Trigonometric ratios: problems questions with answers

Questions

1. Question 1Heights of trees and flagpoles

   From a point $A$ on level ground $30$ m from the foot of a tree, the angle of elevation of the top $T$ is $56^\circ$. Find the height of the tree, $h$ m, correct to 3 significant figures.

   Show the worked solution

   1. Step 1: Label the sides from the angle of elevationFrom the angle of elevation: the height $h$ m is the opposite and the distance $30$ m along the ground is the adjacent.
   2. Step 2: Choose the ratio that uses those two sides (SOH CAH TOA) and write it down$\tan 56^\circ = \dfrac{h}{30}$
   3. Step 3: Make h the subject$h = 30 \tan 56^\circ$$= 44.4768\ldots$$= 44.5$ m (3 s.f.)
2. Question 2Ramps

   A ramp $5$ m long rises to a platform $0.8$ m above level ground. Find the angle the ramp makes with the ground, $x^\circ$, correct to 1 decimal place.

   Show the worked solution

   1. Step 1: Label the two known sides from the angle at the bottom of the rampFrom the angle at the bottom of the ramp: the height $0.8$ m is the opposite and the ramp, $5$ m, is the hypotenuse.
   2. Step 2: Choose the ratio that uses those two sides (SOH CAH TOA) and write it down$\sin x^\circ = \dfrac{0.8}{5}$
   3. Step 3: Two sides known: use the inverse to find the angle$x = \sin^{-1}\left(\dfrac{0.8}{5}\right)$$= 9.2068\ldots$$= 9.2^\circ$ (1 d.p.)
3. Question 3Ladders

   A ladder $4$ m long leans against an upright wall, making an angle of $62^\circ$ with level ground. Find the height, $h$ m, that it reaches up the wall, correct to 3 significant figures.

   Show the worked solution

   1. Step 1: Label the sides from the angle at the foot of the ladderFrom the angle at the foot of the ladder: the height up the wall, $h$ m, is the opposite and the ladder, $4$ m, is the hypotenuse.
   2. Step 2: Choose the ratio that uses those two sides (SOH CAH TOA) and write it down$\sin 62^\circ = \dfrac{h}{4}$
   3. Step 3: Make h the subject$h = 4 \sin 62^\circ$$= 3.5317\ldots$$= 3.53$ m (3 s.f.)
4. Question 4Angles of elevation

   A tree $15$ m tall stands on level ground. A point $A$ on the ground is $50$ m from its foot. Find the angle of elevation, $x^\circ$, of its top $T$ from $A$, correct to 1 decimal place.

   Show the worked solution

   1. Step 1: Label the two known sides from the angle of elevationFrom the angle of elevation: the height $15$ m is the opposite and the distance $50$ m along the ground is the adjacent.
   2. Step 2: Choose the ratio that uses those two sides (SOH CAH TOA) and write it down$\tan x^\circ = \dfrac{15}{50}$
   3. Step 3: Two sides known: use the inverse to find the angle$x = \tan^{-1}\left(\dfrac{15}{50}\right)$$= 16.6992\ldots$$= 16.7^\circ$ (1 d.p.)
5. Question 5Two triangles: find a side, then use it

   In the diagram, $Z$ lies on $XY$ and $WZ$ is perpendicular to $XY$. $WX = 15$ cm, $\angle WXZ = 61^\circ$, $\angle WYZ = 24^\circ$ and $WY = x$ cm. Find $x$, correct to 3 significant figures.

   Show the worked solution

   1. Step 1: Triangle WXZ: label the sides from the 61° angle at XIn triangle $WXZ$, from the 61° angle at X: $WZ$ is the opposite and $WX = 15$ cm is the hypotenuse.
   2. Step 2: Choose the ratio (SOH CAH TOA), then multiply: keep the full calculator value for the next triangle$\sin 61^\circ = \dfrac{WZ}{15}$$WZ = 15 \sin 61^\circ$$= 13.1192\ldots$ cm
   3. Step 3: Triangle WYZ: label the sides from the 24° angle at YIn triangle $WYZ$, from the 24° angle at Y: $WY = x$ cm is the hypotenuse and $WZ$ is the opposite.
   4. Step 4: Choose the ratio (SOH CAH TOA), then divide$\sin 24^\circ = \dfrac{13.1192\ldots}{x}$$x = \dfrac{13.1192\ldots}{\sin 24^\circ}$$= 32.2550\ldots$$= 32.3$ cm (3 s.f.)
6. Question 6Is it safe? Find the angle and decide

   A ramp $3$ m long rises $0.3$ m above level ground. A ramp is safe for wheelchairs when the angle $x^\circ$ it makes with the ground is at most $7^\circ$. Find $x$ and decide whether the ramp is safe for wheelchairs.

   Show the worked solution

   1. Step 1: Label the two known sides from the angle x° at the bottom of the rampFrom the angle x° at the bottom of the ramp: the height, $0.3$ m, is the opposite and the ramp, $3$ m, is the hypotenuse.
   2. Step 2: Choose the ratio that uses those two sides (SOH CAH TOA) and write it down$\sin x^\circ = \dfrac{0.3}{3}$
   3. Step 3: Two sides known: use the inverse to find the angle$x = \sin^{-1}\left(\dfrac{0.3}{3}\right)$$= 5.7391\ldots$$= 5.7^\circ$ (1 d.p.)
   4. Step 4: Compare x° with the limit, 7°$x = 5.7$, at most $7^\circ$, so the ramp is safe for wheelchairs.
7. Question 7Climbing at an angle: speed, time, height

   A cable car moves along a straight cable, rising at $25^\circ$ to the horizontal, at a constant speed of $3$ m/s. Find the height it gains in $3$ minutes, $h$ m, correct to 3 significant figures.

   Show the worked solution

   1. Step 1: Change the time to seconds (× 60), then distance = speed × timeTime $= 3 \times 60$$= 180$ sDistance $= 3 \times 180$$= 540$ m
   2. Step 2: The path is the hypotenuse and the height is opposite the angle: use sin$\sin 25^\circ = \dfrac{h}{540}$$h = 540 \sin 25^\circ$$= 228.2138\ldots$$= 228$ m (3 s.f.)
8. Question 8The top and bottom of a window

   From a point $P$ on level ground in front of a shophouse, the angles of elevation of the top and the bottom of a window in the wall are $48^\circ$ and $37^\circ$. The bottom of the window is $18$ m above the ground. Find the height of the window, $h$ m, correct to 3 significant figures.

   Show the worked solution

   1. Step 1: The bottom of the window is opposite its angle and the distance to the wall (d m) is adjacent: tan gives d$\tan 37^\circ = \dfrac{18}{d}$, so $d = \dfrac{18}{\tan 37^\circ}$$d = 23.8868\ldots$ m
   2. Step 2: The top of the window (T m): the same distance, the larger angleTop: $\tan 48^\circ = \dfrac{T}{23.8868\ldots}$, so $T = 23.8868\ldots \tan 48^\circ$$T = 26.5289\ldots$ m
   3. Step 3: The window is the top height minus the bottom height$h = 26.5289\ldots - 18$$= 8.5289\ldots$$= 8.53$ m (3 s.f.)
9. Question 9Bridge struts: lengths, then a price

   The diagram shows the side view of a footbridge. The deck $AB$ is $12$ m long and $C$ is the point on $AB$ such that $AC : CB = 2 : 3$. Two steel struts, $AD$ and $BD$, support the deck; $D$ is vertically below $C$ and $\angle BAD = 25^\circ$. Steel strut costs $85 per metre when the total length bought is 10 m or less, and $72 per metre for the whole order when more than 10 m is bought. Find the cost of the steel for the two struts, correct to the nearest $100.

   Show the worked solution

   1. Step 1: AC is its share of the deck: the ratio gives the number of equal parts$AC = \dfrac{2}{5} \times 12$$\phantom{AC} = 4.8$ m, so $CB = 12 - 4.8$$CB = 7.2$ m
   2. Step 2: In triangle ACD (right-angled at C): AC is adjacent to the angle at A, AD the hypotenuse, so use cos$AD = \dfrac{4.8}{\cos 25^\circ}$$\phantom{AD} = 5.2962\ldots$ m
   3. Step 3: CD is opposite the angle at A: use tan (keep the full value)$CD = 4.8 \tan 25^\circ$$\phantom{CD} = 2.2382\ldots$ m
   4. Step 4: In triangle BCD (right-angled at C): Pythagoras' theorem for BD$BD = \sqrt{7.2^2 + (2.2382\ldots)^2}$$\phantom{BD} = 7.5398\ldots$ m
   5. Step 5: The total length decides the price per metre (more than 10 m: the lower price for all of it)$\text{Total} = 5.2962\ldots + 7.5398\ldots$$\phantom{\text{Total}} = 12.8361\ldots$ m, which is more than $10$ m, so the price is $72 per metre.
   6. Step 6: The cost, rounded only at the end$\text{Cost} = 12.8361\ldots \times 72$$\phantom{\text{Cost}} = 924.1992\ldots$, so about $900 (to the nearest $100).