Maharashtra Board Class 9 Maths Part 2 Geometry Chapter 8 Set 8.1 Trigonometry Solutions

Question 1. In the given figure, ∠R is the right angle of ∆PQR. Write the following ratios.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज PQR है जहाँ कोण R 90 डिग्री का है। P शीर्ष पर है, Q आधार के दाहिने सिरे पर है, और R समकोण बनाते हुए नीचे बाईं ओर है।
Answer:
(i) sin P
(ii) cos Q
(iii) tan P
(iv) tan Q
Solution:
(i) \( \sin P = \frac{\text{Opposite side of } \angle P}{\text{Hypotenuse}} = \frac{QR}{PQ} \)
(ii) \( \cos Q = \frac{\text{Adjacent side of } \angle Q}{\text{Hypotenuse}} = \frac{QR}{PQ} \)
(iii) \( \tan P = \frac{\text{Opposite side of } \angle P}{\text{Adjacent side of } \angle P} = \frac{QR}{PR} \)
(iv) \( \tan Q = \frac{\text{Opposite side of } \angle Q}{\text{Adjacent side of } \angle Q} = \frac{PR}{QR} \)
In simple words: This question asks for the basic trigonometric ratios (sine, cosine, tangent) for angles P and Q in a right-angled triangle PQR, based on their opposite, adjacent sides, and the hypotenuse.

🎯 Exam Tip: Remember SOH CAH TOA (Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent) to easily write down the ratios.

Question 2. In the right angled ∆XYZ, ∠XYZ = 90° and a, b, c are the lengths of the sides as shown in the figure. Write the following ratios.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज XYZ है जहाँ कोण Y 90 डिग्री का है। X शीर्ष पर है, Z आधार के दाहिने सिरे पर है, और Y समकोण बनाते हुए नीचे बाईं ओर है। भुजाओं की लंबाईयां XY=b, YZ=a और XZ=c हैं।
Answer:
(i) sin x
(ii) tan z
(iii) cos x
(iv) tan x.
Solution:
(i) \( \sin X = \frac{\text{Opposite side of } \angle X}{\text{Hypotenuse}} = \frac{YZ}{XZ} = \frac{a}{c} \)
(ii) \( \tan Z = \frac{\text{Opposite side of } \angle Z}{\text{Adjacent side of } \angle Z} = \frac{XY}{YZ} = \frac{b}{a} \)
(iii) \( \cos X = \frac{\text{Adjacent side of } \angle X}{\text{Hypotenuse}} = \frac{XY}{XZ} = \frac{b}{c} \)
(iv) \( \tan X = \frac{\text{Opposite side of } \angle X}{\text{Adjacent side of } \angle X} = \frac{YZ}{XY} = \frac{a}{b} \)
In simple words: This question applies the basic trigonometric ratios to a right-angled triangle XYZ, using the given side lengths a, b, and c to express sine, cosine, and tangent for angles X and Z.

🎯 Exam Tip: Always identify the hypotenuse, opposite, and adjacent sides relative to the angle in question before writing the ratios to avoid errors.

Question 3. In right angled ∆LMN, ∠LMN = 90°, ∠L = 50° and ∠N = 40°. Write the following ratios.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज LMN है जहाँ कोण M 90 डिग्री का है। L शीर्ष पर 50 डिग्री के कोण के साथ है, N आधार के दाहिने सिरे पर 40 डिग्री के कोण के साथ है, और M समकोण बनाते हुए नीचे बाईं ओर है।
Answer:
(i) sin 50°
(ii) cos 50°
(iii) tan 40°
(iv) cos 40°
Solution:
(i) \( \sin 50^\circ = \frac{\text{Opposite side of } 50^\circ}{\text{Hypotenuse}} = \frac{MN}{LN} \)
(ii) \( \cos 50^\circ = \frac{\text{Adjacent side of } 50^\circ}{\text{Hypotenuse}} = \frac{LM}{LN} \)
(iii) \( \tan 40^\circ = \frac{\text{Opposite side of } 40^\circ}{\text{Adjacent side of } 40^\circ} = \frac{LM}{MN} \)
(iv) \( \cos 40^\circ = \frac{\text{Adjacent side of } 40^\circ}{\text{Hypotenuse}} = \frac{MN}{LN} \)
In simple words: This question requires writing trigonometric ratios for specific angles (50° and 40°) within a right-angled triangle LMN, using the labels of its sides.

🎯 Exam Tip: Note that sin 50° = cos 40° and cos 50° = sin 40° because 50° and 40° are complementary angles (sum to 90°).

Question 4. In the given figure, ∠PQR = 90°, ∠PQS = 90°, ∠PRQ = α and ∠QPS = θ. Write the following trigonometric ratios.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह दो समकोण त्रिभुजों PQR और PQS को दर्शाता है, जो भुजा PQ को साझा करते हैं। कोण PQR 90 डिग्री है और कोण PQS भी 90 डिग्री है। कोण PRQ को अल्फा (α) और कोण QPS को थीटा (θ) से दर्शाया गया है।
Answer:
(i) sin α, cos α, tan α
(ii) sin θ, cos θ, tan θ
Solution:
(i) In ∆PQR,
\( \sin \alpha = \frac{\text{Opposite side of } \alpha}{\text{Hypotenuse}} = \frac{PQ}{PR} \)
\( \cos \alpha = \frac{\text{Adjacent side of } \alpha}{\text{Hypotenuse}} = \frac{RQ}{PR} \)
\( \tan \alpha = \frac{\text{Opposite side of } \alpha}{\text{Adjacent side of } \alpha} = \frac{PQ}{RQ} \)
(ii) In ∆PQS,
\( \sin \theta = \frac{\text{Opposite side of } \theta}{\text{Hypotenuse}} = \frac{QS}{PS} \)
\( \cos \theta = \frac{\text{Adjacent side of } \theta}{\text{Hypotenuse}} = \frac{PQ}{PS} \)
\( \tan \theta = \frac{\text{Opposite side of } \theta}{\text{Adjacent side of } \theta} = \frac{QS}{PQ} \)
In simple words: This question involves identifying trigonometric ratios for two different angles, α and θ, which are part of two interconnected right-angled triangles PQR and PQS, respectively.

🎯 Exam Tip: When dealing with multiple triangles sharing a side, clearly identify which triangle and which angle you are working with to correctly assign opposite, adjacent, and hypotenuse sides.

Maharashtra Board Class 9 Maths Chapter 8 Trigonometry Practice Set 8.1 Intext Questions and Activities

Question 1. In the figure given below, ∆PQR is a right angled triangle. Write the names of sides opposite and adjacent to ∠P and ∠R. (Textbook pg no. 102)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज PQR को दर्शाता है। P शीर्ष पर है, Q समकोण पर है (बाएं नीचे), और R आधार के दाहिने सिरे पर है।
Answer: Solution:
In right angled ∆PQR,
(i) side opposite to ∠P = QR
(ii) side opposite to ∠R = PQ
(iii) side adjacent to ∠P = PQ
(iv) side adjacent to ∠R = QR
In simple words: This question asks to identify the sides opposite and adjacent to angles P and R in a right-angled triangle PQR, helping to understand the basic components for trigonometric ratios.

🎯 Exam Tip: The hypotenuse is always opposite the right angle and does not change; opposite and adjacent sides are relative to the chosen acute angle.