Question

✨Ello ! ! Mates...❤


↪Help me in this question⭐

No useless answers plz❌​

Answer 1

ANSWER:-

Given:

In the figure, given above attachment,

ABC is an isosceles ∆ with BC= 8cm & AB= AC= 5cm.

To find:

⚫sinB

⚫tanC

⚫sin²B+ cos²B

⚫tanC - cotB

Solution:

Here,

AB= BC = 5cm & we draw an altitude AD as we know that isosceles ∆ altitude also a median for not equal side of isosceles ∆.

So,

Given:BC= 8cm

BD= CD= 4cm

Using Pythagoras Theorem:

(Hypotenuse)²=(base)²+(perpendicular)²

In ∆ABD,

=) AB² = AD² + BD²

=) 5² = AD² + 4²

=) 25= AD² + 16

=) AD²= 25 - 16

=) AD² = 9

=) AD = √9

=) AD= 3cm

⚫Sin theta:

$= > \frac{Perpendicular}{Hypotenuse}$

So,

$sinB= \frac{AD}{AB} \\ \\ = > sinB = \frac{3}{5}$

⚫tan theta:

$= > \frac{Perpendicular}{Base}$

So,

$= > tanC = \frac{AD}{CD} \\ \\ = > tanC= \frac{3}{4}$

⚫Cos theta:

$= > \frac{Base}{Hypotenuse}$

So,

$= > cosB = \frac{BD}{AB} \\ \\ = > cosB = \frac{4}{5}$

Then,

${sin}^{2}B + {cos}^{2} B \\ \\ = > ( \frac{3}{5} ) {}^{2} + ( \frac{4}{5} ) {}^{2} \\ \\ = > \frac{9}{25} + \frac{16}{25} \\ \\ = > \frac{9 + 16}{25} \\ \\ = > \frac{25}{25} \\ \\ = > 1$

⚫Cot theta:

$= > \frac{Base}{Perpendicular} \\ \\ = > cotB = \frac{BD}{AD} \\ \\ = > cotB = \frac{4}{3}$

So,

$tanC- cotB \\ \\ = > \frac{3}{4} - \frac{4}{3} \\ \\ = > \frac{ 9 - 16}{12} \\ \\ = > - \frac{7}{12}$

Hope it helps ☺️

Answer 2

Answer:

oof

Step-by-step explanation: