Question

plz solve this question.......... (: ​

Answer 1

SOLUTION :-

•first ,finding sides of triangle.

=> In right ΔACB,

=>(hypotenuse)² = (height)² + (base)²---- (According to Pythagoras theorem)

=> .°. AB² = AC² + BC²

=> (29)²= AC² + (21)²

=> 841 = AC² + 441

=> AC² = 841 - 441

=> AC² = 400

=> √(AC)² = √400 ------( Take square root On both sides.)

=> .°. AC = 20 units.

Now, find sinΘ and cosΘ.

=> SinΘ = side opposite to angle Θ/ hypotenuse

=> .°. SinΘ = AC/AB

=> .°. SinΘ = 20/29.

=> cosΘ = side adjacent to angle Θ / hypotenuse => cosΘ = BC/AB

=> .°. cosΘ = 21/29.

We have to find out,

=> cos²Θ + sin²Θ

putting values.

=> (21/29)² + (20/29)²

=> (21)²+(20)²/(29)²

=> 441+400/841

=> 841/841

=> 1.

.°. cos²Θ + sin²Θ = 1.

Therefore, the value of cos²Θ + sin²Θ is 1.

Answer 2

In right ΔACB,

$=>(hypotenuse)^{2} = (height)^{2} + (base)^{2}$

Using the Pythagorean theorem, we get,

$AB^{2} = AC^{2} + BC^{2} \\=> (29)^{2} = AC^{2} + (21)^{2} \\=> 841 = AC^{2} + 441\\=> AC^{2} = 841 - 441\\=>AC^{2} = 400\\=> AC = 20 \units.$

SinΘ = AC/AB

=>  SinΘ = 20/29

cosΘ = BC/AB

cosΘ = 21/29

To Find, cos²Θ + sin²Θ

$=> (21/29)^{2} + (20/29)^{2} \\=> 441/841 + 400/841\\=>841/841\\=>1$

$\boxed{\boxed{\bold{Therefore, \ the \ value \ is \ 1.}}}$