 – θ)
(B) cos θ = tan (90 – θ)
(C) sin θ = tan (90 – θ)
(D) tan θ = tan (90 – θ)
Answer: (A) sin θ = cos (90 – θ)
In simple words: This question tests the fundamental trigonometric identity relating sine and cosine of complementary angles, where sin θ is equal to cos (90° - θ).

🎯 Exam Tip: Remember the complementary angle identities (sin (90-θ) = cos θ, cos (90-θ) = sin θ, tan (90-θ) = cot θ, etc.) as they are crucial for solving many trigonometry problems.

Question.
(ii) Which of the following is the value of sin 90°?
(A) \( \frac{\sqrt{3}}{2} \)
(B) 0
(C) \( \frac{1}{2} \)
(D) 1
Answer: (D) 1
In simple words: The value of sin 90° is 1, which is a standard trigonometric value to be memorized for specific angles.

🎯 Exam Tip: Memorize the standard trigonometric values for common angles like 0°, 30°, 45°, 60°, and 90° as they frequently appear in problems.

Question.
(iii) 2 tan 45° + cos 45° – sin 45° = ?
(A) 0
(B) 1
(C) 2
(D) 3
Answer: (C) 2
2 tan 45° + cos 45° – sin
\( = 2(1) + \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 2 \)
(C) 2
In simple words: By substituting the known values of tan 45°, cos 45°, and sin 45° into the expression, the terms \( \frac{1}{\sqrt{2}} \) and \( -\frac{1}{\sqrt{2}} \) cancel out, leaving \( 2 \times 1 = 2 \).

🎯 Exam Tip: Accurately substituting standard trigonometric values and performing basic arithmetic operations are essential for these types of questions.

Question.
(iv) \( \frac{\cos 28^\circ}{\sin 62^\circ} \) = ?
(A) 2
(B) -1
(C) 0
(D) 1
Answer: (D) 1
\( \frac{\cos 28^\circ}{\sin 62^\circ} \)
\( = \frac{\sin(90^\circ-28^\circ)}{\sin 62^\circ} \)
...[.: cos θ = sin (90 – θ)]
\( = \frac{\sin 62^\circ}{\sin 62^\circ} \)
= 1
(D) 1
In simple words: Using the complementary angle identity \( \cos \theta = \sin(90^\circ - \theta) \), we can rewrite \( \cos 28^\circ \) as \( \sin(90^\circ - 28^\circ) = \sin 62^\circ \), which then simplifies the fraction to 1.

🎯 Exam Tip: Recognize complementary angle pairs (like 28° and 62°) and apply the appropriate trigonometric identities to simplify expressions efficiently.

Question 2. In right angled ATSU, TS = 5, ∠S = 90°, SU = 12, then find sin T, cos T, tan T. Similarly find sin U, cos U, tan U.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज TSU है जहाँ कोण S 90 डिग्री है। भुजा TS की लंबाई 5 है और भुजा SU की लंबाई 12 है। भुजा TU त्रिभुज का कर्ण है।
Solution:
i. TS = 5, SU = 12 ...[Given]
In ATSU, ∠S = 90° ... [Given]
.:. TU\(^2\) = TS\(^2\) + SU\(^2\) ... [Pythagoras theorem]
= 5\(^2\) + 12\(^2\) = 25 + 144 = 169
.. TU = \( \sqrt{169} \) ...[Taking square root of both sides]
= 13
ii. sin T = \( \frac{\text{Opposite side of ∠T}}{\text{Hypotenuse}} = \frac{\text{SU}}{\text{TU}} = \frac{12}{13} \)
iii. cos T = \( \frac{\text{Adjacent side of ∠T}}{\text{Hypotenuse}} = \frac{\text{TS}}{\text{TU}} = \frac{5}{13} \)
iv. tan T = \( \frac{\text{Opposite side of ∠T}}{\text{Adjacent side of ∠T}} = \frac{\text{SU}}{\text{TS}} = \frac{12}{5} \)
v. sin U = \( \frac{\text{Opposite side of ∠U}}{\text{Hypotenuse}} = \frac{\text{TS}}{\text{TU}} = \frac{5}{13} \)
vi. cos U = \( \frac{\text{Adjacent side of ∠U}}{\text{Hypotenuse}} = \frac{\text{SU}}{\text{TU}} = \frac{12}{13} \)
vii. tan U = \( \frac{\text{Opposite side of ∠U}}{\text{Adjacent side of ∠U}} = \frac{\text{TS}}{\text{SU}} = \frac{5}{12} \)
In simple words: First, use the Pythagorean theorem to find the length of the hypotenuse TU. Then, apply the definitions of sine, cosine, and tangent (SOH CAH TOA) with respect to angles T and U, using the opposite side, adjacent side, and hypotenuse lengths.

🎯 Exam Tip: Always draw and label the right-angled triangle clearly. Remember SOH CAH TOA to correctly identify opposite, adjacent, and hypotenuse sides relative to the angle in question. Pythagorean theorem is often the first step to find missing side lengths.

Question 3. In right angled ∆ΥΧΖ, \( \angle \)X = 90°, XZ = 8 cm, YZ = 17 cm, find sin Y, cos Y, tan Y, sin Z, cos Z, tan Z.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज YXZ है जहाँ कोण X 90 डिग्री है। भुजा XZ की लंबाई 8 है और भुजा YZ (कर्ण) की लंबाई 17 है। भुजा XY अज्ञात है।
Solution:
i. XZ = 8 cm, YZ = 17 cm ...[Given]
In ΔYXZ, \( \angle \)X = 90° ... [Given]
.. YZ\(^2\) = XY\(^2\) + XZ\(^2\) ...[Pythagoras theorem]
.. 17\(^2\) = XY\(^2\) + 8\(^2\)
.. 289 = XY\(^2\) + 64
.. XY\(^2\) = 289 – 64
= 225
.. XY = \( \sqrt{225} \) ...[Taking square root of both sides]
= 15
ii. sin Y = \( \frac{\text{Opposite side of ∠Y}}{\text{Hypotenuse}} = \frac{\text{XZ}}{\text{YZ}} = \frac{8}{17} \)
iii. cos Y = \( \frac{\text{Adjacent side of ∠Y}}{\text{Hypotenuse}} = \frac{\text{XY}}{\text{YZ}} = \frac{15}{17} \)
iv. tan Y = \( \frac{\text{Opposite side of ∠Y}}{\text{Adjacent side of ∠Y}} = \frac{\text{XZ}}{\text{XY}} = \frac{8}{15} \)
v. sin Z = \( \frac{\text{Opposite side of ∠Z}}{\text{Hypotenuse}} = \frac{\text{XY}}{\text{YZ}} = \frac{15}{17} \)
vi. cos Z = \( \frac{\text{Adjacent side of ∠Z}}{\text{Hypotenuse}} = \frac{\text{XZ}}{\text{YZ}} = \frac{8}{17} \)
vii. tan Z = \( \frac{\text{Opposite side of ∠Z}}{\text{Adjacent side of ∠Z}} = \frac{\text{XY}}{\text{XZ}} = \frac{15}{8} \)
In simple words: First, the unknown side XY is found using the Pythagorean theorem. Then, the sine, cosine, and tangent ratios are calculated for angle Y and angle Z by identifying their respective opposite, adjacent, and hypotenuse sides.

🎯 Exam Tip: Pay close attention to which angle (Y or Z) you are calculating the ratios for, as the 'opposite' and 'adjacent' sides will swap roles for each angle. Double-check calculations for square roots and fractions.

Question 4. In right angled ALMN, if \( \angle \)N = θ, \( \angle \)M = 90°, cos θ = \( \frac{24}{25} \), find sin θ and tan θ. Similarly, find (sin\(^2\)\( \theta \)) and (cos\(^2\)\( \theta \)).

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज LMN है जहाँ कोण M 90 डिग्री है और कोण N को θ दर्शाया गया है। त्रिभुज की भुजाओं LM, MN और LN को दर्शाया गया है, जहाँ MN आधार है और LN कर्ण है।
Solution:
i. cos θ = \( \frac{24}{25} \)
In ∆LMN, \( \angle \)M = 90°, \( \angle \)N = θ
cos θ = \( \frac{\text{Adjacent side of θ}}{\text{Hypotenuse}} \)
cos θ = \( \frac{\text{MN}}{\text{LN}} \) ...(ii)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज LMN है जहाँ कोण M 90 डिग्री है और कोण N को θ दर्शाया गया है। भुजा MN की लंबाई 24k है और भुजा LN की लंबाई 25k है, जिससे तीसरी भुजा LM की लंबाई 7k होगी।
.:. \( \frac{\text{MN}}{\text{LN}} = \frac{24}{25} \) ...[From (i) and (ii)]
Let the common multiple be k.
.. MN = 24k and LN = 25k
Now, LN\(^2\) = LM\(^2\) + MN\(^2\) ... [Pythagoras theorem]
.:. (25k)\(^2\) = LM\(^2\) + (24k)\(^2\)
.:. 625 k\(^2\) = LM\(^2\) + 576k\(^2\)
.. LM\(^2\) = 625k\(^2\) – 576k\(^2\)
.:. LM\(^2\) = 49k\(^2\)
.: LM = \( \sqrt{49k^2} \) ...[Taking square root of both sides]
= 7k
ii. sin θ = \( \frac{\text{Opposite side of θ}}{\text{Hypotenuse}} = \frac{\text{LM}}{\text{LN}} = \frac{7k}{25k} = \frac{7}{25} \)
.:. sin\(^2\)\( \theta \) = (sin θ)\(^2\) = \( (\frac{7}{25})^2 = \frac{49}{625} \)
iii. tan θ = \( \frac{\text{Opposite side of θ}}{\text{Adjacent side of θ}} = \frac{\text{LM}}{\text{MN}} = \frac{7k}{24k} = \frac{7}{24} \)
iv. cos θ = \( \frac{24}{25} \) ...[Given]
.. cos\(^2\)\( \theta \) = (cos θ)\(^2\) = \( (\frac{24}{25})^2 = \frac{576}{625} \)
In simple words: By using the given cos θ, we deduce the ratio of adjacent side to hypotenuse. We then use the Pythagorean theorem to find the length of the opposite side in terms of 'k', allowing us to calculate sin θ and tan θ, and finally their squares.

🎯 Exam Tip: Introducing a common multiple 'k' for side lengths derived from a ratio is a good practice to maintain proportionality and accurately calculate the third side using the Pythagorean theorem before finding other ratios.

Question 5. Fill in the blanks.
i. sin 20° = cos _______
ii. tan 30° x tan _______ = 1
iii. cos 40° = sin _______
Solution:
i. sin 20° = cos (90° – 20°) .....[.: sin θ = cos (90 – θ)]
= cos 70°
In simple words: This question tests the understanding of complementary angle identities, where the sine of an angle is equal to the cosine of its complement (90° minus the angle).

🎯 Exam Tip: Practice recognizing and applying the complementary angle identities (sin θ = cos (90-θ) and vice-versa) for quick calculations in fill-in-the-blank questions.

Question 5. (continued)
ii. tan θ x tan (90 – θ) = 1
Substituting θ = 30°,
tan 30° x tan (90 – 30)° = 1
.. tan 30° x tan 60° = 1
In simple words: This fill-in-the-blank uses the identity that the product of tangents of complementary angles is 1, so if one angle is 30°, the other must be 90°-30° = 60°.

🎯 Exam Tip: Remember the identity \( \tan \theta \times \tan(90^\circ - \theta) = 1 \). This allows for quick solutions when one angle is given and its complement is needed to satisfy the product rule.

Question 5. (continued)
iii. cos 40° = sin (90° – 40°) ...[:: COS θ = sin (90 – θ)]
= sin 50°
In simple words: This blank is filled by applying the complementary angle identity, stating that the cosine of an angle is equal to the sine of its complementary angle (90° minus the angle).

🎯 Exam Tip: Consistent application of complementary angle identities is key. If \( \cos A = \sin B \), then \( A + B = 90^\circ \).

Maharashtra Board Class 9 Maths Solutions Chapter 8 Trigonometry Problem Set 8

Question 1. Measuring height of a tree using trigonometric ratios. (Textbook pg. no. 101)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक पेड़ (QR) और एक छड़ी (BC) का चित्रण है, दोनों अपनी छाया (PR और AC) बनाते हैं। सूर्य की किरणें समानांतर दिखाई गई हैं, जो समान कोण बनाते हुए दोनों वस्तुओं से गुजरती हैं।
Solution:
Rays of sunlight are parallel.
So, ∆PQR and ∆ABC are equiangular i.e., similar triangles.
Sides of similar triangles are proportional.
.. \( \frac{\text{QR}}{\text{BC}} = \frac{\text{PR}}{\text{AC}} \)
.. Height of the tree (QR) = \( \frac{\text{BC}}{\text{AC}} \) x PR
Substituting the values of PR, BC and AC in the above equation, we can get length
of QR i.e., the height of the tree.
In simple words: By understanding that the sun's parallel rays create similar triangles (tree and its shadow, stick and its shadow), we can use the proportionality of their sides to calculate the unknown height of the tree.

🎯 Exam Tip: For problems involving heights and shadows, recognize that similar triangles are formed due to parallel sun rays. The proportionality of corresponding sides is the core principle for solving such problems.

Question 2. It is convenient to do the above experiment between 11:30 am and 1:30 pm instead of doing it in the morning at 8'O clock. Can you tell why? (Textbook pg. no. 101)
Solution:
At 8'O clock in the morning, the sunlight is not very bright. At the same time, the
sun is on the horizon and the shadow would by very long. It would be extremely
difficult to measure shadow in that case.
Between 11:30 am and 1:30 pm, the sun is overhead and it would be easier to
measure the length of shadow.
In simple words: Measuring shadows is easier and more accurate when the sun is higher in the sky (mid-day), as shadows are shorter and less distorted, unlike the long, indistinct shadows cast by a low-lying sun in the early morning.

🎯 Exam Tip: Practical reasoning questions require understanding the real-world conditions affecting measurements. Longer shadows due to low sun angles are harder to measure accurately, impacting trigonometric calculations.

Question 3. Conduct the above discussed activity and find the height of a tall tree in your
surrounding. If there is no tree in the premises, then find the height of a pole.
(Textbook pg. no. 101)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक लैंप पोस्ट और एक छड़ी का चित्रण है, दोनों को ज़मीन पर खड़ा दिखाया गया है। यह एक समान सिद्धांत का प्रदर्शन है जहां वस्तुओं की ऊंचाई और उनकी छाया की लंबाई का उपयोग करके माप किया जा सकता है।
Solution:
This question is an activity based question. It is intended to be performed by students.
In simple words: This is a hands-on activity for students to apply the concept of similar triangles by measuring the height of a real-world object like a tree or pole using its shadow and a reference stick.

🎯 Exam Tip: While an activity, understanding the underlying principle (similar triangles and proportional shadows) is crucial. Even if not directly tested, such questions reinforce the practical application of trigonometry.