Right and non-right angled trigonometry, sine and cosine rules, bearings

1StandardCalculator5 marks

In triangle $ABC$, $BC=7.2$ cm, $AC=9.5$ cm and angle $ACB=64^\circ$.

(a) Find the length of $AB$. [3]

(b) Find the area of triangle $ABC$. [2]

Substitutes into the cosine rule: $AB^2=7.2^2+9.5^2-2(7.2)(9.5)\cos64^\circ$M1
$AB^2=82.1$ (or better)A1
$AB=9.06$ cm (accept $9.062$)A1
Uses $\tfrac12ab\sin C$: $\tfrac12(7.2)(9.5)\sin64^\circ$M1
$30.7$ cm$^2$A1
Answer: (a) $9.06$ cm (b) $30.7$ cm$^2$
Examiner tip: The cosine rule needs the angle between the two known sides; here that is $C$, opposite $AB$.

2HardCalculator6 marks

Triangle $ABC$ has $BC=6$ cm, $AC=9$ cm and angle $BAC=35^\circ$. It is possible to draw two different triangles satisfying these conditions.

(a) Find the two possible sizes of angle $ABC$. [4]

(b) Find the smaller of the two possible lengths of $AB$. [2]

Sine rule: $\dfrac{\sin B}{9}=\dfrac{\sin35^\circ}{6}$M1
$\sin B=0.8604$ (accept $0.860$)A1
$B=59.4^\circ$ (accept $59.36^\circ$)A1
$B=120.6^\circ$ (obtuse, from $180^\circ-59.4^\circ$; accept $120.64^\circ$ or $121^\circ$ (3 s.f.))A1
Uses the larger $B$: $C=180-35-120.6=24.4^\circ$ (allow unrounded $C=24.36^\circ$), then $\dfrac{AB}{\sin C}=\dfrac{6}{\sin35^\circ}$ (or cosine rule: $6^2=AB^2+9^2-2(9)(AB)\cos35^\circ$, solving the quadratic and choosing the smaller root)M1
$AB=4.31$ cm (allow $4.32$ from $C=24.4^\circ$; do not accept $4.25$ from prematurely rounded $C=24^\circ$)A1
Answer: (a) $B=59.4^\circ$ or $120.6^\circ$ (b) $4.31$ cm
Examiner tip: In the ambiguous case always ask whether $180^\circ-B$ also fits; the smaller $AB$ comes from the obtuse $B$.

3CoreCalculator4 marks

The diagram shows Mina standing on level ground, $42$ m from the base $B$ of a vertical tower $BT$. Her eye, $E$, is $1.6$ m above the ground. The angle of elevation of the top of the tower, $T$, from her eye is $31^\circ$.

(a) Write down the size of the angle of depression of $E$ from $T$. [1]

(b) Find the height of the tower. [3]

(a) $31^\circ$ (alternate angles with the horizontal)A1
(b) Uses tan in the right-angled triangle above eye level: $\tan31^\circ=\dfrac{h}{42}$ or $h=42\tan31^\circ$M1
$h=25.2$ m (accept $25.24$) for the height above eye levelA1
Height of tower $=25.24+1.6=26.8$ m (accept $26.84$; award only if $1.6$ m is added)A1
Answer: (a) $31^\circ$ (b) $26.8$ m
Examiner tip: Most lost marks come from forgetting to add the $1.6$ m eye height: the right-angled triangle starts at eye level, not at the ground.

Stuck on Right and non-right angled trigonometry, sine and cosine rules, bearings?

Get it sorted in a free 30-minute session

We work through your weakest questions on this topic on Zoom. I reply the same day.

Free 30-min session on this topic Send me 5 questions a day

4CoreNon-calculator5 marks

In triangle $PQR$, $PQ=5$ cm, $PR=8$ cm and angle $QPR=60^\circ$.

(a) Find the length of $QR$. [3]

(b) Find the exact area of triangle $PQR$. [2]

(a) Correct substitution into the cosine rule: $QR^2=5^2+8^2-2(5)(8)\cos60^\circ$M1
Uses $\cos60^\circ=\tfrac12$: $QR^2=49$A1
$QR=7$ cmA1
(b) Correct substitution into $\tfrac12ab\sin C$: $\tfrac12(5)(8)\sin60^\circ$M1
$10\sqrt3$ cm$^2$ (accept $\dfrac{40\sqrt3}{4}$; do not accept $17.3$ as the question asks for an exact value)A1
Answer: (a) $7$ cm (b) $10\sqrt3$ cm$^2$
Examiner tip: On Paper 1 you must know $\sin60^\circ=\tfrac{\sqrt3}{2}$ and $\cos60^\circ=\tfrac12$; a decimal area loses the final mark when an exact answer is asked for.

5StandardCalculator7 marks

A boat sails from port $P$ on a bearing of $052^\circ$ for $14$ km to a buoy $Q$. It then sails on a bearing of $140^\circ$ for $9$ km to a point $R$, as shown in the diagram.

(a) Show that angle $PQR=92^\circ$. [2]

(b) Find the distance $PR$. [2]

(a) Uses parallel north lines: the bearing of $P$ from $Q$ is $052^\circ+180^\circ=232^\circ$ (or angle between $QP$ and south is $52^\circ$ and between south and $QR$ is $40^\circ$)M1
$232^\circ-140^\circ=92^\circ$ (or $52^\circ+40^\circ=92^\circ$) AG; a clear angle argument must be seenA1
(b) $PR^2=14^2+9^2-2(14)(9)\cos92^\circ$M1
$PR=16.9$ km (accept $16.91$)A1
(c) Sine rule (or cosine rule) for angle $QPR$: $\dfrac{\sin QPR}{9}=\dfrac{\sin92^\circ}{16.9}$M1
Angle $QPR=32.1^\circ$ (accept $32.2^\circ$ from rounded $PR$)A1
Bearing $=052^\circ+32.1^\circ=084^\circ$ (accept $084.1^\circ$ or $84.1^\circ$; follow through their angle $QPR$)A1
Answer: (b) $16.9$ km (c) $084^\circ$
Examiner tip: Bearings are measured clockwise from north; in (c) you must add angle $QPR$ to $052^\circ$, not give angle $QPR$ itself as the bearing.

6StandardCalculator5 marks

Point $C$ is at the top of a vertical cliff $CF$ of height $85$ m. Two boats, $X$ and $Y$, are in a straight line with the foot of the cliff $F$, on the same side of it. The angles of depression of $X$ and $Y$ from $C$ are $38^\circ$ and $23^\circ$ respectively.

(a) Find the distance $FX$. [2]

(b) Hence find the distance between the two boats. [3]

(a) Angle $CXF=38^\circ$ used with tan: $\tan38^\circ=\dfrac{85}{FX}$M1
$FX=108.8$ m (accept $109$ m)A1
(b) $\tan23^\circ=\dfrac{85}{FY}$M1
$FY=200.2$ m (accept $200$ m)A1
$XY=200.25-108.80=91.5$ m (accept $91.4$ from $200.2-108.8$; accept answers from the sine rule in triangle $CXY$)A1
Answer: (a) $109$ m ($108.8$) (b) $91.5$ m
Examiner tip: The angle of depression is measured down from the horizontal, so it equals the angle of elevation at the boat; do not use $90^\circ-38^\circ$ with tan the wrong way up.

7HardCalculator6 marks

Triangle $ABC$ has $AB=11$ cm, $AC=7$ cm and area $30$ cm$^2$.

(a) Find the two possible values of angle $BAC$. [4]

(b) Given that $BC$ is the longest side of the triangle, find the length of $BC$. [2]

(a) $\tfrac12(11)(7)\sin A=30$M1
$\sin A=\dfrac{60}{77}$ $(=0.779)$A1
$A=51.2^\circ$A1
$A=128.8^\circ$ (from $180^\circ-51.2^\circ$; accept $129^\circ$)A1
(b) Chooses the obtuse angle in the cosine rule: $BC^2=11^2+7^2-2(11)(7)\cos128.8^\circ$M1
$BC=16.3$ cm (accept $16.33$; the acute angle gives $8.57$, which is not the longest side, M0A0)A1
Answer: (a) $51.2^\circ$ or $128.8^\circ$ (b) $16.3$ cm
Examiner tip: $\sin A$ has two solutions between $0^\circ$ and $180^\circ$; students who stop at $51.2^\circ$ lose a mark in (a) and choose the wrong triangle in (b).

8StandardCalculator8 marks

The diagram shows a field $ABCD$. $AB=120$ m, $BC=95$ m, $AD=70$ m, angle $ABC=104^\circ$ and angle $CAD=48^\circ$.

(a) Find the length of the diagonal $AC$. [2]

(b) Find the area of the field. [4]

(a) $AC^2=120^2+95^2-2(120)(95)\cos104^\circ$M1
$AC=170$ m (accept $170.1$)A1
(b) Area of triangle $ABC=\tfrac12(120)(95)\sin104^\circ$M1
$=5531$ m$^2$ (accept $5530$)A1
Area of triangle $ACD=\tfrac12(170.1)(70)\sin48^\circ$ $(=4425)$ (follow through their $AC$)M1
Total area $=9960$ m$^2$ (accept $9955$ to $9960$; $9952$ from $AC=170$ is acceptable)A1
(c) $CD^2=170.1^2+70^2-2(170.1)(70)\cos48^\circ$ (follow through their $AC$)M1
$CD=134$ m (accept $133.8$)A1
Answer: (a) $170$ m (b) $9960$ m$^2$ (c) $134$ m
Examiner tip: Keep $AC$ unrounded in your calculator for parts (b) and (c); using $AC=170$ is fine, but rounding to $2$ s.f. earlier loses accuracy marks.

9HardCalculator7 marks

In triangle $PQR$, angle $QPR=40^\circ$, $QR=12$ cm and $PR=9$ cm.

(a) Find angle $PQR$. [2]

(b) A student says that there is a second possible value for angle $PQR$. Explain why the student is wrong. [2]

The length of $QR$ is now changed to $k$ cm, with angle $QPR=40^\circ$ and $PR=9$ cm unchanged.

(a) $\dfrac{\sin Q}{9}=\dfrac{\sin40^\circ}{12}$M1
$Q=28.8^\circ$A1
(b) The other solution would be $180^\circ-28.8^\circ=151.2^\circ$, and $151.2^\circ+40^\circ=191.2^\circ$R1
which exceeds $180^\circ$, so it cannot be an angle of the triangle (accept: $QR\gt PR$ so the angle opposite $PR$ must be smaller than $40^\circ$, hence acute)R1
(c) Recognises the boundary as the perpendicular height from $R$: $9\sin40^\circ$M1
$9\sin40^\circ=5.79$A1
$5.79\lt k\lt 9$ (strict inequalities required for A1; accept $9\sin40^\circ\lt k\lt 9$)A1
Answer: (a) $28.8^\circ$ (b) $151.2^\circ+40^\circ\gt 180^\circ$ (c) $5.79\lt k\lt 9$
Examiner tip: Two triangles need the side opposite the given angle to be longer than the height ($9\sin40^\circ$) but shorter than the adjacent given side ($9$).

10StretchCalculator8 marks

Lighthouse $B$ is $18$ km due east of lighthouse $A$. A ship $S$ is on a bearing of $038^\circ$ from $A$ and on a bearing of $323^\circ$ from $B$.

(a) Show that angle $ASB=75^\circ$. [2]

(b) Find the distance $AS$. [2]

The ship then sails due north at a constant speed of $15$ km h$^{-1}$.

(a) Angle $SAB=90^\circ-38^\circ=52^\circ$ and angle $SBA=323^\circ-270^\circ=53^\circ$M1
$180^\circ-52^\circ-53^\circ=75^\circ$ AGA1
(b) $\dfrac{AS}{\sin53^\circ}=\dfrac{18}{\sin75^\circ}$M1
$AS=14.9$ km (accept $14.88$)A1
(c) Angle at $S$ between $SA$ and the ship's northward path $=180^\circ-38^\circ=142^\circ$B1
Cosine rule with distance travelled $d$: $25^2=14.88^2+d^2-2(14.88)d\cos142^\circ$ (or equivalent quadratic, e.g. $d^2+23.45d-403.5=0$)M1
$d=11.5$ km (accept $11.53$; reject the negative root)A1
Time $=\dfrac{11.53}{15}\times60=46.1$ minutes (accept $46.0$ to $46.2$)A1
Answer: (b) $14.9$ km (c) $46.1$ minutes
Examiner tip: Draw the north line through $S$: the angle you need for the cosine rule is $142^\circ$ (co-interior with $38^\circ$), not $38^\circ$.

Stuck on this topic?

Book a free 30-minute session on WhatsApp and we'll find exactly which step is costing you the marks. Or build a timed paper on it.

Free 30-minute session