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AB = h ft = Height of the building
C is the point on the ground from which angle of elevation of building #/_ACB=30^@20'=(30+20/60)^@=(91/3)^@#

D is the point on the ground(50 ft closer to the building from C) from which angle of elevation of building #/_ADB=45^@#
Let BD = x ft

Now #(AB)/(BD)=h/x=tan/_ADB=tan45^@=1#

#=>h/x=1=>x=h#

Again#(AB)/(BC)=(AB)/(BD+DC)=h/(x+50)=tan/_ACB=tan(91/3)^@#

#=>h/(x+50)=0.585#

#=>h/(h+50)=0.585#

#=>h=(h+50)xx0.585#

#=>h(1-0.585)=50xx0.585~~29.25#

#h=29.25/0.415~~70.5 ft#

Answer:

#sin(arccos(x)) = sqrt(1-x^2)#

Explanation:

From Pythagoras, we have:

#sin^2 theta + cos^2 theta = 1#

If #x in [-1, 1]# and #theta = arccos(x)# then:

#theta in [0, pi]#

#sin(theta) >= 0#

Hence:

#sin(arccos(x)) = sin(theta) = sqrt(1 - cos^2 theta) = sqrt(1-x^2)#

Note we can use the non-negative square root since we have already established that #sin(arccos(x)) >= 0#

Answer:

#x=sqrt(3)/3#

Explanation:

#"arccot"(x) + 2arcsin(sqrt(3)/2) = pi#

The inverse sine function #arcsin(x)# is defined as the unique value in the interval #[-pi/2,pi/2]# such that #sin(arcsin(x)) = x#. On that interval, we have #sin(pi/3) = sqrt(3)/2# as a well known angle. Thus #arcsin(sqrt(3)/2) = pi/3#

#=> "arccot"(x) + (2pi)/3 = pi#

#=> "arccot"(x) = pi/3#

#=> cot("arccot"(x)) = cot((pi)/3)#

#=> x = cot((pi)/3)#

#=cos((pi)/3)/sin((pi)/3)#

#=(1/2)/(sqrt(3)/2)#

#=1/sqrt(3)#

#=sqrt(3)/3#

Answer:

The sides measure #4.13" cm."# and #15.35" cm."#.

Explanation:

We know the length of BD and of AC. However, we don't have an angle that is on the exterior (ex BDC, BCD or DBC), so we don't have enough information to use Cosine's Law. We will approach this problem by a different approach.

We know, by the properties of the parallelogram, that diagonals are cut into two equal parts at the point of intersection. Hence, if #AC = 12" cm"#, then #EC = 6" cm"#. Similarly, if #BD = 7" cm"#, then #BE = 3.5" cm"#.

We can now use the Law of Cosines to solve for side #BC#.

#BC^2 = EC^2 + BE^2 - 2(EC)(BE)cos/_BEC#

#BC^2 = 6^2 + 3.5^2 - (2 xx 6 xx 3.5 xx cos(42˚))#

#BC ~= 4.13" cm."#

To find the length of #DC#, we can use the property that #/_BEC# and #/_DEC# are supplementary, or their sum is #180˚#.

#/_BEC + /_DEC = 180˚#

#42˚ + /_DEC = 180˚#

#/_DEC = 138˚#

Since #BE = ED#, we can use the Law of Cosine's once more.

#DC^2 = DE^2 + EC^2 - (2 xx DE xx EC xx cos(/_DEC))#

#DC^2 = 3.5^2 + 6^2 - (2 xx 3.5 xx 6 xx cos(138˚))#

#DC ~= 15.35" cm."#

Hence, the two sides of the parallelogram measure #4.13" cm."# and #15.35" cm."#.

Hopefully this helps!

drawn

Let CB be the cliff . From point A the angle of elevation of the peak C of the cliff is #/_CAB=theta#'. After going up a distance , #AD=k# #m# towards the top of the cliff at an angle,#/_DAE =phi#,it ls found the angle of elevation #/_CDF=alpha#.

DF and DE are perpendiculars drawn from D on CB and AB.

Now #DE =ksinphi and AE =kcosphi#

Let
#h="CB the height of the cliff" and b=BA#

For #DeltaCAB," "(CB)/(BA)=tantheta#

#=>b/h=cottheta=>b=hcottheta#

Now #DF=BE=BA-AE=b-kcosphi#

#CF=CB-FB=CB-DE=h-ksinphi#

For #DeltaCDF," "(CF)/(DF)=tanalpha#

#=>(h-ksinphi)/(b-kcosphi)=tanalpha#

#=>(h-ksinphi)/(hcottheta-kcosphi)=tanalpha#

#=>h-htanalphacottheta=ksinphi-kcosphitanalpha#

#=>h=(k(sinphi-cosphitanalpha))/(1-tanalphacottheta)#

#=>h=(ksintheta(sinphicosalpha-cosphisinalpha))/(sinthetacosalpha-costhetasinalpha)#

#=>h=(ksinthetasin(phi-alpha))/(sin(theta-alpha))#

Given that #f(t)# the value of a stock is modeled as

#f(t)=mt+b+Asin((pit)/6)# where t is in months since Jan1

Let us consider

#"At Jan1 "t= 0 and f(0)=$20.00#

#"At Apr1 "t= 3and f(3)=$37.50#

#"At Jul1 "t= 6 and f(6)=$32.50#

#"At Oct1 "t= 9 and f(6)=$35.00#

#"At Jan1 "t= 12 and f(6)=$50.00#

so
#"At Jan1 "t= 0 and f(0)=$20.00#

#=>mxx0+b+Asin(0)=20=>b=20.......(1)#

#"At Apr1 "t= 3and f(3)=$37.50#

#=>mxx3+20+Asin((pi*3)/6)=37.50#

#=>3m+20+A=37.50#

#=>3m+A=37.50-20.00#

#=>3m+A=17.50............................(2)#

#"At Jul1 "t= 6 and f(6)=$32.50#

#=>6xxm+20+Asin((pi*6)/6)=35.00#

#=>6xxm+20+0=35#

#=>m=15/6=2.5..........................(3)#

From (2) and (3) we get

#=>3xx2.5+A=17.50#

#=>A=10.00#

Hence
(a)
#color(red)(m=2.50" "b=20 and A=10)#
(b)

Now the given equation becomes

#f(t)=2.5t+20+10sin((pit)/6)#

Differentiating this equation w,r to t we get the rate of change of the price of stock as

#f'(t)=d/(dt)(2.5t)+d/(dt)(20)+d/(dt)(10sin((pit)/6))#

#=>f'(t)=2.5+(10pi)/6cos((pit)/6)#
In this equation the value of #cos((pit)/6)# varies with time.
#cos((pit)/6)# has got maximum positive value at t=0 and t =12 as #cos 0 and cos (2pi) = 1#
So rate of positive change of price of stock will occur in this period i.e. Jan every year this means the stock appreciate s most in this period.

(c)

Taking #f'(t)=0# we have

#2.5+(10pi)/6cos((pit)/6)=0#

#=>cos((pit)/6)=-1.5/pi#

#=>t=6/picos^-1(-1.5/pi)~~3.95~~4->color(red)"month of May"#

when t =4

#f''(4)=-(10pi^2)/36sin((pi*4)/6)<0->"indicates maximum"#

Again #t = 6/pi(2pi-cos^-1(-1.5/pi))=12-6/picos^-1(-1.5/pi)#

#=12-3.95=8.05->" indicates month of September" #

when t =8

#f''(8)=-(10pi^2)/36sin((pi*8)/6)>0->color(red)"indicates Minimum"#

So during the period May -September growth of price gets diminished i.e the price of the stock is actually losing in this period.
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