AP 10th Class Maths 8th Chapter Introduction to Trigonometry Exercise 8.1 Solutions

Well-designed AP 10th Class Maths Solutions Chapter 8 Introduction to Trigonometry Exercise 8.1 offers step-by-step explanations to help students understand problem-solving strategies.

Introduction to Trigonometry Class 10 Exercise 8.1 Solutions – 10th Class Maths 8.1 Exercise Solutions

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
Solution:
Given in ∆ABC, ∠B = 90°

By using Pythagoras theorem
AB = 24 cm, BC = 7 cm,
By using Pythagoras Theorem
AC = \(\sqrt{A B^2+B C^2}\) = \(\sqrt{24^2+7^2}\)
= \(\sqrt{576+49}\) = \(\sqrt{625}\) = 25 cm

i)
sin A \(=\frac{\text { Opposite side }}{\text { Hypotenuse }}\) = \(\frac{B C}{A C}\) = \(\frac{7}{25}\)
cos A \(=\frac{\text { Adiacent side }}{\text { Hypotemuse }}\)

ii) sin C \(=\frac{\text { Opposite side }}{\text { Hypotenuse }}\) = \(\frac{A B}{A C}\) = \(\frac{24}{25}\)
cos C \(=\frac{\text { Adiacent side }}{\text { Hypotemuse }}\) = \(\frac{B C}{A C}\) = \(\frac{7}{25}\)

Question 2.
In below figure find tan P – cot R.

Solution:
Given in ∆PQR,
PQ = 12 cm, PR = 13 cm
By using Pythagoras theorem,
PR² = PQ² + QR²

OR = \(\sqrt{P R^2-P Q^2}\)
= \(\sqrt{13^2-12^2}\)
= \(\sqrt{169-144}\)

Question 3.
If sin A = \(\frac{3}{4}\), calculate cos A and tan A.
Solution:
Given sin A = \(\frac{3}{4}\) = \(\frac{\text { Opposite side }}{\text { Hypotenuse }}\)
In ∆ABC, ∠B = 90°, BC = 3 AC = 4
By usng Pythagoras Theorem,
AB = \(\sqrt{A C^2}\)
cos A \(=\frac{\text { Adjacent side }}{\text { Hypotenuse }}\) = \(\frac{\mathrm{BA}}{\mathrm{AC}}\) = \(\frac{\sqrt{7}}{4}\)
tan A \(=\frac{\text { Opposite side }}{\text { Adjacent side }}\) = \(\frac{\mathrm{BC}}{\mathrm{AB}}\) = \(\frac{3}{\sqrt{7}}\)

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

Question 5.
Given sec θ = \(\frac{13}{12}\), calculate all other trigonometric ratios.
Solution:
Given sec θ \(\frac{13}{12}\) \(=\frac{\text { Hypotenuse }}{\text { Adjacent side }}\) = \(\frac{A C}{A B}\)
In ΔABC, ∠B = 90°, AC = 13, AB = 12
By using Pythagoras theorem,

Question 6.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution:
Given ∠A and ∠B are acute angles and cos A = cos B.
In ΔABC, ∠C = 90°

Angles opposite to the equal sides are equal.
So, ∠A = ∠B.

Question 7.
If cot θ = \(\frac{7}{8}\), evaluate:
i) \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\)
ii) cot²θ
Solution:
Given cot θ = \(\frac{7}{8}\) \(=\frac{\text { Adjacent side }}{\text { Opposite side }}\) = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)
By using Pythagoras theorem
In ΔABC, ∠B = 90°,

i)

ii)
cot²θ
cot θ = \(\frac{7}{8}\)
⇒ cot²θ = \(\left(\frac{7}{8}\right)^2\) = \(\frac{49}{64}\)

Question 8.
If 3 cot A = 4, check whether \(\frac{1-\tan ^2 A}{1+\tan ^2 A}\) = cos²A – sin²A or not.
Solution:
Given, 3 cot A = 4
cot A = \(\frac{4}{3}\) \(=\frac{\text { Adjacent side }}{\text { Opposite side }}\) = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)
In ∆ABC, ∠B = 90°
By using Pythagoras theorem,

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
Solution:
Given triangle ABC,

∠B = 90° and tan A = \(\frac{1}{\sqrt{3}}\)
By using Pythagoras theorem,

i) sin A cos C + cos A sin C
= \(\frac{1}{2} \cdot \frac{1}{2}\) + \(\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2}\) = \(\frac{1}{4}\) + \(\frac{3}{4}\) = \(\frac{4}{4}\) = 1
∴ sin A cos C + cos A sin C = 1

ii) Cos A cos C – sin A sin C
= \(\frac{\sqrt{3}}{2} \cdot \frac{1}{2}\) – \(\frac{1}{2} \cdot \frac{\sqrt{3}}{2}\) = \(\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}\) = 0
∴ cos A cos C – sin A sin C = 0

Question 10.
In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Solution:
Given in ΔPQR, ∠Q = 90°
PR + QR = 25 cm and PQ = 5 cm
Let QR = x then PR = 25 – QR = 25 – x
By using Pythagoras theorem

Question 11.
State whether the following are true or false. Justify your answer.
i) The value of tan A is always less than 1.
ii) sec A = \(\frac{12}{5}\) for some value of angle A.
iii) cos A is the abbreviation used for the cosecant of angle A.
iv) cot A is the product of cot and A.
v) sin θ = \(\frac{4}{3}\) for some angle θ.
Solution:
i) False. Because tan0 values increases from θ to ∝ (not determinate).
tan 45° = 1, tan 60° = \(\sqrt{3}\) = 1.732
These all are greater than 1.
tan P = \(\frac{12}{5}\) = 2.4 > 1.

ii) True.
cos A is always less than 1.
But sec A is always ≥ 1.

iii) False.
cos A is the abbreviation used for cosine of angle A.

iv) False.
cot A is the ratio of two sides.
But A is the angle.
So, that is not the product of cot and A.

v) False.
Because sin θ = \(\frac{4}{3}\) ≤ 1