KSEEB Solutions for Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.1

Students can Download Class 10 Maths Chapter 11 Introduction to Trigonometry Ex 11.1 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 11.1

Question 1.
In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) sin A, cos A
(ii) sin C, cos C
Answer:
In ∆^le ABC ∠B = 90°
AC² = AB² + BC² [pythagoras theorem]

AC² = (24)² + (7)²
= 576 + 49 = 625
AC = \(\sqrt{625}\)
AC = 25 cm.

Question 2.
In Fig.11.3 find tan P – cot R.

Answer:
In ∆^le PQR ∠Q = 90°
(PR)² = (PQ)² + (QR)² [pythagoras theorem]
(13)² = (12)² + (QR)²
169= 144 +(QR)²
(QR)² = 169 – 144
(QR)² = 25
QR = \(\sqrt{25}\) = 5cm

Question 3.
If sin A = \(\frac{3}{4}\) calculate cos A and tan A
Answer:

In ∆^le ABC ∠B = 90°
AC² = AB² + BC² [pythagoras theorem]
(4)² = AB² + (3)²
16 = AB² + 9
AB² = 16 – 9 = 7

Question 4.
Given 15 cot A = 8, find sin A and sec A.
Answer:

15 cot A = 8

In ∆^le ABC ∠B = 90°
AC² = AB² + BC²
AC²= (8)² + (15)²
AC² = 64 + 225 – 289

Question 5.
Given sec 0 = \(\frac{13}{12}\) calculate all other trigonometric ratios.
Answer:

sec θ = \(\frac{\text { hyp }}{\text { adj }}=\frac{13}{12}\)
In ∆ ABC ∠B = 90°
AC² = AB² + BC²
(13)² = AB² + (12)²
AB² = 169 – 144 = 25
AB = \(\sqrt{25}\)
AB = 5
Trigonometric ratios

Question 6.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A= ∠B.
Answer:

\(\frac{A C}{B C}\) = 1
∴ AC = BC.
In a triangle two sides are equal then opposite angles are also equal
∴ ∠A = ∠B.

Question 7.
If cot θ = \(\frac{7}{8}\) evaluate:

Answer:

In ∆^le ABC ∠B = 90°
AC² = AB² + BC²
AC² = (8)² + (7)²
AC² = 64 + 49 = 113
AC = \(\sqrt{113}\)

Question 8.
If 3 cotA = 4, check whether \(\frac{1-\tan ^{2} A}{1+\tan ^{2} A}\) = cos² A – sin² A or not.
Answer:

In ∆^le ABC ∠B = 90°
AC² = AB² + BC² [pythagoras theorem]
AC² = (4k)² + (3k)²
AC² = 16k² + 9k²
AC² = 25k²
AC = \(\sqrt{25 k^{2}}\) = 5k.
AC = 5k.

∴ LHS = RHS
\(\frac{1-\tan ^{2} A}{1+\tan ^{2} A}\) = cos²A – sin²A

Question 9.
In triangle ABC, right-angled at B, if tan A = \(\frac{1}{\sqrt{3}}\), find the value of:
i) sin A cos C + cos A sin C
ii) cos A cos C – sin A sin C
Answer:

In ∆^le ABC ∠B = 90°
AC² = AB² + BC² [pythagoras theorem]
AC² = (\(\sqrt{3}\)k)² + (1k)²
AC² = 3k² + k² = 4k²
AC² = \(\sqrt{4}\)k² = 2k

i) sin A cos C + cos A sin C

ii) cos A cos C – sin A sin C

Question 10.
In ∆ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P. Determine the values of sin P, cos P and tan P.
Answer:

PR + QR = 25 cm and PQ = 5 cm
Let PR = x cm and QR = (25 – x)
In ∆^le PQR ∠Q = 90°
PR² = PQ² + QR²
(x)² = (5)² + (25 – x)²
x² = 25 + 625 – 50x + x²
x² – x² – 50x + 650 = 0
50x = 650
x = \(\frac{650}{50}\)
x = 13 cm
PR = x = 13cm and QR = 25 – x = 25 – 13 = 12 cm, QR = 12 cm