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1. In any triangle ABC, prove that a sin (B – C) + b sin (C – A) + c sin (A – B) = 0.2. Find the radius of the circle in which a central angle of 60° intercepts an arc of length 37.4 cm (use π = 22/7).3. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?4. Find the value of √3 cosec 20° – sec 20°.5. Show that tan 3x tan 2x tan x = tan 3x – tan 2x – tan x. |
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Answer» Q1. Let sinA/a = sinB/b = sinC/c = k (By sine rule) sinA = ak, sinB = bk, sinC = ck sin(A - B) = sinAcosB - cosAsinB = akcosB - bkcosA = k(acosB - BcosA) sin (B - C) = sinBcosC - cosBsinC = bkcosC - ckcosB = k(bcosC - c.cosB) sin (C - A) = sinCcosA - cosCsinA = ckcosA - akcosC = k(c.cosA - acosC) Now, a sin (B – C) + b sin (C – A) + c sin (A – B) = ak(bcosC - c.cosB) + bk(c.cosA - acosC) + ck(c.cosA - acosC) = k(bc cosA - bc cosA) + k(ab cosC - ab cosC) + k(ac cosB - ac cosB) = 0 = RHS Hence, proved Q2. Let r be the radius. Area of sector = lr/2 (l is length and r is the radius) lr/2 = θπr2/360 37.4r / 2 = 22× r2 / 42 1.7/2 = r/42 r = 1.7 × 21 r = 35.7 cm Q3. The wheel makes 360 revolutions in 60 seconds So, the wheel will make 6 revolution in 1 second. Angle made in 1 revolution by wheel = 2π Angle made in 6 revolutions = 12π Therefore, the wheel turns 12π radians in one second. Q4. √3/sin20 - 1/cos20 = (√3cos20 - sin20) / (sin20. cos20) = 2(√3/2cos20 - 1/2. sin20) / (sin20. cos20) = 2(sin60cos20 - cos60.sin20) / (sin20. cos20) = 2(sin40) / (sin20. cos20) = 2{2 × (sin20. cos20)} / (sin20. cos20) = 2 × 2 = 4 Q5. tan3x = tan(2x + x) or tan3x = (tan2x + x) / (1 - tan2x. tanx) or tan3x - tan3x.tan2x.tanx = (tan2x + x) or tan3x - tan2x - tanx = tan3x.tan2x.tanx or tan3x.tan2x.tanx = tan3x - tan2x - tanx LHS = RHS. Hence, proved.
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