Some Applications of Trigonometry — CBSE Class 10 Maths Important Questions

Answer: (a) $10\sqrt3$103 m

$\tan30^\circ=\dfrac{h}{30}\Rightarrow h=30\tan30^\circ=\dfrac{30}{\sqrt3}=10\sqrt3$30^=h/30 h=3030^=30/3=103 m.

Answer: (b) $45^\circ$45^

If height $=$= shadow $=h$=h, then $\tan\theta=\dfrac{h}{h}=1\Rightarrow\theta=45^\circ$=h/h=1=45^.

Answer: (a) $\dfrac{15\sqrt3}{2}$153/2 m

The ladder is the hypotenuse, so $\sin60^\circ=\dfrac{\text{height}}{15}\Rightarrow\text{height}=15\sin60^\circ=15\times\dfrac{\sqrt3}{2}=\dfrac{15\sqrt3}{2}$60^=height/15=1560^=15×3/2=153/2 m.

Answer: (c) $60^\circ$60^

$\tan\theta=\dfrac{\text{height}}{\text{distance}}=\dfrac{50\sqrt3}{50}=\sqrt3\Rightarrow\theta=60^\circ$=height/distance=503/50=3=60^.

Answer: (a) Both A and R are true and R is the correct explanation of A.

If the tower has height $h$h and the Sun's altitude is $\theta$, then $\tan\theta=\dfrac{h}{\text{shadow}}$=h/shadow, so shadow $=h\cot\theta$=h. As $\theta$ decreases, $\cot\theta$ increases, hence the shadow lengthens. R correctly explains A.

Let the required distance be $d$d. Then $\tan45^\circ=\dfrac{15}{d}$45^=15/d.

Since $\tan45^\circ=1$45^=1, we get $1=\dfrac{15}{d}\Rightarrow d=15$1=15/d d=15 m.

The point is 15 m from the foot of the tower.

The string is the hypotenuse of the right triangle, and 60 m is the side opposite the $60^\circ$60^ angle. Let the length of the string be $L$L.

$\sin60^\circ=\dfrac{60}{L}\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{60}{L}\Rightarrow L=\dfrac{120}{\sqrt3}=40\sqrt3$60^=60/L 3/2=60/L L=120/3=403 m.

The length of the string is $40\sqrt3\approx69.2$40369.2 m.

Let the height of the tower be $h$h and the shorter shadow (at $60^\circ$60^) be $x$x. The longer shadow (at $30^\circ$30^) is then $x+40$x+40.

At $60^\circ$60^: $\tan60^\circ=\dfrac{h}{x}\Rightarrow h=\sqrt3\,x$60^=h/x h=3\,x.

At $30^\circ$30^: $\tan30^\circ=\dfrac{h}{x+40}\Rightarrow h=\dfrac{x+40}{\sqrt3}$30^=h/x+40 h=x+40/3.

Equating: $\sqrt3\,x=\dfrac{x+40}{\sqrt3}\Rightarrow 3x=x+40\Rightarrow 2x=40\Rightarrow x=20$3\,x=x+40/3 3x=x+40 2x=40 x=20 m.

Therefore $h=\sqrt3\times20=20\sqrt3=20\times1.73=34.6$h=3×20=203=20×1.73=34.6 m.

The height of the tower is $20\sqrt3\approx34.6$20334.6 m.

Let $B$B be the foot of the tree, $D$D the point where it broke, and $A$A the point on the ground where the top strikes, with $AB=8$AB=8 m and $\angle A=30^\circ$A=30^. The broken part $DA$DA (now leaning) equals the upper part of the tree.

Standing part: $\tan30^\circ=\dfrac{BD}{AB}\Rightarrow BD=8\tan30^\circ=\dfrac{8}{\sqrt3}$30^=BD/AB BD=830^=8/3 m.

Broken (leaning) part: $\cos30^\circ=\dfrac{AB}{DA}\Rightarrow DA=\dfrac{8}{\cos30^\circ}=\dfrac{8}{\sqrt3/2}=\dfrac{16}{\sqrt3}$30^=AB/DA DA=8/30^=8/3/2=16/3 m.

Original height $=BD+DA=\dfrac{8}{\sqrt3}+\dfrac{16}{\sqrt3}=\dfrac{24}{\sqrt3}=8\sqrt3=8\times1.73=13.84$=BD+DA=8/3+16/3=24/3=83=8×1.73=13.84 m.

The tree was $8\sqrt3\approx13.84$8313.84 m tall.

Let the horizontal distance between the building and the tower be $d$d. From the level of the building top, the foot of the tower is 7 m below and the line of sight to it is depressed $45^\circ$45^.

Depression to foot: $\tan45^\circ=\dfrac{7}{d}\Rightarrow d=7$45^=7/d d=7 m.

Elevation to top: let the part of the tower above the building-top level be $p$p. Then $\tan60^\circ=\dfrac{p}{d}\Rightarrow p=d\tan60^\circ=7\sqrt3$60^=p/d p=d60^=73 m.

Total height of tower $=7+p=7+7\sqrt3=7(1+\sqrt3)=7(1+1.73)=7\times2.73=19.11$=7+p=7+73=7(1+3)=7(1+1.73)=7×2.73=19.11 m.

The cable tower is $7(1+\sqrt3)\approx19.11$7(1+3)19.11 m high.

Let the common height be $h$h and let the point $P$P be $x$x m from the foot of the pole seen at $60^\circ$60^. Then $P$P is $(80-x)$(80-x) m from the other pole.

From the $60^\circ$60^ pole: $\tan60^\circ=\dfrac{h}{x}\Rightarrow h=\sqrt3\,x$60^=h/x h=3\,x.

From the $30^\circ$30^ pole: $\tan30^\circ=\dfrac{h}{80-x}\Rightarrow h=\dfrac{80-x}{\sqrt3}$30^=h/80-x h=80-x/3.

Equating the two expressions for $h$h:

$\sqrt3\,x=\dfrac{80-x}{\sqrt3}\Rightarrow 3x=80-x\Rightarrow 4x=80\Rightarrow x=20$3\,x=80-x/3 3x=80-x 4x=80 x=20 m.

So the point is 20 m from one pole and $80-20=60$80-20=60 m from the other.

Height $h=\sqrt3\times20=20\sqrt3\approx34.64$h=3×20=20334.64 m.

The poles are $20\sqrt3\approx34.64$20334.64 m high, and the point is 20 m and 60 m from the two poles.

Let the lighthouse be $LM$LM with $L$L at the top, $LM=75$LM=75 m. Let the nearer ship (depression $45^\circ$45^) be at distance $d_1$d_1 and the farther ship (depression $30^\circ$30^) at distance $d_2$d_2 from the foot $M$M. The angles of depression equal the corresponding angles of elevation at the ships.

Nearer ship: $\tan45^\circ=\dfrac{75}{d_1}\Rightarrow d_1=75$45^=75/d_1 d_1=75 m.

Farther ship: $\tan30^\circ=\dfrac{75}{d_2}\Rightarrow d_2=75\sqrt3$30^=75/d_2 d_2=753 m.

Distance between the ships $=d_2-d_1=75\sqrt3-75=75(\sqrt3-1)=75(1.73-1)=75\times0.73=54.75$=d_2-d_1=753-75=75(3-1)=75(1.73-1)=75×0.73=54.75 m.

The two ships are $75(\sqrt3-1)\approx54.75$75(3-1)54.75 m apart.

(i) From $A$A: $\tan30^\circ=\dfrac{h}{OA}\Rightarrow OA=\dfrac{h}{\tan30^\circ}=h\sqrt3$30^=h/OA OA=h/30^=h3. From $B$B: $\tan60^\circ=\dfrac{h}{OB}\Rightarrow OB=\dfrac{h}{\tan60^\circ}=\dfrac{h}{\sqrt3}$60^=h/OB OB=h/60^=h/3.

(ii) Since $A$A is farther, $OA-OB=40$OA-OB=40:

$h\sqrt3-\dfrac{h}{\sqrt3}=40\Rightarrow \dfrac{3h-h}{\sqrt3}=40\Rightarrow \dfrac{2h}{\sqrt3}=40\Rightarrow h=20\sqrt3$h3-h/3=40 3h-h/3=40 2h/3=40 h=203.

Then $OB=\dfrac{h}{\sqrt3}=\dfrac{20\sqrt3}{\sqrt3}=20$OB=h/3=203/3=20 m.

(iii) $h=20\sqrt3=20\times1.73=34.6$h=203=20×1.73=34.6 m.

The temple is $20\sqrt3\approx34.6$20334.6 m high (and $B$B is 20 m, $A$A is 60 m from its foot).