Worked example · Easy
Find a hypotenuse
A right-angled triangle has shorter sides 9 cm and 12 cm. Find its longest side.
\[c^2=9^2+12^2=81+144=225\]\[c=\sqrt{225}=\boxed{15\text{ cm}}\]
Chapter 41
Pythagoras’ theorem and trigonometry
From right-angled triangles to three-dimensional geometry: master Pythagoras, trigonometric ratios, obtuse angles, the sine and cosine rules, and angles between lines and planes.
By the end of this chapter, you should be able to
- calculate missing sides using Pythagoras’ theorem;
- use sine, cosine and tangent in right-angled triangles to find sides and angles;
- solve two-dimensional applications, including bearings and angles of elevation and depression;
- use sine and cosine for obtuse angles and solve simple trigonometric equations in a given interval;
- use sine rule, cosine rule and the formula \(\tfrac12ab\sin C\) in any triangle;
- find lengths and angles within three-dimensional figures;
- calculate angles between a line and a plane using perpendicular projections.
1
Subtopic 1
Pythagoras’ theorem
In a right-angled triangle, the longest side is the hypotenuse, opposite the right angle. Pythagoras’ theorem states that the square of the hypotenuse equals the sum of the squares of the other two sides.
\[\boxed{c^2=a^2+b^2}\qquad\boxed{c=\sqrt{a^2+b^2}}\]
Here \(c\) is the hypotenuse; \(a\) and \(b\) are the shorter sides. To find a shorter side, subtract the other shorter side squared from the hypotenuse squared.
Worked example · Medium
Find a shorter side
The hypotenuse is 17 m and one shorter side is 8 m. Find the remaining side.
\[x^2=17^2-8^2=289-64=225\]\[x=\sqrt{225}=\boxed{15\text{ m}}\]
Worked example · Exam
A ladder against a wall
A 6.5 m ladder leans against a vertical wall. Its foot is 2.5 m from the wall. How high up the wall does it reach?
The wall and ground meet at 90°, so the ladder is the hypotenuse.
\[h^2=6.5^2-2.5^2=42.25-6.25=36\]\[h=\boxed{6\text{ m}}\]Problem-solving method: sketch the right-angled triangle, label the right angle and known sides, decide whether to add or subtract squares, then take the positive square root. A hypotenuse must always be the longest side. In geometry problems, the perpendicular from a point to a line is its shortest distance to that line.
2
Subtopic 2
Trigonometry
Trigonometry links the acute angles in a right-angled triangle to ratios of its side lengths. Choose an acute angle \(\theta\), then label the sides relative to that angle: H is the hypotenuse, O is opposite \(\theta\), and A is adjacent to \(\theta\) (but not the hypotenuse).
\[\boxed{\sin\theta=\frac{O}{H}}\qquad\boxed{\cos\theta=\frac{A}{H}}\qquad\boxed{\tan\theta=\frac{O}{A}}\]
Use the calculator in degree mode for all questions in this chapter. Side ratios have no units.
Choosing the method: identify the two relevant sides and select the ratio containing them. To find a side in the numerator, multiply; to find one in the denominator, divide. To find an angle, use an inverse function: \(\sin^{-1}\), \(\cos^{-1}\), or \(\tan^{-1}\).
Worked example · Easy
Finding an opposite side
A right triangle has hypotenuse 14 cm and an acute angle of 30°. Find the opposite side.
\[\sin30^\circ=\frac{x}{14}\]\[x=14\sin30^\circ=\boxed{7\text{ cm}}\]
Worked example · Medium
Finding a hypotenuse
A right triangle has adjacent side 12 cm to a 40° angle. Find its hypotenuse.
\[\cos40^\circ=\frac{12}{h}\]\[h=\frac{12}{\cos40^\circ}=\boxed{15.7\text{ cm}}\ \text{(3 s.f.)}\]
Worked example · Medium
Finding an angle with an inverse ratio
The opposite and adjacent sides to angle \(\theta\) are 9 cm and 12 cm. Find \(\theta\).
\[\tan\theta=\frac9{12}=0.75\]\[\theta=\tan^{-1}(0.75)=\boxed{36.9^\circ}\ \text{(3 s.f.)}\]
Isosceles triangles: if two sides are equal, draw the line of symmetry from the top vertex to the midpoint of the base. This forms two right triangles, allowing Pythagoras or SOH–CAH–TOA to be used.
Worked example · Medium
An isosceles triangle
An isosceles triangle has equal sides of 10 cm and a base of 12 cm. Find its vertical height.
The altitude bisects the 12 cm base into two 6 cm sections.
\[h^2=10^2-6^2=64\]\[h=\boxed{8\text{ cm}}\]Angles of elevation and depression: the elevation angle is measured upwards from a horizontal line, and the depression angle downwards. Alternate angles between parallel horizontal lines can help transfer a depression angle to a ground-level triangle.
Worked example · Exam
Angle of elevation with eye height
A student whose eyes are 1.6 m above the ground observes the top of a 19.6 m building at an angle of elevation of 35°. Find the horizontal distance to the building.
The vertical rise from eye level is \(19.6-1.6=18\) m.
\[\tan35^\circ=\frac{18}{d}\]\[d=\frac{18}{\tan35^\circ}=\boxed{25.7\text{ m}}\ \text{(3 s.f.)}\]Bearings: measure clockwise from North and write them as three digits. A journey on bearing \(\beta\) makes an angle \(\beta\) from North; for an acute bearing, northward displacement is \(d\cos\beta\) and eastward displacement is \(d\sin\beta\). For other quadrants, draw a compass sketch and select the correct signs/directions.
Worked example · Exam
Ship displacement from a bearing
A vessel sails 120 km on a bearing of 045°. Find how far north and east it travels.
\[N=120\cos45^\circ=\boxed{84.9\text{ km}}\]\[E=120\sin45^\circ=\boxed{84.9\text{ km}}\]
Both displacements are positive on this bearing.
3
Subtopic 3
The sine and cosine functions for obtuse angles
The definitions of sine and cosine extend beyond acute angles. On a circle of radius 1 centred at the origin, the point reached by an angle \(\theta\) measured anticlockwise from the positive x-axis has coordinates \((\cos\theta,\sin\theta)\).
\[\boxed{\sin(180^\circ-\theta)=\sin\theta}\qquad\boxed{\cos(180^\circ-\theta)=-\cos\theta}\]
For \(90^\circ<\theta<180^\circ\), the reference angle \(180^\circ-\theta\) is acute. Tangent is negative in this quadrant, but the textbook section focuses on sine and cosine.
Worked example · Easy
Sine and cosine of an obtuse angle
Find the exact values of \(\sin150^\circ\) and \(\cos150^\circ\).
\[180^\circ-150^\circ=30^\circ\]\[\sin150^\circ=\sin30^\circ=\boxed{\frac12}\]\[\cos150^\circ=-\cos30^\circ=\boxed{-\frac{\sqrt3}{2}}\]
Worked example · Medium
Two solutions of a sine equation
Solve \(\sin x=0.6\) for \(0^\circ\le x\le180^\circ\).
\[x_1=\sin^{-1}(0.6)=36.9^\circ\]\[x_2=180^\circ-36.9^\circ=143.1^\circ\]\[\boxed{x=36.9^\circ\ \text{or}\ 143.1^\circ}\]
Worked example · Exam
Negative cosine in the stated interval
Solve \(\cos x=-0.4\), where \(0^\circ\le x\le180^\circ\).
Cosine is negative in the second quadrant.
\[x=\cos^{-1}(-0.4)=\boxed{113.6^\circ}\]This is the only solution in the stated interval.
Remember: when solving \(\sin x=k\) for a range up to 180°, check both an acute solution and its supplementary angle if \(0
4
Subtopic 4
Non-right-angled triangles
For a triangle \(ABC\), the angles use capital letters \(A,B,C\), and the lengths opposite them are \(a,b,c\). The three angles sum to 180°. Right-triangle trigonometric ratios by themselves are not enough when there is no 90° angle.
Sine rule
Use the sine rule when an angle and its opposite side are known, plus one other side or angle. Write the version that makes the unknown easier to isolate.
\[\boxed{\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}}\quad\text{or}\quad\boxed{\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}}\]
Worked example · Easy
Find a side using the sine rule
In triangle ABC, \(A=40^\circ\), \(B=75^\circ\) and \(a=12\) cm. Find \(b\).
\[\frac b{\sin75^\circ}=\frac{12}{\sin40^\circ}\]\[b=\frac{12\sin75^\circ}{\sin40^\circ}=\boxed{18.0\text{ cm}}\ \text{(3 s.f.)}\]
Worked example · Medium
Find an angle using the sine rule
In triangle ABC, \(A=42^\circ\), \(a=9\) cm and \(b=7\) cm. Find the acute angle \(B\).
\[\frac{\sin B}{7}=\frac{\sin42^\circ}{9}\]\[\sin B=\frac{7\sin42^\circ}{9}\approx0.5204\]\[B=\sin^{-1}(0.5204)=\boxed{31.4^\circ}\ \text{(3 s.f.)}\]
The ambiguous case of the sine rule
If two sides and an angle not between them are given (SSA), there may be two possible triangles, one, or none. When an inverse sine gives an acute angle \(B\), also test \(180^\circ-B\). Both are valid only if the remaining triangle angle is positive and the supplied lengths are consistent.
Worked example · Exam
Investigate both possible triangles
In triangle ABC, \(A=30^\circ\), \(a=6\) cm and \(b=10\) cm. Find the possible values of angle \(B\).
\[\sin B=\frac{10\sin30^\circ}{6}=\frac56\]\[B_1=\sin^{-1}(5/6)\approx56.4^\circ\]\[B_2=180^\circ-56.4^\circ\approx123.6^\circ\]
Both are allowed: \(30+56.4<180\) and \(30+123.6<180\).
\[\boxed{B=56.4^\circ\ \text{or}\ 123.6^\circ}\]Cosine rule
Use the cosine rule for two sides and the included angle (SAS), or for three known sides (SSS). It also works for obtuse angles.
\[\boxed{a^2=b^2+c^2-2bc\cos A}\qquad\boxed{\cos A=\frac{b^2+c^2-a^2}{2bc}}\]
The matching angle is always opposite the matching lower-case side. For an obtuse included angle, \(\cos A\) is negative.
Worked example · Medium
Find a side using the cosine rule
Two sides of a triangle are 7 cm and 9 cm, with an included angle of 60°. Find the third side.
\[a^2=7^2+9^2-2(7)(9)\cos60^\circ\]\[a^2=49+81-63=67\]\[a=\sqrt{67}=\boxed{8.19\text{ cm}}\ \text{(3 s.f.)}\]
Worked example · Exam
Find an angle from three sides
The sides of a triangle are 7 cm, 8 cm and 9 cm. Find the angle opposite the 9 cm side.
\[\cos C=\frac{7^2+8^2-9^2}{2(7)(8)}=\frac{32}{112}\]\[C=\cos^{-1}\left(\frac{2}{7}\right)=\boxed{73.4^\circ}\ \text{(3 s.f.)}\]
The general formula for the area of any triangle
If you know two sides and their included angle, calculate the area directly without finding a perpendicular height.
\[\boxed{\text{Area}=\frac12 ab\sin C=\frac12 bc\sin A=\frac12 ac\sin B}\]
An included angle of 120° or 150° is fine: its sine is positive. Write area in square units.
Worked example · Medium
Area from two sides and the included angle
Find the area of a triangle with sides 8 cm and 11 cm, enclosing an angle of 120°.
\[\text{Area}=\frac12(8)(11)\sin120^\circ\]\[=44\sin120^\circ=\boxed{38.1\text{ cm}^2}\ \text{(3 s.f.)}\]
Selection checklist: right angle? → Pythagoras or SOH–CAH–TOA; opposite side–angle pair? → sine rule; SAS or SSS? → cosine rule; two sides with included angle and area required? → \(\tfrac12ab\sin C\).
5
Subtopic 5
Finding lengths and angles in three dimensions
Many three-dimensional problems are solved by identifying a two-dimensional right-angled triangle inside the solid. The useful triangle may lie on the base, on a face, or diagonally through the solid. Draw that triangle separately, label known lengths, and solve one stage at a time.
\[\boxed{d=\sqrt{l^2+w^2+h^2}}\]
For a cuboid only: \(l,w,h\) are three mutually perpendicular edges; \(d\) is its space diagonal. This formula is Pythagoras applied twice.
Worked example · Easy
Space diagonal of a cuboid
A rectangular box is 3 cm wide, 4 cm deep and 12 cm high. Find its longest diagonal.
\[d=\sqrt{3^2+4^2+12^2}=\sqrt{9+16+144}\]\[d=\sqrt{169}=\boxed{13\text{ cm}}\]
Worked example · Medium
Angle made by the space diagonal
For the same 3 cm × 4 cm × 12 cm cuboid, find the angle between the space diagonal and the base.
The projection of the diagonal on the base is its 3 cm × 4 cm diagonal.
\[b=\sqrt{3^2+4^2}=5\text{ cm}\]\[\tan\theta=\frac{12}{5}\]\[\theta=\tan^{-1}(12/5)=\boxed{67.4^\circ}\]Worked example · Exam
Height and angle in a 3-D setting
A tree stands 40 m north of a point P, and a second observation point Q is 30 m east of P. Its top is 20 m above the ground. Find the angle of elevation of the top from Q.
First find the horizontal distance from Q to the foot of the tree:
\[r=\sqrt{40^2+30^2}=50\text{ m}\]Then use the right triangle containing the tree height:
\[\tan\theta=\frac{20}{50}=0.4\]\[\theta=\tan^{-1}(0.4)=\boxed{21.8^\circ}\]Accuracy: retain exact squares and roots in intermediate steps (for example, use \(\sqrt{45}\), not 6.71) and round only at the end. When points lie on horizontal ground, first use their ground distances, then combine with a vertical height.
6
Subtopic 6
The angle between a line and a plane
The angle between a line and a plane is the acute angle between the line and its perpendicular projection onto that plane. To identify the right triangle, drop a perpendicular from the raised endpoint to the plane. The projection is the horizontal/planar side; the height is perpendicular to the plane.
When the plane is horizontal, the familiar angle of elevation is the angle between the line of sight and the plane. For a vertical face, project the line onto that face instead: the correct projection may differ from the one on the base.
Worked example · Easy
Diagonal of a cuboid above its base
A cuboid measures 8 cm by 6 cm on its base and has height 5 cm. Find the angle between its space diagonal and its base.
First calculate the diagonal of the base.
\[p=\sqrt{8^2+6^2}=10\text{ cm}\]\[\tan\theta=\frac{5}{10}\]\[\theta=\tan^{-1}(0.5)=\boxed{26.6^\circ}\]Worked example · Medium
An angle with a vertical face
A cuboid has width 8 cm, depth 6 cm and height 5 cm. Find the angle between its space diagonal and the vertical face of width 8 cm and height 5 cm.
The perpendicular distance from the far corner to that face is the depth, 6 cm. The projection of the diagonal onto the face is
\[p=\sqrt{8^2+5^2}=\sqrt{89}\]\[\tan\theta=\frac{6}{\sqrt{89}}\]\[\theta=\tan^{-1}\left(\frac6{\sqrt{89}}\right)=\boxed{32.5^\circ}\]Worked example · Exam
Sloping edge of a pyramid
A right pyramid has a rectangular base measuring 6 cm × 8 cm. Its apex is vertically above the centre of the base, 12 cm above it. Find the angle between an edge from the apex to a base corner and the base.
The projection of the sloping edge goes from the corner to the centre of the rectangle. The centre-to-corner ground distance is
\[r=\sqrt{3^2+4^2}=5\text{ cm}\]\[\tan\theta=\frac{12}{5}\]\[\theta=\tan^{-1}(12/5)=\boxed{67.4^\circ}\]Final check: clearly name the required plane, mark the foot of the perpendicular, and find the angle between the line and its projection—not between the line and an arbitrary edge of the solid.
Build confidence → apply skills → exam-style reasoning
Chapter 41 practice
Try each question before opening Show Answer. Solutions include the method, substitution and final result. Questions here are newly written in the style of the chapter, not reproductions of past papers.
A. Pythagoras and basic trigonometry · Confidence builders
1Find the longest side[3 marks]
A right-angled triangle has sides 5 cm and 12 cm meeting at 90°. Find the hypotenuse.
Show Answer
\[c^2=5^2+12^2=25+144=169\]\[c=\sqrt{169}=\boxed{13\text{ cm}}\]
2Find a shorter side[3 marks]
The hypotenuse of a right triangle is 10 cm and one shorter side is 6 cm. Find the other shorter side.
Show Answer
\[a^2=10^2-6^2=100-36=64\]\[a=\sqrt{64}=\boxed{8\text{ cm}}\]
3Recognise the opposite side[3 marks]
Relative to an acute angle, the opposite side is 6 cm and the hypotenuse is 10 cm. Write the sine ratio and find the angle.
Show Answer
\[\sin\theta=\frac6{10}=0.6\]\[\theta=\sin^{-1}(0.6)=\boxed{36.9^\circ}\]
4Calculate an opposite side[3 marks]
A triangle has a right angle, hypotenuse 18 cm and acute angle 25°. Find the opposite side.
Show Answer
\[\sin25^\circ=\frac{x}{18}\]\[x=18\sin25^\circ=\boxed{7.61\text{ cm}}\]
5Calculate an adjacent side[3 marks]
A right triangle has hypotenuse 20 cm and an acute angle of 60°. Find the adjacent side.
Show Answer
\[\cos60^\circ=\frac{x}{20}\]\[x=20\cos60^\circ=\boxed{10\text{ cm}}\]
6Find a missing angle[3 marks]
An opposite side is 5 cm and adjacent side is 12 cm. Find the acute angle.
Show Answer
\[\tan\theta=\frac5{12}\]\[\theta=\tan^{-1}(5/12)=\boxed{22.6^\circ}\]
7Solve an obtuse-angle identity[2 marks]
Without a calculator, given \(\sin35^\circ=0.574\), find \(\sin145^\circ\).
Show Answer
\[145^\circ=180^\circ-35^\circ\]\[\sin145^\circ=\sin35^\circ=\boxed{0.574}\]
8Negative cosine[2 marks]
Given \(\cos50^\circ=0.643\), write down \(\cos130^\circ\).
Show Answer
\[\cos130^\circ=-\cos50^\circ=\boxed{-0.643}\]
B. Applying ratios, angles and triangle rules
9A diagonal footpath[4 marks]
A rectangular plot is 24 m long and 18 m wide. Find the length of a straight diagonal footpath.
Show Answer
\[d=\sqrt{24^2+18^2}=\sqrt{576+324}\]\[d=\sqrt{900}=\boxed{30\text{ m}}\]
10Distance from a wall[4 marks]
A ladder 8 m long rests against a wall. Its foot is 1.8 m from the wall. Find the height reached, to 3 significant figures.
Show Answer
\[h^2=8^2-1.8^2=64-3.24=60.76\]\[h=\sqrt{60.76}=\boxed{7.79\text{ m}}\]
11Tangent in a slope[4 marks]
A ramp rises 1.5 m over a horizontal run of 8 m. Find its angle of inclination.
Show Answer
\[\tan\theta=\frac{1.5}{8}=0.1875\]\[\theta=\tan^{-1}(0.1875)=\boxed{10.6^\circ}\]
12Distance with a depression angle[4 marks]
A lookout is 30 m above sea level. The angle of depression to a boat is 20°. Find the boat’s horizontal distance from the foot of the lookout.
Show Answer
The boat sees the same 20° angle of elevation.
\[\tan20^\circ=\frac{30}{d}\]\[d=\frac{30}{\tan20^\circ}=\boxed{82.4\text{ m}}\]13A bearing and displacement[4 marks]
A walker travels 15 km on bearing 060°. Find the distance travelled east and north.
Show Answer
\[E=15\sin60^\circ=\boxed{13.0\text{ km}}\]\[N=15\cos60^\circ=\boxed{7.50\text{ km}}\]
14Sine rule: find a length[4 marks]
A triangle has \(A=50^\circ\), \(B=70^\circ\) and \(a=10\) cm. Find \(b\).
Show Answer
\[\frac{b}{\sin70^\circ}=\frac{10}{\sin50^\circ}\]\[b=\frac{10\sin70^\circ}{\sin50^\circ}=\boxed{12.3\text{ cm}}\]
15Third angle and a side[5 marks]
Triangle ABC has \(A=35^\circ\), \(B=95^\circ\) and \(a=8\) cm. Find \(C\) and \(c\).
Show Answer
\[C=180^\circ-35^\circ-95^\circ=\boxed{50^\circ}\]\[\frac c{\sin50^\circ}=\frac8{\sin35^\circ}\]\[c=\frac{8\sin50^\circ}{\sin35^\circ}=\boxed{10.7\text{ cm}}\]
16Cosine rule: find a side[4 marks]
Two sides of a triangle are 6 cm and 10 cm with included angle 45°. Find the third side.
Show Answer
\[c^2=6^2+10^2-2(6)(10)\cos45^\circ\]\[c=\sqrt{136-120\cos45^\circ}=\boxed{7.15\text{ cm}}\]
17Cosine rule: find an angle[5 marks]
The three sides of a triangle measure 5 cm, 7 cm and 8 cm. Find the angle opposite the 8 cm side.
Show Answer
\[\cos C=\frac{5^2+7^2-8^2}{2(5)(7)}=\frac{10}{70}=\frac17\]\[C=\cos^{-1}(1/7)=\boxed{81.8^\circ}\]
18Area without a height[4 marks]
Two sides of a triangle measure 9 cm and 12 cm, and the included angle is 50°. Find its area.
Show Answer
\[A=\frac12(9)(12)\sin50^\circ\]\[A=54\sin50^\circ=\boxed{41.4\text{ cm}^2}\]
19Two sine solutions[4 marks]
Solve \(\sin x=0.5\) for \(0^\circ\le x\le180^\circ\).
Show Answer
\[x_1=\sin^{-1}(0.5)=30^\circ\]\[x_2=180^\circ-30^\circ=150^\circ\]\[\boxed{x=30^\circ\ \text{or}\ 150^\circ}\]
C. Three-dimensional skills
20Cuboid space diagonal[4 marks]
A cuboid measures 9 cm by 12 cm by 8 cm. Calculate its space diagonal.
Show Answer
\[d=\sqrt{9^2+12^2+8^2}=\sqrt{81+144+64}\]\[d=\sqrt{289}=\boxed{17\text{ cm}}\]
21Find a 3-D angle[4 marks]
A cuboid is 6 cm by 8 cm on its base and 10 cm high. Find the angle between its space diagonal and the base.
Show Answer
\[p=\sqrt{6^2+8^2}=10\]\[\tan\theta=\frac{10}{10}=1\]\[\theta=\boxed{45^\circ}\]
22Projection of a line[5 marks]
A point is 9 m vertically above a level ground point O. A point P is 12 m horizontally from O. Find the distance from P to the raised point and the angle of elevation.
Show Answer
\[d=\sqrt{9^2+12^2}=\boxed{15\text{ m}}\]\[\theta=\tan^{-1}\left(\frac9{12}\right)=\boxed{36.9^\circ}\]
23Angle with a different face[5 marks]
A cuboid is 5 cm wide, 12 cm deep and 4 cm high. Find the angle between its space diagonal and the vertical face that is 12 cm by 4 cm.
Show Answer
The line projects onto that face to make diagonal \(p\).
\[p=\sqrt{12^2+4^2}=\sqrt{160}\]\[\tan\theta=\frac5{\sqrt{160}}\]\[\theta=\boxed{21.6^\circ}\]D. Exam-style multi-step challenges
24A two-stage tower problem[6 marks]
A tower is 36 m high. From a point on level ground its top has an angle of elevation of 28°. Find (i) the horizontal distance to the tower and (ii) the straight-line distance to the top.
Show Answer
(i) Use tangent with opposite 36 m:
\[d=\frac{36}{\tan28^\circ}=\boxed{67.7\text{ m}}\](ii) Use sine, avoiding rounding the intermediate value:
\[L=\frac{36}{\sin28^\circ}=\boxed{76.7\text{ m}}\]25Isosceles geometry[6 marks]
An isosceles triangle has equal sides 13 cm and base 10 cm. Find (i) its perpendicular height and (ii) its apex angle.
Show Answer
Symmetry divides the base into two 5 cm halves.
\[h=\sqrt{13^2-5^2}=\sqrt{144}=\boxed{12\text{ cm}}\]The half-apex angle \(\alpha\) has opposite 5, hypotenuse 13.
\[\alpha=\sin^{-1}(5/13)\approx22.62^\circ\]\[\text{Apex angle}=2\alpha=\boxed{45.2^\circ}\]26Ambiguous sine rule[7 marks]
For triangle ABC, \(A=35^\circ\), \(a=8\) cm and \(b=10\) cm. Determine both possible values of \(B\), and explain why both are valid.
Show Answer
\[\sin B=\frac{10\sin35^\circ}{8}\approx0.71697\]\[B_1\approx45.8^\circ,\qquad B_2=180^\circ-B_1\approx134.2^\circ\]
Both are valid because adding \(A=35^\circ\) leaves a positive third angle in each case: about 99.2° or 10.8°.
\[\boxed{B=45.8^\circ\ \text{or}\ 134.2^\circ}\]27Triangle survey problem[7 marks]
Two survey points A and B are 80 m apart. A landmark C is observed with \(\angle BAC=42^\circ\) and \(\angle ABC=68^\circ\). Find AC and the area of triangle ABC.
Show Answer
\[C=180^\circ-42^\circ-68^\circ=70^\circ\]
Use sine rule; AB is opposite C.
\[AC=\frac{80\sin68^\circ}{\sin70^\circ}\approx\boxed{78.9\text{ m}}\]For area, use unrounded AC:
\[A=\frac12(80)(AC)\sin42^\circ\approx\boxed{2110\text{ m}^2}\ \text{(3 s.f.)}\]28A 3-D observation[6 marks]
A vertical mast is 18 m tall. Its foot is 24 m north and 10 m east of an observer. Find (i) the horizontal distance to the mast and (ii) the angle of elevation to its top.
Show Answer
\[p=\sqrt{24^2+10^2}=\sqrt{676}=\boxed{26\text{ m}}\]\[\tan\theta=\frac{18}{26}\]\[\theta=\tan^{-1}(18/26)=\boxed{34.7^\circ}\]
29Line-to-base angle of a pyramid[7 marks]
A pyramid has a horizontal square base of side 10 cm, and its apex is 12 cm vertically above the centre. Find (i) the sloping edge length to a base corner and (ii) the angle of that edge with the base.
Show Answer
Projection of the edge onto the base is half the base diagonal.
\[p=\sqrt{5^2+5^2}=5\sqrt2\]\[L=\sqrt{12^2+(5\sqrt2)^2}=\sqrt{194}=\boxed{13.9\text{ cm}}\]\[\theta=\tan^{-1}\left(\frac{12}{5\sqrt2}\right)=\boxed{59.5^\circ}\]30A 3-D cuboid and two planes[8 marks]
A cuboid has length 12 cm, width 5 cm and height 9 cm. Find the angle its space diagonal makes with (i) the base and (ii) the vertical 12 cm by 9 cm face.
Show Answer
(i) The base projection has length \(\sqrt{12^2+5^2}=13\) cm.
\[\theta=\tan^{-1}(9/13)=\boxed{34.7^\circ}\](ii) The projection onto the stated face has length \(\sqrt{12^2+9^2}=15\) cm; the perpendicular offset is 5 cm.
\[\phi=\tan^{-1}(5/15)=\boxed{18.4^\circ}\]Chapter 41 high-grade checklist
Pythagoras
I identify the hypotenuse and use addition for it, subtraction for a shorter side.
SOH–CAH–TOA
I label sides relative to the angle, choose the correct ratio and check degree mode.
Inverse trigonometry
I use inverse sine/cosine/tangent when an angle is unknown.
Bearings and heights
I sketch north and horizontal lines before selecting ratios.
Obtuse angles
I remember sine stays positive and cosine becomes negative in the second quadrant.
Sine rule
I match each side with its opposite angle and check SSA for two solutions.
Cosine rule
I use it for SAS or SSS, including obtuse triangles.
Triangle area
I choose the included angle in \(\tfrac12ab\sin C\).
Three dimensions
I select and draw the correct 2-D right triangle within the solid.
Line and plane
I find the perpendicular projection of the line onto the specified plane.
Working and accuracy
I keep enough working, correct units and unrounded intermediate values.