Question

any mods please help​

Answer 1

$\Huge\rm\red{Answer}$

19) Sin theta= 7/25 and we know that sin theta Perpendicular/ Hypotenuse

Now, cos theta= Base/ Hypotenuse.

So for finding the base we use Pythagoras Theroem:

P²+B²= H²

7²+B²= 25²

49+B²= 625

B²= 625-49

B²=576

B= 24

Now, cos theta= 24/25

And tan theta = Perpendicular/ Base tan theta= 7/24

• I Hope it's Helpful My Friend.

Answer 2

Answer:

Q.19)

We know the Identity:

• Sin²Ф + Cos²Ф = 1

Now we are given that,

→ SinФ = 7/25

We are required to find the value of CosФ.

Substituting the known values in the identity, we get:

$\implies (\dfrac{7}{25})^2 + Cos^2\theta = 1\\\\\\\implies \dfrac{49}{625} + Cos^2\theta = 1\\\\\\\implies Cos^2\theta = 1 - \dfrac{49}{625}\\\\\\\implies Cos^2\theta = \dfrac{ 625 - 49}{625}\\\\\\\implies Cos^2\theta = \dfrac{576}{625}$

Taking Square roots on both sides we get:

$\implies \sqrt{Cos^2\theta} = \sqrt{ \dfrac{576}{625} }\\\\\\\implies \boxed{ \bf{Cos\:\theta = \dfrac{24}{25}}}$

Hence the value of Cos Ф is 24/25.

Hence Option (d) is the right answer.

Q.20)

We know that, the relation between areas of similar triangles to their sides is given as:

$\boxed{ \bf{\dfrac{ \Delta(1)}{\Delta(2)} = (\dfrac{S_1}{S_2})^2 }}$

That is,

The area of two similar triangles are in the ratio of the squares of their corresponding sides.

According to the question,

$\dfrac{A\:\Delta(1)}{A\:\Delta(2)} = \dfrac{3}{1}\\\\\\\implies (\dfrac{S_1}{S_2})^2 = \dfrac{3}{1}\\\\\\\text{Taking Square root on both sides we get:}\\\\\\\implies \boxed{ \bf{ \dfrac{S_1}{S_2} = \dfrac{\sqrt{3}}{1}}}$

Hence Option (c) is the correct answer.