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SL 3.2—2d and 3d trig, sine rule, cosine rule, area - IB Questionbank | RevisionDojo

Practice SL 3.2—2d and 3d trig, sine rule, cosine rule, area with authentic IB Mathematics Analysis and Approaches (AA) exam questions for both SL and HL students. This question bank mirrors Paper 1, 2, 3 structure, covering key topics like functions and equations, calculus, complex numbers, sequences and series, and probability and statistics. Get instant solutions, detailed explanations, and build exam confidence with questions in the style of IB examiners.

Paper

Level

Question 1

SLPaper 2

A research drone is tethered by two cables, $AP$ and $BP$ , attached to points $A$ and $B$ on level ground. The points $A$ and $B$ are 80 m apart. The drone $P$ is directly above the point $P_0$ on the ground. In the horizontal triangle $ABP_0$ , the angle at $P_0$ between the lines $P_0A$ and $P_0B$ is $60^\circ$ .

At a certain instant, the angle of elevation of the drone from $A$ is $28^\circ$ and from $B$ is $35^\circ$ .

Draw a fully-labelled 3-D diagram showing $A, B, P_0, P$ and all known angles.

[2]

Let the horizontal distances $AP_0 = x$ m and $BP_0 = y$ m. Show that $x \tan 28^\circ = y \tan 35^\circ.$

[2]

Using the cosine rule in triangle $ABP_0$ , show that $80^2 = x^2 + y^2 - 2xy \cos 60^\circ.$

[2]

Use your results from part 2 and 3 to find the height $h$ of the drone above the ground. Give your answer to the nearest metre.

[3]

The drone then moves horizontally so that its projection $P_0$ traces out a circular path of radius 80 m centred at $A$ . Calculate the angle swept by $P_0$ in radians and the length of the arc of this path when the bearing of $P_0$ from $A$ changes from $050^\circ$ to $130^\circ$ .

[3]

The tension in each cable is proportional to its length. If the tension in $AP$ is $T_A = k|AP|$ and in $BP$ is $T_B = k|BP|$ , find the ratio $\dfrac{T_A}{T_B}$ at this instant.

[3]

Question 2

SLPaper 2

Given the triangle ABC where $AC=5 \text{ cm}, BC=4 \text{ cm}, AB=6 \text{ cm}$ .

Find $\angle B$ .

[3]

Find area of the triangle $ABC$ .

[3]

Question 3

SLPaper 2

A rescue helicopter $P$ hovers above a mountain valley. It is connected by two winch cables to huts $A$ and $B$ on level ground. The distance between the huts is $90 \text{ m}$ . The helicopter is vertically above point $P_0$ , and in the horizontal triangle $ABP_0$ , $\angle AP_0B = 70^\circ$ .

The angle of elevation of the helicopter is $32^\circ$ from $A$ and $45^\circ$ from $B$ .

Sketch a fully-labelled 3-D diagram showing $A, B, P_0, P$ and all given angles.

[2]

Let $AP_0=x$ m and $BP_0=y$ m. Show that $x\tan32^\circ = y\tan45^\circ.$

[2]

Using the cosine rule in triangle $ABP_0$ , write down an equation relating $x$ and $y$ .

[2]

Use your results to find the height $h$ of the helicopter above the ground, correct to the nearest metre.

[2]

The point $P_0$ later moves horizontally around a circular path of radius $90 \text{ m}$ centred at $A$ . Find the length of the arc, in metres and radians, when its bearing from $A$ changes from $010^\circ$ to $100^\circ$ .

[3]

If tensions are $T_A = k|AP|$ and $T_B = k|BP|$ , find $\dfrac{T_A}{T_B}$ .

[3]

Question 4

SLPaper 2

A cyclist can choose between using two ramps. The equations of the ramps with height $y$ at horizontal distance $x$ are $y = 3x$ and $y = 2x$ .

By finding the angle they make with the ground, find the angle created between the two ramps.

[4]

Explain which ramp the cyclist should take if they are trying to reach a higher level in the shortest horizontal distance.

[1]

Question 5

SLPaper 2

A hot-air balloon $P$ is anchored by two ropes to points $A$ and $B$ on level ground. The distance $AB$ is 75 m. The balloon is vertically above $P_0$ , and $\angle AP_0B = 60^\circ$ in the horizontal triangle $ABP_0$ .

The angles of elevation of the balloon are $28^\circ$ from $A$ and $37^\circ$ from $B$ .

Draw a fully-labelled 3-D diagram showing $A,B,P_0,P$ and all the given angles.

[2]

Let $AP_0=x$ m and $BP_0=y$ m. Show that

x\tan28^\circ = y\tan37^\circ.

[2]

Using the cosine rule in triangle $ABP_0$ , write an equation relating $x$ and $y$ .

[2]

Find the height $h$ of the balloon, to the nearest metre.

[2]

The projection $P_0$ moves horizontally on a circle of radius $75\text{ m}$ centered at $A$ . Find the angle subtended by the arc in radians and the length of the arc in metres when the bearing of $P_0$ from $A$ changes from $045^\circ$ to $115^\circ$ .

[3]

If $T_A = k|AP|$ and $T_B = k|BP|$ , find $\dfrac{T_A}{T_B}$ .

[3]

Question 6

SLPaper 2

For triangle $ABC$ , $\angle A=30^{\circ}$ , $BC=4$ , and $AC=6$ . Find the possible values of angle $B$ and the corresponding lengths of side $AB$ .

[7]

Question 7

SLPaper 2

A satellite dish $D$ is positioned vertically above a point $D_0$ on the ground. Two control posts, $C_1$ and $C_2$ , are 40 m apart. The bearing of $C_2$ from $C_1$ is $090^\circ$ . The bearing of the dish $D_0$ from $C_1$ is $010^\circ$ . At a certain instant, the angle of elevation of the dish from $C_1$ is $60^\circ$ and from $C_2$ is $45^\circ$ .

Draw a fully-labelled 3D diagram showing $C_1,C_2,D_0,D$ and all known angles.

[2]

Let the horizontal distances $C_1 D_0 = x$ m and $C_2 D_0 = y$ m. Show that

x\tan60^\circ = y\tan45^\circ .

[2]

Using the cosine rule in triangle $C_1C_2D_0$ , express $y$ in terms of $x$ and known constants.

[3]

Combine your results from parts 2 and 3 to find the height $h$ of the dish. Give your answer to the nearest tenth of a metre.

[3]

The dish's projection $D_0$ describes a circular path of radius $x$ m. The path subtends an angle of $2.5$ radians. Calculate the arc length in metres.

[3]

Find the ratio of the horizontal distance from $C_2$ to the 3D distance $|C_2 D|$ at this instant.

[3]

Question 8

SLPaper 1

The following diagram shows a quadrilateral $ABCD$ with $AD=5$ , $AB=10$ , and $\angle BAD = 30^\circ$ . $BC=6$ and $\angle BCD = 45^\circ$ .

Find the exact area of the triangle $ABD$ .

[2]

Find the exact length of $BD$ .

[3]

Find the exact value of $\sin \angle CDB$ .

[3]

Question 9

SLPaper 2

A rescue beacon $B$ is positioned vertically above a point $B_0$ on the ground. Two observers, $O_1$ and $O_2$ , are 500 m apart on level ground. The bearing of $O_2$ from $O_1$ is $045^\circ$ . The bearing of the beacon $B_0$ from $O_1$ is $075^\circ$ . At a certain instant, the angle of elevation of the beacon from $O_1$ is $15^\circ$ and from $O_2$ is $25^\circ$ .

Draw a fully-labelled 3-D diagram showing $O_1,O_2,B_0,B$ and all known angles.

[2]

Let the horizontal distances $O_1 B_0 = x$ m and $O_2 B_0 = y$ m. Show that $x\tan15^\circ = y\tan25^\circ .$

[2]

Using the cosine rule in triangle $O_1O_2B_0$ , express $y$ in terms of $x$ and known constants.

[3]

Combine your results from parts 2 and 3 to find the height $h$ of the beacon, given that $O_1B_0 > 500\text{ m}$ . Give your answer to the nearest metre.

[5]

In a different scenario, the beacon's projection $B_0$ moves along a circular path of radius 600 m centered at $O_1$ . Calculate, in metres, the arc length of the path when the bearing from $O_1$ to $B_0$ changes from $075^\circ$ to $135^\circ$ .

[3]

Using the values found in part 4, let the length of the tether cable attached at $O_1$ be $L_1 = |O_1B|$ and the length of the cable attached at $O_2$ be $L_2 = |O_2B|$ . Find the ratio $\dfrac{L_1}{L_2}$ at that instant.

[3]

Question 10

SLPaper 2

Given the triangular farm $ABC$ with $AB=12$ , $AC=20$ and $\angle BAC=35^{\circ}$ . Now there lies a point $X$ on $AC$ such that $AX:XC$ is $1:4$ as shown in the figure. The fencing has to be installed around its perimeter.

Find the total length of the fencing required.

[3]

Now $B$ connects to $X$ making a new farm. Calculate the area of triangle $BXC$ .

[4]

SL 3.2—2d and 3d trig, sine rule, cosine rule, area Questionbank