Question

In the given figure, angle B= 90° and angle ADB = x°. Find:
(i) sin²C + cos²C
(ii) tan x° - cos x° + 3sin x°.​

Answer 1

hope it helps you.......

Answer 2

Given : angle B= 90° and angle ADB = x°

To Find :   sin²C + cos²C

tan x° - cos x° + 3sin x°.​

   Sin ∠CAB

Solution:

AC  = 20

AB = 12

BC² = AC² - AB²

=> BC² = 20² - 12²

=>  BC² = 16²

=> BC = 16

DB² = AD² - AB²

=> DB²  = 15² - 12²

=> DB² = 9²

=> DB  = 9

DC = BC - BD = 16 - 9 = 7

 Sin ∠CAB  = BC/AC  =  16/20

=>   Sin ∠CAB  =  4/5

sin²C + cos²C = 1

(12/20)² + (16/20)²  = 1

tan x° - cos x° + 3sin x°.​

x = ∠ADB

tanx  =  12/9  = 4/3

Cosx = 9/15  = 3/5

Sinx  = 12/15  = 4/5

4/3  - 3/5  + 3(4/5)

= (20 - 9  + 36)/15

= 47/15

tan x° - cos x° + 3sin x°  = 47/15

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