[NCERT] Class 10 Ex 8.1 Introduction to Trigonometry Q1 Solution - In Triangle ABC, right-angled at B AB = 24cm, BC = 7cm.

 Excercise 8.1

Question1: In $\Δ$ABC, right-angled at B, AB = 24cm, BC = 7cm. Determine

(i) sinA,cosA

(ii) sinC, CosC

Solution:

Given ABC is a right angled triangle at B

and AB = 24cm, BC = 7cm

In $\Δ$ABC, 

${AC}^2$ =$ {AB}^2 + {BC}^2$

${AC}^2 = {24}^2 + 7^2$

${AC}^2 = 576 + 49$

$AC = \√{625}$

$AC = 25 cm$

For angle A, the Base will be AB and the Perpendicular will be BC

$sinA = {Perpendicular}/{Hypotenous} = {BC}/{AC} = {7}/{25}$

$cosA = {Base}/{Hypotenous} = {AB}/{AC} = {24}/{25}$

For angle C, the base will be BC and the perpendicular will be AB

 $sinC = {Perpendicular}/{Hypotenous} = {AB}/{AC} = {24}/{25}$

$cosC = {Base}/{Hypotenous} = {BC}/{AC} = {7}/{25}$

• Ex 8.1 Question 2 Solution
  • Ex 8.1 Question 3 Solution
    • Ex 8.1 Question 4 Solution
      • Ex 8.1 Question 5 Solution
        • Ex 8.1 Question 6 Solution