Question

In Fig. 5.48, AD = 4 cm, BD = 3 cm and CB = 12 cm, find the cot θ.
(a)[tex]\frac{12}{5}[/tex
(b)[tex]\frac{5}{12}[/tex
(c)[tex]\frac{13}{12}[/tex
(d)[tex]\frac{12}{13}[/tex

Answer 1

Given : In Fig. 5.48 , AD = 4cm , BD = 3 cm , CB = 12 cm

 

To find : the value of cot θ

 

Solution :  

In ∆ADB,  

AB² = AD² + BD²

[By using Pythagoras theorem]

AB² = 4² + 3²

AB² = 16 + 9

AB = √16 + 9  

AB = √25  

AB = 5 cm

Then,

cot θ = B/P  

cot θ = CB/AB  

cot θ = 12/5

Hence the value of cot θ is 12/5 .

Among the given options option (A) 12/5 is  correct.

Hope this answer will help you…

 

Some more similar questions from this chapter :  

In Fig 5.47, the value of $cos\Phi$ is

(a)$\frac{5}{4}$

(b)$\frac{5}{3}$

(c)$\frac{3}{}$

(d)$\frac{4}{5}$

https://brainly.in/question/17101513

If $cos\Theta=\frac{2}{3}$, then $2sec^{2}\Theta+2tan^{2}\Theta-7$ is equal to

(a)1

(b)0

(c)3

(d)4

https://brainly.in/question/7533973

Answer 2

Answer:

Step-by-step explanation:

cott=adjacent side/opposite side

AB^2=BD^2+AD^2

AB^2=9+16

AB=sqrt(25)=5

now cot t= BC/AB=12/5