# 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.
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