Math, asked by shilpagundkal6, 5 months ago


84xy + 196x - 36y - 84

Answers

Answered by Vaanyapopli
1

Answer:

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: }}}}}

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}}sinC=

hypotenuse

perpendicular

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

AC

AB

\sf{ \sin48 \degree30' = \dfrac{AB}{5}}sin48°30

=

5

AB

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}}0.7490=

5

AB

\sf5 \times 0.7490 = AB5×0.7490=AB5×0.7490=AB5×0.7490=AB

★\sf{AB = 3.7450 \: cm}AB=3.7450cm

For finding BC i.e, base

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

hypotenuse

base

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

AC

BC

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

=

5

BC

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}0.6626=

5

BC

\sf{0.6626 \times 5 = BC}0.6626×5=BC

★ \sf{BC = 3.313 \: cm}BC=3.313cm

Now let's find out area

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

Areaofrighttriangle=

2

1

×b×h

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

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

2

1

×3.313×3.7450

\sf = 1.6565 \times 3.7450=1.6565×3.7450=1.6565×3.7450=1.6565×3.7450

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

2

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

≈6.204cm

2

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

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}cosθ=

AC

CD

=

15

12

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

AC

AD

=

12

9

In ∆ BCD, we have:

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

BC

CD

=

20

12

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

BC

BD

=

20

16

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}=

15

12

×

20

12

15

9

×

20

16

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

300

144

300

144

\sf \therefore \cos(\theta + \phi )∴cos(θ+ϕ)

We know cos 90° = 0

\sf \cos( \theta + \phi) = \cos90 \degreecos(θ+ϕ)cos(θ+ϕ)=cos90\degreecos(θ+ϕ)

\sf{ \theta + \phi = 90 \degree}θ+ϕ=90°

Hence Proved!!

In right-angled triangle ACD, By using Pythagoras theorem

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

2

=AD

2

+DC

2

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

2

=12

2

+9

2

\sf225 = 144 + 81225=144+81

\sf{255 = 255}255=255

∴ ∆ ACD is a right-angled triangle.

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

NOW,

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

hypotenuse

perpendicular

\sf \sin\alpha = \dfrac{CD}{AC}sinα=

AC

CD

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

15

12

=

5

4

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

hypotenuse

base

\sf \cos\beta = \dfrac{BD}{BC}cosβ=

BC

BD

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

20

16

=

5

4

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

base

perpendicular

\sf\tan \phi = \dfrac{BD}{DC}tanϕ=

DC

BD

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

12

16

=

3

4

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