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• प्रश्न ४.खालील वाक्प्रचारांचा
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Answers
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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