Question
Complete the following activity, to find the two-digit numbers which are divisible by 6.
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Similar questions
The following table shows the daily supply of electricity to different places in a town. To show the information by a pie diagram, measures of central angles of sectors are to be decided. Complete the following activity to find the measures :
Write the correct number in the given boxes from the following $A. P.$
$70, 60, 50, 40, . . .$
Here $t _1=\square, t _2=\square, t _3=\square, \ldots$
$\therefore a =\square, d =\square$
→$70, 60, 50, 40, . . .$
Here $t _1=\square, t _2=\square, t _3=\square, \ldots$
$\therefore a =\square, d =\square$
Complete the following activity to find the number of natural numbers from 1 to 171 which are divisible by 5 .
If $\sec \theta+\tan \theta=\sqrt{3}$, complete the activity to find the value of $\sec \theta-\tan \theta$
Activity:
$ \square=1+\tan ^2 \theta \quad \ldots . . .[\text { Fundamental trigonometric identity] }$
$\square-\tan ^2 \theta=1$
$(\sec \theta+\tan \theta) \cdot(\sec \theta-\tan \theta)=\square$
$\sqrt{3} \cdot(\sec \theta-\tan \theta)=1$
$(\sec \theta-\tan \theta)=\square $
→→→Activity:
$ \square=1+\tan ^2 \theta \quad \ldots . . .[\text { Fundamental trigonometric identity] }$
$\square-\tan ^2 \theta=1$
$(\sec \theta+\tan \theta) \cdot(\sec \theta-\tan \theta)=\square$
$\sqrt{3} \cdot(\sec \theta-\tan \theta)=1$
$(\sec \theta-\tan \theta)=\square $
Prove that $\cos ^2 \theta \cdot\left(1+\tan ^2 \theta\right)=1$. complete the activity given below.
Activity:
$ \text { L.H.S }=\square$
$=\cos ^2 \theta \times \square \ldots . .\left[1+\tan ^2 \theta=\square\right]$
$=(\cos \theta \times \square)^2$
$=1^2$
$=1$
$=\text { R.H.S } $
→Activity:
$ \text { L.H.S }=\square$
$=\cos ^2 \theta \times \square \ldots . .\left[1+\tan ^2 \theta=\square\right]$
$=(\cos \theta \times \square)^2$
$=1^2$
$=1$
$=\text { R.H.S } $
Complete the following activity to find the value of x.
Complete the following activity to find the 19th term of the A.P. 10, 15, 20,..........
In the adjoining figure BP ⊥ AC, CQ ⊥ AB, A – P- C, A- Q- B , then prove that △ APB and △ AQC are similar.