Right-angled triangles

Unit 8 Right-angled triangles

Key unit competence: Apply Pythagoras' theorem and trigonometric ratios in finding unknown sides and angles in right-angled triangles and solve real-world problems.

Key ideas

• In right triangle $ABC$ with the right angle at $A$, let $H$ be the foot of the altitude from $A$ on the hypotenuse $BC$; it splits $BC$ into $BH=p$ and $HC=q$.
• Median theorem: the median from the right angle to the hypotenuse equals half the hypotenuse: $AM=\frac{BC}{2}$ ($M$ is the centre of the circumscribed circle).
• Altitude (height) theorem: $h^2=p\times q$, where $h=AH$.
• Legs theorem: each leg squared equals the hypotenuse times the adjacent segment: $A{B}^2=BC\times BH$ and $A{C}^2=BC\times HC$.
• Trigonometric ratios for an acute angle $\theta$: $\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}}$, $\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}$, $\tan \theta =\frac{\text{opposite}}{\text{adjacent}}$ (SOH CAH TOA).
• To find an angle use the inverse functions: $\theta ={\sin }^{-1}(\dots )$, etc. Angle of elevation is measured up from the horizontal; angle of depression down from the horizontal.
• Special values: $\sin 30^{\circ }=\frac{1}{2}$, $\cos 60^{\circ }=\frac{1}{2}$, $\tan 45^{\circ }=1$, $\sin 60^{\circ }=\cos 30^{\circ }=\frac{\sqrt{3}}{2}$, $\sin 45^{\circ }=\cos 45^{\circ }=\frac{\sqrt{2}}{2}$.

Formulas to know

Common misconceptions

• "The opposite side is always the vertical side." No: it is the side opposite the angle you are using.
• "$\sin \theta$ can be greater than 1." No: the opposite side is shorter than the hypotenuse, so $0<\sin \theta <1$ for acute angles.
• "$h^2=p+q$." No: the altitude theorem uses a product: $h^2=pq$.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. In right triangle $PQR$ (right angle at $P$), the altitude $PS$ meets $QR$ with $QS=2$ and $SR=8$. Find $PS$, $PQ$ and $PR$.

   Show answer

   $P{S}^2=16$, $PS=4$; $P{Q}^2=10\times 2=20$, $PQ=2\sqrt{5}$; $P{R}^2=10\times 8=80$, $PR=4\sqrt{5}$.

   Common mistake: Using $QS+SR$ for the altitude.

2. Use the legs theorem to prove Pythagoras' theorem.

   Show answer

   $A{B}^2=BC\cdot BH$ and $A{C}^2=BC\cdot HC$. Adding: $A{B}^2+A{C}^2=BC(BH+HC)=BC\cdot BC=B{C}^2$.

   Common mistake: Assuming what you want to prove.

3. In a right triangle the hypotenuse is 26 cm. Find the length of the median to the hypotenuse.

   Show answer

   13 cm.

   Common mistake: Using Pythagoras unnecessarily or with wrong sides.

4. Show that in a right triangle with legs $a$, $b$ and altitude $h$ to the hypotenuse $c$: $ab=ch$.

   Show answer

   Both $\frac{1}{2}ab$ and $\frac{1}{2}ch$ give the area of the triangle, so $ab=ch$.

   Common mistake: Confusing $h$ with one of the legs.

5. Find the angle $\theta$ in a right triangle with opposite side 5 cm and adjacent side 12 cm.

   Show answer

   $\tan \theta =\frac{5}{12}$, $\theta \approx 22.6^{\circ }$.

   Common mistake: Using sine with the adjacent side.

6. Without a calculator, find $x$ in a right triangle with hypotenuse $x$ and a $30^{\circ }$ angle opposite a side of 7 cm.

   Show answer

   $\sin 30^{\circ }=\frac{7}{x}$, $x=14$ cm.

   Common mistake: Writing $x=7\sin 30^{\circ }$.

7. A ladder 6 m long leans against a wall making $70^{\circ }$ with the ground ($\cos 70^{\circ }\approx 0.342$). How far is its foot from the wall?

   Show answer

   $6\cos 70^{\circ }\approx 2.05$ m.

   Common mistake: Using $\sin 70^{\circ }$ (that gives the height).

8. From the top of a 30 m building, the angle of depression of a car is $25^{\circ }$ ($\tan 25^{\circ }\approx 0.466$). How far is the car from the building?

   Show answer

   $\frac{30}{\tan 25^{\circ }}\approx 64.4$ m.

   Common mistake: Multiplying 30 by $\tan 25^{\circ }$.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. The altitude to the hypotenuse divides it into 4 cm and 9 cm. The altitude is:	a. 15
   2. A ladder 8 m long makes $60^{\circ }$ with the ground. How high up the wall does it reach?	b. $4\sqrt{3}$ m
   3. The hypotenuse is 25 and the segment adjacent to leg $AB$ is 9. Find $AB$.	c. ${\cos }^{-1}(0.5)=60^{\circ }$
   4. To find the angle whose cosine is 0.5, we compute:	d. 6 cm

   Show answer

   1 → d (6 cm) · 2 → b ($4\sqrt{3}$ m) · 3 → a (15) · 4 → c (${\cos }^{-1}(0.5)=60^{\circ }$)

   • $h^2=4\times 9=36$.
   • $8\sin 60^{\circ }=8\times \frac{\sqrt{3}}{2}$.
   • $A{B}^2=25\times 9=225$.
   • Use the inverse cosine.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   the right-angle vertex  ·  the midpoint of the hypotenuse  ·  $45^{\circ }$  ·  32
   1. A ramp rises 3 m over a horizontal distance of 3 m. Its angle with the ground is .
   2. In a right triangle, the circumcentre is .
   3. In right triangle $ABC$ (right angle at $A$), $BC=20$ and median $AM$ is drawn. Triangle $ABM$ has $AB=12$. Its perimeter is .
   Show answer

   1: $45^{\circ }$ · 2: the midpoint of the hypotenuse · 3: 32

   • $\tan \theta =1$.
   • This follows from the median theorem.
   • Step 1. $AM=BM=10$.
     Step 2. So $12+10+10$.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. From the top of a 20 m cliff the angle of depression of a boat is $45^{\circ }$. How far is the boat from the foot of the cliff?	a. 20 m
   2. In right triangle $ABC$ with right angle at $C$, $AC=5$, $BC=12$. Find $AB$.	b. 13
   3. A 6 m ladder has its foot 2 m from a wall. Find the angle it makes with the ground (1 d.p.).	c. $\frac{3}{4}$
   4. In the same triangle, $\tan \theta$ is:	d. $70.5^{\circ }$

   Show answer

   1 → a (20 m) · 2 → b (13) · 3 → d ($70.5^{\circ }$) · 4 → c ($\frac{3}{4}$)

   • Step 1. $\tan 45^{\circ }=1$.
     Step 2. So the distance equals the height.
   • Pythagoras.
   • Step 1. $\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{2}{6}=\frac{1}{3}$.
     Step 2. $\theta ={\cos }^{-1}\frac{1}{3}\approx 70.5^{\circ }$.
   • Opposite 3, adjacent 4.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   5 cm  ·  $30^{\circ }$  ·  $45^{\circ }$  ·  $\frac{4}{5}$
   1. In the same triangle, $\cos \theta$ is .
   2. In a right triangle the hypotenuse is 10 cm. The median to the hypotenuse is .
   3. The opposite side is 7 and the hypotenuse 14. The angle is .
   Show answer

   1: $\frac{4}{5}$ · 2: 5 cm · 3: $30^{\circ }$

   • Adjacent 4, hypotenuse 5.
   • Median theorem: half the hypotenuse.
   • $\sin \theta =\frac{1}{2}$.

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