Question
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Answer
Roots are distinct and real when $b^2-4 a c=5$, not real when $b^2-4 a c=-5$.
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From given figure, In ∆ABC, If AC = 12 cm. then AB =?
Activity: From given figure, In $\triangle ABC , \angle ABC =90^{\circ}, \angle ACB =30^{\circ}$
$\therefore \angle BAC =\square$
$\therefore \triangle ABC$ is $30^{\circ}-60^{\circ}-90^{\circ}$ triangle
$\therefore$ In $\triangle ABC$ by property of $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.
$ \therefore A B=\frac{1}{2} A C \text { and } \square=\frac{\sqrt{3}}{2} A C$
$\therefore \square=\frac{1}{2} \times 12 \text { and } B C=\frac{\sqrt{3}}{2} \times 12$
$\therefore \square=6 \text { and } B C=6 \sqrt{3} $
→Activity: From given figure, In $\triangle ABC , \angle ABC =90^{\circ}, \angle ACB =30^{\circ}$
$\therefore \angle BAC =\square$
$\therefore \triangle ABC$ is $30^{\circ}-60^{\circ}-90^{\circ}$ triangle
$\therefore$ In $\triangle ABC$ by property of $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.
$ \therefore A B=\frac{1}{2} A C \text { and } \square=\frac{\sqrt{3}}{2} A C$
$\therefore \square=\frac{1}{2} \times 12 \text { and } B C=\frac{\sqrt{3}}{2} \times 12$
$\therefore \square=6 \text { and } B C=6 \sqrt{3} $
In ∆ABC, ∠C is an acute angle, seg AD ⊥ seg BC. Prove that:
AB² = BC² + AC² - 2BC×DC
AB² = BC² + AC² - 2BC×DC
In fig. BP ⊥ AC, CQ ⊥ AB, A−P−C, and A−Q−B then show that ΔAPB and ΔAQC are similar.
In ΔAPB and ΔAQC
In ΔAPB and ΔAQC
∠APB = [ ]° ......(i)
∠AQC = [ ]° ......(ii)
∠APB ≅ ∠AQC .....[From (i) and (ii)]
∠PAB ≅ ∠QAC .....[______]
ΔAPB ~ ΔAQC .....[______]
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 $
There are 9 tickets in a box, each bearing one of the numbers from 1 to 9 . One ticket is drawn at random from the box.
Event A : Ticket shows an even number.
Complete the following activity :
Event A : Ticket shows an even number.
Complete the following activity :
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 $
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 $t _1=\square, t _2=\square, t _3=\square, \ldots$
$\therefore a =\square, d =\square$