Two buildings are facing each other on a road of width 12 metre. From the top of the first building, which is 10 metre high, the angle of elevation of the top of the second is found to be 60°. What is the height of the second building?
Let AB represent height of first building and CD represent height of second building.BD is the width of the road.Draw AMCDGiven Angle of elevation CAM = 60˚Given AB = 10BD = 12In AMDB , D = B = 90˚Since AMCD , M = 90˚A = 90˚ [Angle sum property of quadrilateral]Since each angle equal to […]
From the top of a lighthouse, an observer looking at a ship makes angle of depression of 60°. If the height of the lighthouse is 90 metre, then find how far the ship is from the lighthouse. (√3 =1.73)
Let C represent position of ship and AB represent height of the light houseGiven AB = 90mAngle of depression DAC = 60˚Here BCAD.BCA = DAC [Alternate interior angles]BCA = 60˚In ABC tan60 = AB/BC√3 = 90/BCBC = 90/√3 = 90√3/3= 30√3= 30 ×1.73= 51.9Hence the ship is 51.9 m away from light house.
A person is standing at a distance of 80m from a church looking at its top. The angle of elevation is of 45°. Find the height of the church.
Let C represent position of person and AB represent height of the church.Angle of elevation = C = 45˚BC = 80mIn right angled triangle ABC , tan = tan 45˚ = AB/BC1 = AB/80AB = 80Hence height of the church is 80m.
Prove that:
(1) sin2/ cos + cos = (sin2+cos2)/cos= 1/cos [sin2+cos2 = 1]= sec [1/cos = sec]Hence proved. (2) cos2 (1 + tan2 ) = cos2 +sin2 [cos2 ×tan2 = cos2 ×sin2/cos2 = sin2]= 1 [sin2+cos2 = 1]Hence proved. (3) √[(1-sin)/(1+ sin)] = √[(1-sin)/(1+ sin)]×√[(1- sin)/(1- sin)] [rationalizing denominator]= √[(1-sin)2/(1-sin2)]= √[(1-sin)2/cos2 [1-sin2 = cos2 ]= (1-sin)/cos [taking square root]= (1/cos)-(sin/cos)= sec-tan [1/cos = sec , sin/cos = tanHence proved. (4) […]
If tan = 1 then, find the values of (sin+ cos)/( sec+ cosecθ ).
Given tan = 1We know that tan 45˚ = 1= 45˚sin 45 = 1/√2cos45 = 1/√2sec45 = √2cosec45 = √2(sin+ cos)/( sec+ cosecθ ) = (sin45+ cos45)/( sec45+ cosec45 )= [(1/√2)+( 1/√2)]÷[√2+√2]= (2/√2)÷2√2= (2/√2)×(1/2√2)= 1/2Hence (sin+ cos)/( sec+ cosecθ ) = 1/2
If 5sec- 12cosec = 0, find the values of sec, cos and sin.
Given 5sec- 12cosec = 05sec = 12cosec5/cos = 12/sin [sec = 1/cos and cosec = 1/sin]5/12 = cos/sinsin/cos = 12/5tan = 12/5We know that 1+tan2 = sec21+(12/5)2 = sec21+(144/25) = sec2(25+144)/25 = sec2169/25 = sec2Taking square root on both sidessec = 13/5cos = 1/sec = 5/13We know that sin2+cos2=1sin2+(5/13)2 = 1sin2 = 1-(5/13)2sin2 = 1-(25/169)sin2 = (169-25)/169sin2 = 144/169Taking square root on […]
If cot = 40/9 , find the values of cosec and sin.
Given cot = 40/9We have 1+cot2 = cosec21+(40/9)2 = cosec21+(1600/81) = cosec2(81+1600)/81 = cosec21681/81 = cosec2cosec2 = 1681/81Taking square root on both sidescosec = 41/9We have sin = 1/cosecsin = 1÷(41/9)sin = 9/41Hence cosec = 41/9 and sin = 9/41.
If tan = 3/4 , find the values of sec and cos.
Given tan = 3/4We have 1+tan2= sec21+(3/4)2 = sec21+(9/16) = sec2sec2 = (16+9)/16 = 25/16Taking square root on both sidessec = 5/4We have cos = 1/seccos = 1÷(5/4)cos = 4/5Hence sec = 5/4 and cos = 4/5.
If sin = 7/25 , find the values of cos and tan.
Given sin = 7/25We have sin2+cos2 = 1(7/25)2+ cos2 = 1(49/625)+ cos2 = 1cos2 = 1-(49/625)cos2 = (625-49)/625cos2 = 576/625Taking square root on both sidescos = 24/25tan = sin/cos= (7/25) ÷(24/25)= (7/25) ×(25/24)= 7/24Hence cos = 24/25 and tan = 7/24.
Show that points P(1,-2), Q(5,2), R(3,-1), S(-1,-5) are the vertices of a parallelogram.
ProofGiven points are P(1,-2), Q(5,2), R(3,-1), S(-1,-5).By distance formula PQ = √[(x2-x1)2+(y2-y1)2]PQ = √[(5-1)2+(2-(-2))2]= √[(4)2+(4)2]= √[(16+16)]= √32PQ = √32 ………………(i)By distance formula QR = √[(x2-x1)2+(y2-y1)2]QR = √[(3-5)2+(-1-2)2]= √[(-2)2+(-3)2]= √[(4+9)]= √13QR = √13 ………………(ii)By distance formula RS = √[(x2-x1)2+(y2-y1)2]RS = √[(-1-3)2+(-5-(-1))2]= √[(-4)2+(-4)2]= √[(16+16)]= √32RS = √32 ………………(iii)By distance formula PS = √[(x2-x1)2+(y2-y1)2]PS = √[(-1-1)2+(-5-(-2))2]= √[(-2)2+(-3)2]= √[(4+9)]= […]