Question
Write the values of the following trigonometric ratios.
$\cos 45^{\circ}=\frac{⬜}{⬜}$
$\cos 45^{\circ}=\frac{⬜}{⬜}$
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In $\Delta LMN , I =5, m =13, n =12$ then complete the activity to show that whether the given triangle is right angled triangle or not.
*(l, $m , n$ are opposite sides of $\angle L , \angle M , \angle N$ respectively)
Activity: In $\triangle LMN , I =5, m =13, n =\square$
$
\therefore 1^2=\square, m^2=169, n^2=144 \text {. }
$
$
\therefore 1^2+n^2=25+144=\square
$
$
\therefore \square+1^2=m^2
$
$\therefore$ By Converse of Pythagoras theorem, $\triangle LMN$ is right angled triangle.
→*(l, $m , n$ are opposite sides of $\angle L , \angle M , \angle N$ respectively)
Activity: In $\triangle LMN , I =5, m =13, n =\square$
$
\therefore 1^2=\square, m^2=169, n^2=144 \text {. }
$
$
\therefore 1^2+n^2=25+144=\square
$
$
\therefore \square+1^2=m^2
$
$\therefore$ By Converse of Pythagoras theorem, $\triangle LMN$ is right angled triangle.
Write the correct number in the given boxes from the following A. P.
– 3, – 8, – 13, – 18, . . .
$\text { Here } t _3=\square, t _2=\square, t _4=\square, t _1=\square $
$ t _2- t _1=\square, t _3- t _2=\square $
$ \therefore a =\square, d =\square$
→– 3, – 8, – 13, – 18, . . .
$\text { Here } t _3=\square, t _2=\square, t _4=\square, t _1=\square $
$ t _2- t _1=\square, t _3- t _2=\square $
$ \therefore a =\square, d =\square$
Complete the following activity to find the 19th term of the A.P. 10, 15, 20,..........
Find distance between point $A(-1,1)$ and point $B(5,-7)$ :
Solution: Suppose $A\left(x_1, y_1\right)$ and $B\left(x_2, y_2\right)$
$x_1=-1, y_1=1 \text { and } x_2=5, y_2=-7$
Using distance formula,
$ d(A, B)=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
$\therefore d(A, B)=\sqrt{\square+[(-7)+\square]^2}$
$\therefore d(A, B)=\sqrt{\square}$
$\therefore d(A, B)=\square $
→Solution: Suppose $A\left(x_1, y_1\right)$ and $B\left(x_2, y_2\right)$
$x_1=-1, y_1=1 \text { and } x_2=5, y_2=-7$
Using distance formula,
$ d(A, B)=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
$\therefore d(A, B)=\sqrt{\square+[(-7)+\square]^2}$
$\therefore d(A, B)=\sqrt{\square}$
$\therefore d(A, B)=\square $
In figure 3.52 , chords $\mathrm{PQ}$ and $\mathrm{RS}$ intersect at $\mathrm{T}$.
(i) Find $m$ (arc SQ) if $m \angle \mathrm{STQ}=58^{\circ}, m \angle \mathrm{PSR}=24^{\circ}$.
(ii) Verify,
$\angle \mathrm{STQ}=\frac{1}{2}[m(\operatorname{arc} \mathrm{PR})+m(\operatorname{arcSQ})]$
(iii) Prove that :
$\angle \mathrm{STQ}=\frac{1}{2}[m(\operatorname{arc} \mathrm{PR})+m(\operatorname{arcSQ})]$ for any measure of $\angle \mathrm{STQ}$.
(iv) Write in words the property in (iii).
→(i) Find $m$ (arc SQ) if $m \angle \mathrm{STQ}=58^{\circ}, m \angle \mathrm{PSR}=24^{\circ}$.
(ii) Verify,
$\angle \mathrm{STQ}=\frac{1}{2}[m(\operatorname{arc} \mathrm{PR})+m(\operatorname{arcSQ})]$
(iii) Prove that :
$\angle \mathrm{STQ}=\frac{1}{2}[m(\operatorname{arc} \mathrm{PR})+m(\operatorname{arcSQ})]$ for any measure of $\angle \mathrm{STQ}$.
(iv) Write in words the property in (iii).
Some sequences are given below. Show the positions of the terms by $t_1, t_2, t_3, \ldots$
i. $7,7,7,7, \ldots$ Here $t_1=7, t_2=$⬜, $t_3=$ ⬜, ...
ii. $-2,-6,-10,-14, \ldots$ Here $t _1=-2, t _2=$⬜, $t_3=$⬜, ....
→i. $7,7,7,7, \ldots$ Here $t_1=7, t_2=$⬜, $t_3=$ ⬜, ...
ii. $-2,-6,-10,-14, \ldots$ Here $t _1=-2, t _2=$⬜, $t_3=$⬜, ....
Observe the figure and complete the following activity
In fig, ∠B = 75°, ∠D = 75°
∠B ≅ [ ______ ] ...[each of 75°]
∠C ≅ ∠C ...[ ______ ]
ΔABC ~ Δ [ ______ ] ...[ ______ similarity test]
Fill in the blanks with correct number
$\left|\begin{array}{ll}3 & 2 \\ 4 & 5\end{array}\right|=3 \times \square-\square \times 4=\square-8=\square$
→$\left|\begin{array}{ll}3 & 2 \\ 4 & 5\end{array}\right|=3 \times \square-\square \times 4=\square-8=\square$
Complete the following activity to determine the nature of the root of the quadratic equation $x^2 + 2x + 9 = 0$.
Compare $x^2+ 2x + 9 = 0$ with $ax^2 + bx + c = 0$.
→Compare $x^2+ 2x + 9 = 0$ with $ax^2 + bx + c = 0$.