Some Applications of Trigonometry — CBSE Class 10 Maths Important Questions

Answer: (a) $10\sqrt3$10√3 m

$\tan30^\circ=\dfrac{h}{30}\Rightarrow h=30\tan30^\circ=\dfrac{30}{\sqrt3}=10\sqrt3$tan 30°=h/30⇒ h=30 tan 30°=30/√3=10√3 m.

Answer: (b) $45^\circ$45°

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

Answer: (a) $\dfrac{15\sqrt3}{2}$15√3/2 m

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

Answer: (c) $60^\circ$60°

$\tan\theta=\dfrac{\text{height}}{\text{distance}}=\dfrac{50\sqrt3}{50}=\sqrt3\Rightarrow\theta=60^\circ$tan θ=height/distance=50√3/50=√3⇒ θ=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 $\theta$θ, then $\tan\theta=\dfrac{h}{\text{shadow}}$tan θ=h/shadow, so shadow $=h\cot\theta$=h cot θ. As $\theta$θ decreases, $\cot\theta$cot θ increases, hence the shadow lengthens. R correctly explains A.

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

Since $\tan45^\circ=1$tan 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.

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

The length of the string is $40\sqrt3\approx69.2$40√3≈69.2 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°: $\tan60^\circ=\dfrac{h}{x}\Rightarrow h=\sqrt3\,x$tan 60°=h/x⇒ h=√3 x.

At $30^\circ$30°: $\tan30^\circ=\dfrac{h}{x+40}\Rightarrow h=\dfrac{x+40}{\sqrt3}$tan 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=20√3=20×1.73=34.6 m.

The height of the tower is $20\sqrt3\approx34.6$20√3≈34.6 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: $\tan30^\circ=\dfrac{BD}{AB}\Rightarrow BD=8\tan30^\circ=\dfrac{8}{\sqrt3}$tan 30°=BD/AB⇒ BD=8 tan 30°=8/√3 m.

Broken (leaning) part: $\cos30^\circ=\dfrac{AB}{DA}\Rightarrow DA=\dfrac{8}{\cos30^\circ}=\dfrac{8}{\sqrt3/2}=\dfrac{16}{\sqrt3}$cos 30°=AB/DA⇒ DA=8/(cos 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=8√3=8×1.73=13.84 m.

The tree was $8\sqrt3\approx13.84$8√3≈13.84 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: $\tan45^\circ=\dfrac{7}{d}\Rightarrow d=7$tan 45°=7/d⇒ d=7 m.

Elevation to top: let the part of the tower above the building-top level be $p$p. Then $\tan60^\circ=\dfrac{p}{d}\Rightarrow p=d\tan60^\circ=7\sqrt3$tan 60°=p/d⇒ p=d tan 60°=7√3 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+7√3=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 pedestal be $h$h m high and the point be $d$d m from its foot.

Top of the pedestal: $\tan45^\circ=\dfrac{h}{d}\Rightarrow d=h$tan 45°=h/d⇒ d=h.

Top of the statue: $\tan60^\circ=\dfrac{h+1.6}{d}\Rightarrow \sqrt3\,h=h+1.6$tan 60°=(h+1.6)/d⇒ √3 h=h+1.6.

$h(\sqrt3-1)=1.6\Rightarrow h=\dfrac{1.6}{\sqrt3-1}=\dfrac{1.6(\sqrt3+1)}{2}=0.8(\sqrt3+1)$h(√3-1)=1.6⇒ h=1.6/(√3-1)=(1.6(√3+1))/2=0.8(√3+1).

$h=0.8\times2.73\approx2.18$h=0.8×2.73≈2.18 m.

The pedestal is $0.8(\sqrt3+1)\approx2.18$0.8(√3+1)≈2.18 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: $\tan60^\circ=\dfrac{h}{x}\Rightarrow h=\sqrt3\,x$tan 60°=h/x⇒ h=√3 x.

From the $30^\circ$30° pole: $\tan30^\circ=\dfrac{h}{80-x}\Rightarrow h=\dfrac{80-x}{\sqrt3}$tan 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=20√3≈34.64 m.

The poles are $20\sqrt3\approx34.64$20√3≈34.64 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 $d_1$d₁ and the farther ship (depression $30^\circ$30°) at distance $d_2$d₂ from the foot $M$M. The angles of depression equal the corresponding angles of elevation at the ships.

Nearer ship: $\tan45^\circ=\dfrac{75}{d_1}\Rightarrow d_1=75$tan 45°=75/d₁⇒ d₁=75 m.

Farther ship: $\tan30^\circ=\dfrac{75}{d_2}\Rightarrow d_2=75\sqrt3$tan 30°=75/d₂⇒ d₂=75√3 m.

Distance between the ships $=d_2-d_1=75\sqrt3-75=75(\sqrt3-1)=75(1.73-1)=75\times0.73=54.75$=d₂-d₁=75√3-75=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.

The angle of depression from the top equals the angle of elevation from the car (alternate angles).

(i) $\tan30^\circ=\dfrac{h}{d_1}\Rightarrow d_1=h\sqrt3$tan 30°=h/d₁⇒ d₁=h√3.

(ii) $\tan60^\circ=\dfrac{h}{d_2}\Rightarrow d_2=\dfrac{h}{\sqrt3}$tan 60°=h/d₂⇒ d₂=h/√3.

(iii) Distance covered in 6 s $=h\sqrt3-\dfrac{h}{\sqrt3}=\dfrac{2h}{\sqrt3}$=h√3-h/√3=2h/√3. Speed $=\dfrac{2h}{\sqrt3}\div6=\dfrac{h}{3\sqrt3}$=2h/√3÷6=h/3√3 per second.

(iv) Time $=d_2\div\text{speed}=\dfrac{h}{\sqrt3}\div\dfrac{h}{3\sqrt3}=3$=d₂÷ speed=h/√3÷ h/3√3=3 seconds.

The car reaches the foot of the tower 3 seconds later.