Question

now, give me this answer..​

Answer 1

Answer:

im triangle adc using pythagoras theorem

DC^2 + AD^2=AC^2

5^2+AD^2=13^2

25+AD^2=169

AD^2=169-25

AD^2=144

AD=√144

AD=12

therefore tan x=5 by 12

in triangle ABD

BD =21-5=16

AD=12

using pythagoras theorem

AB^2= BD^2+AD^2

AB^2= 16^2+12^2

AB^2= 256+144

AB=√400

AB=20

there fore cos y=base by hypotenous

=16 by 20

and sin y

=perpendicular by hypotenous

12 by20

Answer 2

Answer:

The values for i) tanx=5/12,   ii) cosy=4/5 and,    iii) siny=3/5

Step-by-step explanation:

First let us calculate the unknown sides.

This can be done by using Pythagoras theorem i.e., Hypotenuse²=Adjacent²+Opposite²

Consider ΔADC.

Here we know AC=13 and DC=5.

Then from Pythagoras theorem,

AC²=AD²+DC²

⇒13²=AD²+5²

⇒AD²=13²-5²

⇒AD²=169-25

⇒AD²=144

⇒AD=√144

⇒AD=12

Now consider the ΔABD

Here AD=12, BD=BC-DC=21-5=16

then from Pythagoras theorem

AB²=AD²+BD²

⇒AB²=12²+16²

⇒AB²=144+256

⇒AB²=400

⇒AB=√400

⇒AB=20

wkt

for a right angled triangle

sinФ=$\frac{opposite}{hypotenuse}$, cosФ=$\frac{adjacent}{hypotenuse}$ and tanФ=$\frac{opposite}{adjacent}$

(i) tanx=$\frac{opposite}{adjacent}$=$\frac{DC}{AD}$=$\frac{5}{12}$

(ii) cosy=$\frac{adjacent}{hypotenuse}$=$\frac{BD}{AB}$=$\frac{16}{20}$=$\frac{4}{5}$

(iii) siny=$\frac{opposite}{hypotenuse}$=$\frac{AD}{AB}$=$\frac{12}{20}$=$\frac{3}{5}$