Question

in triangle abc angle C is equal to 90 degree ab is equals to 29 and BC is equals to 21 find the value of following first cos ²b + sin ² B second cos²B-sin²B​

Answer 1

Step-by-step explanation:

In ∆ABC C = 90°

AB= 29CM

BC= 21 CM

then AC^2 = AB^2 -BC^2

AC^2 = 29^2 - 21^2

AC^2 = 841-441

AC^2 = 400

AC= 20

1)

COS^2 B +SIN^2 B

(21/29)^2 + (20/29)^2

441/841 + 400/841

841/841= 1

2)

COS^2 B - SIN^2 B

(21/29)^2 - (20/29)^2

441/841 - 400/841

41/841

Answer 2

Use this thing to solve