Home
Class 10
MATHS
In a right angled triangle ABC, right an...

In a right angled triangle ABC, right angled at B find `cos A-sin A` if BC= 9 cm,
AC - AB =1 cm.

Promotional Banner

Topper's Solved these Questions

  • ALGEBRIC METHOD OF SOLVING A PAIR OF LINEAR EQUATIONS

    KALYANI PUBLICATION|Exercise EXERCISE|39 Videos

  • AREA OF SIMILAR TRIANGLES

    KALYANI PUBLICATION|Exercise EXERCISE|20 Videos

Similar Questions

Explore conceptually related problems

In a right angled triangle ABC, right angled at B find cos A-sin A if BC= 20 cm, AC - AB =10 cm.

In a right angled triangle ABC, right angled at B find cos A-sin A if BC= 80 cm, AC - AB =2 cm.

ABC is a right angle triangle right angled at C. If sin A = 3/5 find tanB.

In the figure ABC is a right angled triangle, right angled at A. Find the shaded region if AB=6cm , BC=10cm and O is the centre of the in circle of ABC. ( pi=3.14 )

ABC is a right angle triangle right angled at C. If angle A= 60° and AC= 6cm . Find CB and AB.

∆ABC is a right angled triangle right angled at A and AD _|_ BC . Prove that AD^2 = BD.CD .

ABC is a right angle triangle right angled at C. If AC =3 and AB:BC = 13:5, then find sinA and tanB.

ABC a right angle triangle, right angled at B. AD and CE are the medians drawn from A and C respectively. If AC = 5 m and AD = (3sqrt5)/2 m , find the length of CE.

∆ABC is a right angled at B and D is mid-point of BC. Prove that AC^2 = 4AD^2 - 3AB^2 .

In a right triangle ABC, angleB = 90^@ If AC = 13 cm, BC = 5 cm, Find AB.

KALYANI PUBLICATION-ALGEBRIC METHOD OF SOLVING A PAIR OF LINEAR EQUATIONS-EXERCISE
  1. In a right angled triangle ABC, right angled at B find cos A-sin A if ...

    Text Solution

    |

  2. Solve by the method of elimination: 5x-3y=1, 2x+5y=19

    Text Solution

    |

  3. Solve by the method of elimination: 3x+4y=7, 5x-8y=8

    Text Solution

    |

  4. Solve by the method of elimination: 5x-3y=16, 3x-y=12

    Text Solution

    |

  5. Solve by the method of substitution: 2x+3y=31, 17x-11y=8

    Text Solution

    |

  6. Solve by the method of substitution: ax+by=1, bx+ay=(a+b)^2/(a^2+b^2...

    Text Solution

    |

  7. Solve by the method of substitution: x+y=a+b, ax-by=a^2-b^2

    Text Solution

    |

  8. Solve by the method of cross multiplication: 8x+3y=1, 7x+4y=-6

    Text Solution

    |

  9. Solve by the method of cross multiplication: 2x+y=35, 3x+4y=65

    Text Solution

    |

  10. Solve by the method of cross multiplication: ax+by=a-b, bx-ay=a+b

    Text Solution

    |

  11. Solve by any method: (xy)/(x+y)=6/5, (xy)/(y-x)=6

    Text Solution

    |

  12. Solve by any method: (x+y)/(xy)=1, (x-y)/(xy)=65

    Text Solution

    |

  13. Solve by any method: x+2y=1.3, 3/(2x+5y)=1

    Text Solution

    |

  14. Solve by any method: 31x+43y=117, 43x+31y=105

    Text Solution

    |

  15. Solve by any method: 148x+231y=527, 231x+148y=610

    Text Solution

    |

  16. Solve by any method: ax+by=c, bx+ay=1+c

    Text Solution

    |

  17. Solve by any method: x+5y=36, (x+y)/(x-y)=5/3

    Text Solution

    |

  18. Solve by any method: x-y=0.9, 11/(2(x+y))=1

    Text Solution

    |

  19. Solve by any method: x/a+y/b=a+b, x/a^2+y/b^2=2

    Text Solution

    |

  20. Solve by any method: x/a+y/b=2, ax-by=a^2-b^2

    Text Solution

    |

  21. Word Problem of Numbers A number between 10 and 100 is equal to eigh...

    Text Solution

    |