Question

Two forces of magnitudes k cos A and k cos B act along CA, CB of a triangle ABC . Prove that their resultant is k sin C. (B.Sc.86 Calicut; B.Sc. 79, 8 M.K.U.) [Hint:- Use the result that in a triangle cos^ 2 A+cos^ 2 B+cos^ 2; C=1-2 cos A cos B cos C]​

Answer 1

Answer:

Answer:

If x=cos1cos2cos3..cos89 and y=cos2cos6cos10…cos86, then what is the nearest integer to 2/7log (base2) (y/x)?

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Given,

x=cos(1)cos(2)cos(3)…cos(87)cos(88)cos(89)

and y=cos(2)cos(6)cos(10)…cos(82)cos(86)

As we know,

cos(A)cos(B)=cos(A+B)+cos(A−B)2

So,

cos(1)cos(89)=cos(89+1)+cos(89−1)2=cos(90)+cos(88)2

As, cos(90)=0

cos(1)cos(89)=cos(88)2

Similarly,

cos(2)cos(88)=cos(88+2)+cos(88−2)2=cos(86)2

and so on.

So,

x=cos(1)cos(2)cos(3)…cos(87)cos(88)cos(89)

=cos(88)cos(86)cos(84)…cos(4)cos(2)cos(45)244

=cos(88)cos(86)cos(84)…cos(4)cos(2)2442–√

[As cos(45)=12√ ]

Now,

cos(2)cos(88)=cos(86)2 [Similar analogy as above.]

So,

x=cos(88)cos(86)cos(84)…cos(4)cos(2)2442–√

=cos(86)cos(82)cos(78)…cos(6)cos(2)2662–√

=y2662–√

=y266+12

=y21332

So,

2log2(yx)7

=2log2(21332)7

=2×133)7×2

=19

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

Two forces of magnitudes k cos A and k cos B act along CA, CB of a triangle ABC.

Let resultant be R which makes an angle x with CA.

From figure,

R, k cosA and k cosB are in equilibrium.

By Lami's Theorem,

$\rm :\longmapsto\:\dfrac{k \: cosA}{sin(\pi - (C - x))} = \dfrac{k \: cosB}{sin(\pi - x)} = \dfrac{R}{sinC}$

$\rm :\longmapsto\:\dfrac{k \: cosA}{sin(C - x)} = \dfrac{k \: cosB}{sinx} = \dfrac{R}{sinC} - - - (1)$

Taking first and second member, we get

$\rm :\longmapsto\:\dfrac{k \: cosA}{sin(C - x)} = \dfrac{k \: cosB}{sinx}$

$\rm :\longmapsto\:\dfrac{\: cosA}{sin(C - x)} = \dfrac{\: cosB}{sinx}$

$\rm :\longmapsto\:cosAsinx = cosBsin(C - x)$

$\rm :\longmapsto\:cosAsinx = cosB[sinCcosx - sinxcosC]$

$\rm :\longmapsto\:cosAsinx = cosBsinCcosx - sinxcosCcosB$

$\rm :\longmapsto\:cosAsinx + sinxcosCcosB = cosBsinCcosx$

$\rm :\longmapsto\:sinx(cosA + cosCcosB) = cosBsinCcosx$

$\rm :\longmapsto\:cosA + cosCcosB = cosBsinC \times \dfrac{cosx}{sinx}$

$\rm :\longmapsto\:cos[\pi - (B + C)] + cosCcosB = cosBsinCcotx$

$\rm :\longmapsto\: - cos(B + C) + cosCcosB = cosBsinCcotx$

$\rm :\longmapsto\: - [cosBcosC - sinBsinC] + cosCcosB = cosBsinCcotx$

$\rm :\longmapsto\: - cosBcosC + sinBsinC + cosCcosB = cosBsinCcotx$

$\rm :\longmapsto\: sinBsinC = cosBsinCcotx$

$\rm :\longmapsto\: sinB = cosB \: cotx$

$\rm :\longmapsto\: tanB \: = \: cotx$

$\rm :\longmapsto\: tanB \: = \: tan\bigg[\dfrac{\pi}{2} - x\bigg]$

$\bf :\longmapsto\: B \: = \: \dfrac{\pi}{2} - x$

Now, From equation (1), Taking second and third member,

$\rm :\longmapsto\: \dfrac{k \: cosB}{sinx} = \dfrac{R}{sinC}$

On substituting the value of B, we get

$\rm :\longmapsto\: \dfrac{k \: cos\bigg[\dfrac{\pi}{2} - x\bigg]}{sinx} = \dfrac{R}{sinC}$

$\rm :\longmapsto\: \dfrac{k \: sinx}{sinx} = \dfrac{R}{sinC}$

$\rm :\longmapsto\: k= \dfrac{R}{sinC}$

$\bf\implies \:R = k \: sinC$

Hence, Proved

Identities used

$\boxed{ \bf{ \: sin(\pi - x) = sinx}}$

$\boxed{ \bf{ \: sin(x - y) = sinxcosy - sinycosx}}$

$\boxed{ \bf{ \: cos(\pi - x) \: = \: - \: cosx}}$

$\boxed{ \bf{ \: cos(x + y) = cosxcosy - xsiny}}$