In figure 1- AD- 3 cm- BD - 4 cm and CB - 12 cm- then the value of tan theta equalsdfrac-3-4-dfrac-5-12-dfrac-4-3-dfrac-12-5-

Question

In figure 1- AD- 3 cm- BD - 4 cm and  CB - 12 cm- then the value of tan theta  equalsdfrac-3-4-dfrac-5-12-dfrac-4-3-dfrac-12-5-

Toppr Admin · Accepted Answer

-x21D2-tan-x3B8-ABBCNow- in -x25B3-ADB-x21D2-AB2-BD2-AD2-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0- -42-32-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0- -25-x21D2-AB-5cm-x2234-tan-x3B8-512Hence- the answer is 512-

In figure 1, $A D = 3 c m, B D = 4 c m$ and $C B = 12 c m$ , then the value of tan $θ$ equals

$\frac{3}{4}$
$\frac{5}{12}$
$\frac{4}{3}$
$\frac{12}{5}$

$\Rightarrow tan θ = \frac{A B}{B C}$
Now, in $△ A D B,$
$\Rightarrow A B^{2} = B D^{2} + A D^{2}$
$= 4^{2} + 3^{2}$
$= 25$
$\Rightarrow A B = 5 c m$
$∴ tan θ = \frac{5}{12}$
Hence, the answer is $\frac{5}{12} .$

In figure 1, AD=3 cm, BD=4 cm and CB=12 cm, then the value of tan θ equals3451243125

⇒tanθ=ABBCNow, in △ADB,⇒AB2=BD2+AD2 =42+32 =25⇒AB=5cm∴tanθ=512Hence, the answer is 512.

In figure 1, $A D = 3 c m, B D = 4 c m$ and $C B = 12 c m$ , then the value of tan $θ$ equals

$\frac{3}{4}$
$\frac{5}{12}$
$\frac{4}{3}$
$\frac{12}{5}$

$\Rightarrow tan θ = \frac{A B}{B C}$
Now, in $△ A D B,$
$\Rightarrow A B^{2} = B D^{2} + A D^{2}$
$= 4^{2} + 3^{2}$
$= 25$
$\Rightarrow A B = 5 c m$
$∴ tan θ = \frac{5}{12}$
Hence, the answer is $\frac{5}{12} .$