- Two identical blocks each of mass 3 Kg placed on a frictionless table as shown in the
figure, are pushed by a force of 30 N. calculate.
Answers
Answer:
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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: }}}}[/tex]
In ∆ ACD, we have:
\sf\cos\theta = \dfrac{CD}{AC} = \dfrac{12}{15}cosθ=ACCD=1512
\sf \sin \theta = \dfrac{AD}{AC} = \dfrac{9}{12}sinθ=ACAD=129
In ∆ BCD, we have:
\sf\cos \phi = \dfrac{CD}{BC} = \dfrac{12}{20}cosϕ=BCCD=2012
\sf\sin \phi = \dfrac{BD}{BC} = \dfrac{16}{20}sinϕ=BCBD=2016
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}=1512×2012−159×2016
\sf = \dfrac{144}{300} - \dfrac{144}{300}=300144−300144
\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} }AC2=AD2+DC2
\sf{15}^{2} = {12}^{2} + {9}^{2}152=122+92
\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=hypotenuseperpendicular
\sf \sin\alpha = \dfrac{CD}{AC}sinα=ACCD
\sf \sin\alpha = \dfrac{12}{15} = \dfrac{4}{5}sinα=1512=54
▩ \sf \cos\beta = \dfrac{base}{hypotenuse}cosβ=hypotenusebase
\sf \cos\beta = \dfrac{BD}{BC}cosβ=BCBD
\sf\cos\beta = \dfrac{16}{20} = \dfrac{4}{5}cosβ=2016=54
▩ \sf \tan \phi = \dfrac{perpendicular}{base}tanϕ=baseperpendicular
\sf\tan \phi = \dfrac{BD}{DC}tanϕ=DCBD
\sf\tan\phi = \dfrac{16}{12} = \dfrac{4}{3}tanϕ=1216=34