Math, asked by Anonymous, 8 months ago

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Q. 44 ) Find the area of the right angled triangle with hypotenuse 5 cm and one of acute angle is 48°38`.

Q. 45 ) From the given figure, prove that θ + Φ = 90° . Also prove that there are two other right angled triangles. Find sin α , cos β and tan Φ.

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Answered by Anonymous
33

 \purple{\large{\underline{\underline{ \rm{Problem: }}}}}

Find the area of the right angled triangle with hypothenuse 5 cm and one of the acute angle is 48°30'.

 \purple{\large{\underline{\underline{ \rm{Solution: }}}}}

Here, first we need to find AB and then BC. Therefore, by finding both height and breadth, we will apply the formula of area of right triangle.

From the figure,

For finding AB i.e, height (perpendicular)

 \sf{ \sin C =  \dfrac{perpendicular}{hypotenuse}}

 \sf \sin C =  \dfrac{AB}{AC}

 \sf{ \sin48 \degree30' =  \dfrac{AB}{5}}

N.B :-

sin 48° = 0.7431

But we have sin 48°30'

Let's convert 30' into degrees, we have:

30' = 0.5°

Now 48° + 0.5° = 48.5°

∴ sin 48.5° = 0.7490

Let's continue.......

 \sf{0.7490 =  \dfrac{AB}{5}}

 \sf5 \times 0.7490 = AB

 \sf{AB = 3.7450 \: cm}

For finding BC i.e, base

 \sf  \cos C =  \dfrac{base}{hypotenuse}

 \sf{ \cos C  =  \dfrac{BC}{AC} }

 \sf \cos48 \degree 30'  =  \dfrac{BC}{5}

N.B :-

cos 48° = 0.6691

But we have cos 48°30'

30' = 0.5°

Now 48° + 0.5° = 48.5°

∴ cos 48.5° = 0.6626

Let's continue......

 \sf0.6626 =  \dfrac{BC}{5}

 \sf0.6626 \times 5 = BC

 \sf{BC = 3.313 \: cm}

Now let's find out area

 \underline{ \boxed{ \sf{Area \: of \: right \: triangle =  \frac{1}{2}  \times b \times h}}}

 \sf =  \dfrac{1}{2}  \times BC \times AB

 \sf =  \dfrac{1}{2}  \times 3.313 \times 3.7450

 \sf = 1.6565 \times 3.7450

 \sf = 6.2035925 \:  {cm}^{2}

Area of the right angled triangle  \green{ \underline{ \boxed{ \sf{ \gray{  \approx 6.204 \:  {cm}^{2} }}}}}

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 \purple{\large{\underline{\underline{ \rm{Problem: }}}}}

From the given figure, prove that θ + Φ = 90°. Prove also that there are two other right-angles in the figure. Find: (i) sin α (ii) cos β (iii) tan Φ.

 \purple{\large{\underline{\underline{ \rm{Solution: }}}}}

In ACD, we have:

  \sf\cos\theta =  \dfrac{CD}{AC} =  \dfrac{12}{15}

 \sf \sin \theta =  \dfrac{AD}{AC}  =  \dfrac{9}{12}

In BCD, we have:

  \sf\cos \phi   =  \dfrac{CD}{BC}  =  \dfrac{12}{20}

  \sf\sin \phi =  \dfrac{BD}{BC}  =  \dfrac{16}{20}

We know,

cos ( A + B ) = cosA cosB - sinA sinB

∴ cos (θ + Φ) = cos θ cos Φ - sin θ sin Φ

 \sf =  \dfrac{12}{15}  \times  \dfrac{12}{20}  -  \dfrac{9}{15}  \times  \dfrac{16}{20}

 \sf =  \dfrac{144}{300}  -  \dfrac{144}{300}  = 0

 \sf  \therefore \cos(\theta +  \phi )= 0

We know cos 90° = 0

 \sf \cos( \theta  + \phi)  =  \cos90 \degree

 \sf{ \theta +  \phi = 90 \degree}

Hence Proved!!

In right-angled triangle ACD, By using Pythagoras theorem

 \sf{ {AC}^{2}  =  {AD}^{2}  +  {DC}^{2} }

  \sf{15}^{2}  =  {12}^{2}  +  {9}^{2}

 \sf225 = 144 + 81

 \sf{255 = 255}

∴ ∆ ACD is a right-angled triangle.

Similarly, ∆ BCD is also a right-angled triangle.

NOW,

  \sf\sin\alpha  =  \dfrac{perpendicular}{hypotenuse}

 \sf \sin\alpha  =  \dfrac{CD}{AC}

 \sf \sin\alpha  =  \dfrac{12}{15} =  \dfrac{4}{5}

 \sf \cos\beta  =  \dfrac{base}{hypotenuse}

 \sf \cos\beta  =  \dfrac{BD}{BC}

  \sf\cos\beta  =  \dfrac{16}{20}  =  \dfrac{4}{5}

 \sf \tan \phi =  \dfrac{perpendicular}{base}

  \sf\tan \phi =  \dfrac{BD}{DC}

  \sf\tan\phi =  \dfrac{16}{12} =  \dfrac{4}{3}

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Answered by llAbdulkadarll
9

44 ) \huge\underline{Question}

Find the area of the right angled triangle with hypotenuse 5 cm and one of acute angle is 48°38`.

\huge\underline{Answer}

____________________________

From the figure ,

= sin θ =  \frac{AB}{AC}

= sin 48°20' =  \frac{AB}{5}

= 0.7490 =  \frac{AB}{AC}

= 5 × 0.7490 = AB

= AB = 3.7450 cm.

____________________________

Then,

= cos θ =  \frac{BC}{AC}

= cos 48°30' =  \frac{BC}{5}

= 0.6626 =  \frac{BC}{5}

= 0.6626 × 5 = BC

= BC = 3.313 cm.

____________________________

Area of right triangle =  \frac{1}{2} bh

=  \frac{1}{2} bh × 3.3130 × 3.7450

= 1.6565 × 3.7450

= 6.2035925 cm^2

_____________________

45 ) \huge\underline{Question}

From the given figure, prove that θ + Φ = 90° . Also prove that there are two other right angled triangles. Find sin α , cos β and tan Φ.

\huge\underline{Answer}

____________________________

In ∆ACD ,

= cos θ =  \frac{CD}{AC} =  \frac{12}{15}

= sin θ =  \frac{AD}{AC} =  \frac{9}{12}

____________________________

In ∆BCD ,

= cos ϕ =  \frac{CD}{BC} =  \frac{12}{20}

= sin ϕ =  \frac{AD}{AC} =  \frac{16}{20}

____________________________

= cos ( A + B ) = cos A cos B - sin A sin B

= cos ( θ + ϕ ) = cos θ cos ϕ - sin θ sin ϕ

=  \frac{12}{15} ×  \frac{12}{20} -  \frac{9}{15} ×  \frac{16}{20}

=  \frac{144}{300} -  \frac{144}{300} = 0

= cos ( θ + ϕ ) = 0

= cos 90° = 0

= cos ( θ + ϕ ) = cos 90°

= θ + ϕ = 90°

____________________________

Using Pythagoras theorom :

=  {AC}^{2} = {AD}^{2} + {DC}^{2}

=  {15}^{2} = {12}^{2} + {9}^{2}

= 225 = 144 + 81

= 225 = 225

So, It is a right angled triangle.

____________________________

= sin α =  \frac{perpendicular}{hypotenuse}

= sin α =  \frac{CD}{AC}

= sin α =  \frac{12}{15} =  \frac{4}{5}

____________________________

= cos β =  \frac{BD}{BC}

= cos β =  \frac{16}{20}

= cos β =  \frac{4}{5}

____________________________

= tan ϕ =  \frac{perpendicular}{Base}

= tan ϕ =  \frac{BD}{DC}

= tan ϕ =  \frac{16}{12}

= tan ϕ =  \frac{4}{3}

____________________________

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