Question

from the figure given, find the value of (1) tanx (2) cos y . (3) sin y.
maths ​

Answer 1

Answer:

In triangle ADC:

By Pythagoras Theorem:

AD^2= (13)^2-(5)^2

AD^2=169-25

AD^2=144

AD= 12

We know that ,

Tan(x)=DC/AD

(Perpendicular/base)

Tan(x) =5/12

BD=BC-DC

BD=21-5=16

In traingle ABD,

By Pythagoras theorem

AB^2=(16)^2+(12)^2

AB^2= 256+144

AB^2=400

AB= 20

We know that,

cos(y)= base/hypotenuse=16/20=4/5

sin(y)= Perpendicular/hypotenuse= 12/20=3/5

Answer 2

$\sf \: In \: ∆ABC \\ \sf \: By \: Pythagoras \: theorem \\ \sf \: (AC) {}^{2} = (CD) {}^{2} + (AD) {}^{2} \\ \sf \: {13}^{2} = {5}^{2} + (AD) {}^{2} \\ \sf \:169 = 25 + (AD) {}^{2} \\ \sf \:169 - 25 = (AD) {}^{2} \\ \sf \: \sqrt{144} = (AD) \\ \sf \:12 = AD \\ \\ \sf \: as \: we \: know \\ \sf{ \red{ \tan(0) =\frac{opposite}{adjacent}}} \\ \tan(x) = \frac{5}{12} \\ \\ \sf \: In ∆ABD \\ \sf \: BD=BC-CD \\ \sf \: BD=21-5=16 \\ \\ \sf \: By \: Pythagoras \: theorem \\ \sf \: AB {}^{2} =AD {}^{2} + BD {}^{2} \\ \sf \: AB {}^{2} = {12}^{2} + {16}^{2} \\ \sf \: AB {}^{2} = 144 + 256\\ \sf \: AB= \sqrt{400} \\ \sf \: = 20\\ \\ \\ \sf \: as \: we \: know \: that \\ \sf{ \orange{ \cos(0) = \frac{adjacent}{hypotenuse}}} \\ \sf \: \cos(y) = \frac{16}{20} \\ \\ \\ \sf { \red{ \sin(0) = \frac{opposite}{hypotenuse}}} \\ \sf \: \sin(y) = \frac{12}{20} \\ \\ \sf \: @WildCat7083$