## Thursday, September 24, 2020

• September 24, 2020
• • , ,
• 3 comments

## Mathematics trigonometry Exercise - 11.1

Hi friends and my dear students! In this post, I have covered Andhra Pradesh SSC(10th class) Mathematics trigonometry Exercise - 11.1 Solutions. After Reading Mathematics trigonometry Exercise, Please do share it with your friends. You can Learn Maths for All Classes here.

1. In right angle triangle ABC, 8 cm, 15 cm and 17 cm are the lengths of AB,BC and CA respectively. Then, find out sin A, Cos A, and tan A.

Sol: Given in right angle triangle ABC, 8 Cm, 15 Cm and 17 Cm are the length of AB, BC and CA respectively

i) sin A= Opposite side /Hyptenuse =BC/AC=15/17

ii) Cos A = Adjacent side/Hyptenuse = AB/AC =8/17

iii) Tan A =Opposite side/Adjacent side=BC/AB=15/8

2. The sides of a Right angle triangle PQR are PQ = 7 cm, QR = 24 cm and √Q = 90 respectively. Then find, tan P, tan R.

Sol:  Given the sides of a right angle triangle PQR are PQ = 7 cm, QR = 24 cm and Q=90

BY Pythagoras theorem

PR 2 = PQ 2+ QR 2

PR 2= (7) 2+ (25) 2

PR 2= 49+ 576

PR 2= 625

PR =√625

PR =25

Tan R= QP/QR=7/24

Tan p -Tan R = 24/7 - 7/24

=  24 × 24 - 7 × 7

-----------------

168

= 527/168

Introduction To Trigonometry

3. In a right angle triangle ABC with right angle at B, in which a= 24 units, b = 25 units andBAC =Ø Then, find CosØ and tanØ.

Sol: Given in a right angle triangle ABC with right angle at B, in which

a = 24 units, b= 25 units and BAC=Ø

BY Pythagoras theorem

AC 2 = BC 2+ AB 2

(25) 2= (24) 2+ (AB) 2

625 = 576 + (AB) 2

(AB) 2= 625-576

(AB) 2 =49

(AB)  = √49

AB = 7

I ) Cos Ø= AB/AC = 7/ 25

ii) Tan Ø =BC/AB = 24/7

4. If Cos A =12/13 than find sin A tan A.

Sol: Given Cos A = Adjacent side/Hypotenuse=12/13

BY Pythagoras theorem

AC 2 = BC 2 + AB 2

13 =12 2+ BC 2

BC 2 =13 2-12 2

BC 2 =169-144

BC=√25

BC=√5×5

BC=5

Now

Sin A = BC/AC =5/13

Tan A = BC/AB =5/12

5. If 3 tan A = 4 than find sin A Cos A

Sol: Given tan A = Opposite side/Adjacent side= 4/3

BY Pythagoras theorem

AC 2= AB 2 + BC 2

AC 2= 32+42

AC 2=9+16

AC 2=25

AC=√25

AC=5

Now    Sin A = BC/AC =4/5

Cos A = AC/AB =3/5

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