Question
Write the values of the following trigonometric ratios.
$\cos 45^{\circ}=\frac{⬜}{⬜}$
$\cos 45^{\circ}=\frac{⬜}{⬜}$
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From the given figure, in $\triangle A B C$, if $A D \perp B C, \angle C=45^{\circ}, A C=8 \sqrt{2}, B D=5$, then for finding value of $A D$ and $BC$, complete the following activity.
Activity: In $\triangle ADC$, if $\angle ADC =90^{\circ}, \angle C =45^{\circ}$ [Given]
$\therefore \angle DAC =\square \quad$..... [Remaining angle of $\triangle ADC ]$
By theorem of $45^{\circ}-45^{\circ}-90^{\circ}$ triangle,
$ \therefore \square=\frac{1}{\sqrt{2}} AC \text { and } \square=\frac{1}{\sqrt{2}} AC$
$\therefore AD =\frac{1}{\sqrt{2}} \times \square \text { and } DC =\frac{1}{\sqrt{2}} \times 8 \sqrt{2}$
$\therefore AD =8 \text { and } DC =8$
$\therefore BC = BD + DC$
$=5+8$
$=13 $
→Activity: In $\triangle ADC$, if $\angle ADC =90^{\circ}, \angle C =45^{\circ}$ [Given]
$\therefore \angle DAC =\square \quad$..... [Remaining angle of $\triangle ADC ]$
By theorem of $45^{\circ}-45^{\circ}-90^{\circ}$ triangle,
$ \therefore \square=\frac{1}{\sqrt{2}} AC \text { and } \square=\frac{1}{\sqrt{2}} AC$
$\therefore AD =\frac{1}{\sqrt{2}} \times \square \text { and } DC =\frac{1}{\sqrt{2}} \times 8 \sqrt{2}$
$\therefore AD =8 \text { and } DC =8$
$\therefore BC = BD + DC$
$=5+8$
$=13 $
| Draw a circle with center O and radius 3 cm |
| ↓ |
| Take any point P on the circle |
| ↓ |
| Draw ray OP |
| ↓ |
| Draw perpendicular to ray OP from point P |
In ∆ABC, ∠C is an acute angle, seg AD ⊥ seg BC. Prove that:
AB² = BC² + AC² - 2BC×DC
AB² = BC² + AC² - 2BC×DC
.......... (I) theorem of angle bisector.Write the values of the following trigonometric ratios.
$\sin 30^{\circ}=\frac{1}{⬜}$
→→$\sin 30^{\circ}=\frac{1}{⬜}$
From given figure, In $\triangle ABC , AD \perp BC$, then prove that $AB ^2+ CD ^2= BD ^2+ AC ^2$ by completing activity.
Activity: From given figure, In $\triangle A C D$, By pythagoras theorem
$ A C^2=A D^2+\square$
$\therefore A D^2=A C^2-C D^2 $
Also, In $\triangle ABD$, by pythagoras theorem,
$ A B^2=\square+B D^2$
$\therefore A D^2=A B^2-B D^2 \quad \ldots . .(I I)$
$\therefore \square-B D^2=A C^2-\square$
$\therefore A B^2+C D^2=A C^2+B D^2 $
→Activity: From given figure, In $\triangle A C D$, By pythagoras theorem
$ A C^2=A D^2+\square$
$\therefore A D^2=A C^2-C D^2 $
Also, In $\triangle ABD$, by pythagoras theorem,
$ A B^2=\square+B D^2$
$\therefore A D^2=A B^2-B D^2 \quad \ldots . .(I I)$
$\therefore \square-B D^2=A C^2-\square$
$\therefore A B^2+C D^2=A C^2+B D^2 $
To check the rule for the terms of the sequence look at the arrangements and fill the empty boxes suitably.
3,3,3,3,…
3,3,3,3,…
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 $ax^2 + bx + c = 0$.
Find $\frac{A(\triangle A B C)}{A(\triangle A P Q)}$
Solution :
In ∆ABC, BC is the base and AR is the height.
In ∆APQ, PQ is the base and AR is the height.
[The ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
$\therefore \frac{ A (\triangle ABC )}{ A (\triangle APQ )}=\frac{ ⬜ \times ⬜ }{ ⬜ \times ⬜ }$
$\therefore \frac{ A (\triangle ABC )}{ A (\triangle APQ )}=\frac{ ⬜ }{⬜}$
→Solution :
In ∆ABC, BC is the base and AR is the height.
In ∆APQ, PQ is the base and AR is the height.
[The ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
$\therefore \frac{ A (\triangle ABC )}{ A (\triangle APQ )}=\frac{ ⬜ \times ⬜ }{ ⬜ \times ⬜ }$
$\therefore \frac{ A (\triangle ABC )}{ A (\triangle APQ )}=\frac{ ⬜ }{⬜}$