Andhra pradesh SSC Mathematics Trigonometry Do This & Try this Solutions Class 10th

Mathematics trigonometry DO THIS & TRY THIS Sums Solutions

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DO THIS

1. Identify  "Hypotenuse", "Opposite side" and "Adjacent side" for the given triangles

Sol.  ln the triangle PQR 

       Opposite side =PQ                          

       Adjacent side =QR           

      Hypotenuse = PR

  
2.1) for angle X

    2) For angle Y   

Sol. In the triangle XYZ             

1)     För angle X 

       Opposite side = YZ 

       Adjacent side= XZ

       Hypotenuse= XY 

 2. For angle Y 

 Opposite side =XZ

 Adjacent side =YZ 

 Hypotenuse = XY

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SSC(10th class) Trigonometry Exercise - 11.1 Solutions

3. In triangle XYZ, angle Y is right angle, XY=17 cm and YZ=15 cm, then find (i) sin X (ii) cos Z.  (iil )tanX

Sol:   Given  triangle XYZ,

∠Y is right angle. 

 By Pythagoras theorem 

XZ ²  = YZ ² +XY ²       

 

 XY ^2 = 17²-15²

XY ² = 289-225

XY ² = 64

XY = √64

XY = 8

i)  sin X =  opposite side to ∠x / Hypotenuse

                                              =  YZ / XZ=15/17

 ii) cos Z = Adjacent side to ∠x / Hypotenuse

                                             = YZ / XZ=15/17 

 iii) tan X = Opposite side to ∠X/ Adjacent side to ∠X

                                           = YZ/XZ=15/ 8 

TRY THIS:

1. Write length of "Hypotenuse", "Opposite side" and "Adjacent side" for the given angles in the given triangles. 

 i)  For angle C 

ii)  For angle A

 Sol.     By Pythagoras theorem

 AC ² = AB ² + BC ²                           

 (5)² = AB ² +4²

25 =AB²+16

 AB ² = 25 –16

AB ² =9

AB = V 9

AB    = 3

 For angle C:

 Opposite side = AB = 3 cm

 Adjacent side = BC = 4 cm

Hypotenuse = AC = 5 cm

  For angle A:

Opposite side = BC = 4 cm

Adjacent side =AB = 3 cm

Hypotenuse AC = 5 cm

 2. In a right angle triangle ABC, right angle is at C. BC + CA = 23 cm and BC- CA = 7cm  than find sin A and tan B

  Sol. In a right angle triangle ABC, right angle is at C.

Given that

BC+CA = 23……... (1) 

BC-CA =7………. .(2) 

  Solve equation 1 and 2

 BC+CA = 23……... (1) 

 BC-CA =7………. .(2) 

    ................................

 Add     2 BC = 30

BC = 30/2

BC =15 

Substituting BC =15 in equation (1)

 BC + CA = 23

 CA = 23 – BC

 CA = 23 - 15 

 CA = 8 

 By Pythagoras theorem

   AB ² = AC ² + BC ² 

         = 8² +15²

         = 64+22

     AB ²= 289 

    AB =√ 289

          =√17 X 17

          =17

sin A =BC/AB = 15/17           

tan B = AC/BC= 8/ 15