(x - 1)(x + 2)(2x - 7) {}^{2} < = 0
Answers
Answer:
Problem:
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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