Students can Download Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.
Karnataka State Syllabus Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.3
1. Evaluate:
i) \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)
Answer:
ii) \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\)
Answer:
iii) cos 48° – sin 42°
Answer:
cos 48° – sin 42°
= cos 48° – sin (90 – 48°)
= cos 48° – cos 48°
= 0
iv) cosec 31° – sec 59°
Answer:
cosec 31° – sec 59°
= cosec 31° – sec (90 – 31°)
= cosec 31° – cosec 31° = 0
2. Show that:
i) tan 48° tan 23° tan 42° tan 67° = 1
Answer:
= tan 48° tan 23° tan 42° tan 67°
= tan 48° tan23 tan(90 – 48°) tan (90 – 23)
= tan 48° tan 23 cot48° cot 23°
= tan 48° tan 23. \(\frac{1}{\tan 48^{\circ}} \cdot \frac{1}{\tan 23}\)
= 1.
ii) cos 38° cos 52° – sin 38° sin 52° = 0
Answer:
LHS = cos38° cos 52° – sin 38° sin 52°
= cos 38° × cos (90 – 38) – sin 38° × sin (90 – 38)
= cos 38° × sin 38° – sin 38° × cos 38°
= 0
Question 3.
If tan 2A = cot(A – 18°), where 2A is an angle, find the value of A.
Answer:
tan 2A = cot (A – 18°)
cot (90 – 2A) = cot (A – 18)
90 – 2A = A – 18
3A = 108°
∴ A = 36°
Question 4.
If tan A = cot B, prove that A + B = 90°.
Answer:
tan A = cot B
cot(90 – A) = cot B
90 – A =B
90 = A + B
A + B = 90°.
Question 5.
If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Answer:
sec 4A = cosec (A – 20°)
cosec (90 – 4A) = cosec (A – 20)
90 – 4A = A – 20
5A = 110°
∴ A = 22°.
Question 6.
If A, B and C are interior angles of a triangle ABC, then show that
Answer:
Since A, B and C are interior angles of ∆ ABC
Therefore, A + B + C = 180° ÷ by 2
Question 7.
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Answer:
sin 67° + cos 75°
= sin (90 – 23) + cos (90 – 15)
= cos 23° + sin 15°.