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Grade 12th passTrigonometry

1)A wire of length 18 m had been tied with electric pole at an angle of elevation 30º with
the ground. Because it was convering a long distance, it was cut and tied at an angle of
elevation 60º with the ground. How much length of the wire was cut? ANS(7.608)
2)Ganesh observes the top and foot of an opposite building from a window at first floor of
his house which is at 8 meter height. The angles of elevation and depression of the top
and the foot of the building are 60º and 30º respectively. Find the height of the building ANS(24)
3) Find x so that x, x + 2, x + b are conseecitive terms of a geometric progression ans(-4)
4) An iron pillar consists of a cylindrical portion of 2.8 m. height and 20 cm. in diameter and
a cone of 42 cm. height surmounting it. Find the weight of the pillar if 1 cm3 of iron weighs
7.5 g .ANS(693KG)
5) A toy is made in the form of hemisphere surmounted by a right cone whose circular base is
joined with the plane surface of the hemisphere. The radius of the base of the cone is
7 cm. and its volume is
3/2 of the hemisphere. Calculate the height of the cone and thesurface area of the toy correct to 2 places of decimal (22/7) ANS(H=22.05CM;TOY=793)

6) the height of a solid cylinder is10 cm and diameter is 7cm. Two equal conical holesof radius 3cm and height 4 cm are cut off as shown the figure. Find the volume of the remaining solid ANS(309.57)
7) A solid metallic sphere of diameter 28 cm is melted and recast into a number of smaller
cones, each of diameter 4+ 2/3 cm and height 3cm. Find the number of cones so formed ANS(224)

rajesh

11 Years agoGrade 12th pass

1 Answer

Sumit Majumdar

11 Years ago

Dear student,
(1) If we assume the height at which the wire has been joined is y m and the new length of the wire is x m. Then, using trigonometry, we would have:
$y=18sin30^{\circ}$ , $x=\frac{18sin30^{\circ}}{sin60^{\circ}}=6\sqrt{3}$
Hence the length of the wire that was cut is equal to: $18-6\sqrt{3}$ m
(2) The total height of the builiding would be equal to the height above the window .
So, this comes out to be:
$\frac{8}{tan30^{\circ}}tan60^{\circ}=24 m$
(3) For b = 3, we would have:
$\left ( x+2 \right )^2=x(x+3)\Rightarrow x=-4$
(4) The total volume would be equal to:
$V=\pi r^2h+\pi r^2 \frac{h'}{3}=\pi (10^2)(280)+\pi (10^2) \frac{42}{3}=9240000 cm^3$
Hence the required weight would be equal to:
$V\times 7.5=69300000 gm=69300 kg$
(5) Volume of a hemisphere is equal to
$V=\frac{2}{3}\pi r^3$
Hence, the volume of the cone would be equal to
$V_c=\pi r^3=\frac{22}{7}\times 7^3=1078 cm^3$
Hence, the height of the cone would be equal to:
$h=3\times r=21 cm$
Thus the total surface area of the toy would be equal to:
$S=\pi r \left ( r+\sqrt{r^2+h^2} \right )+2\pi r^2$

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