## Mathematics trigonometry Exercise **- 11.1**

**Andhra Pradesh SSC(10th class) Mathematics trigonometry Exercise - 11.1 Solutions.**After Reading

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**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

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 p= QR/QP=24/7

Tan R= QP/QR=7/24

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

= 24 × 24 - 7 × 7

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

168

= 527/168

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**3. In a right angle triangle ABC
with right angle at B, in which a= 24 units, b = 25 units and∠BAC =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

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 2 =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

** ****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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