Question

please help me with my homework ​

Answer 1

$\huge\boxed{\fcolorbox{red}{yellow}{Answer}}$

______________________________________

In ∆ABD ,

Hypotenuse = 13cm

Base = 5cm

Height = x cm

______________________________________

By Pythagoras theorem ,

AB^2 = BD^2 + DA^2

13^2 = X^2 + 5^2

$\sqrt{{13}^{2} - {5}^{2} } = {x}^{2}$

169 - 25 = x^2

144 = x ^2

12 = x.

so , height = 12cm

______________________________________

(i)Sin B = x

We know that ,sin = height/hypotenuse = p/h

Height = 12 cm

Hypotenuse = 13 cm.

so ,sin B = 12/13 cm

______________________________________

In triangle ABC,

Height = 12 cm

Base = 16 cm

______________________________________

(ii) tanC = y

we know that ,tan = height/base = p/b

tanC = 12/16 = 4/3cm

______________________________________

In ∆ABD,

Height = 12 cm

hypotenuse = 13cm

Base = 5 cm

sec B = h/b

sec B = 13/5.

tan B = p/b

tan B = 12/5

______________________________________

(iii)sec^2B - tan^2B

(13/5)^2 - (12/5)^2

169/25 - 144/25

(169 - 144)/25

25/25

= 1.

sec^2B - tan^2B = 1

______________________________________

In ∆ABC,

In ∆ABC,height = 12 cm

In ∆ABC,height = 12 cmbase = 16 cm

In ∆ABC,height = 12 cmbase = 16 cmhypotenuse = 20 cm [by Pythagoras theorem]

sin C = p/h = 12/20 = 3/5 cm

cos B = b/h = 16/20 = 4/5 cm

______________________________________

(iv)sin^2C + cos^2C

(3/5)^2 + (4/5)^2

9/25 + 16/25

(9 + 16)/25

25/25

1 .

sin^2C + cos^2C = 1.

______________________________________$\huge\sf\bold\pink{hope\: it\: helps\: you}$

${\huge{\overbrace{\underbrace{\green{thanks}}}}}$