Pythagoras and trigonometry

19 cm is the hypotenuse.

$CB^2 = 19^2 - 10^2 = 361 - 100 = 261$

$CB = \sqrt{261} = 16.155\ldots \approx 16.2$ cm

The hypotenuse is 8.5 cm.

$x^2 = 8.5^2 - 4^2 = 72.25 - 16 = 56.25$

$x = \sqrt{56.25} = 7.5$

(a) The right angle is at $B$, so the longest side is $AC$. Sarah treated $BC$ as the hypotenuse. Correct: $BC^2 = 8^2 - 6^2 = 28$, $BC = 5.29$ cm.

(b) Count squares from $C$: $CR = 4$ but $CZ = 10$, so $10 \div 4 = 2.5$. Roy enlarged by 2.5, not $1\frac{1}{2}$.

$AC = BC$: $2x + 11 = 4x - 4$, so $2x = 15$ and $x = 7.5$.

Sides: $AC = BC = 26$, $AB = 20$ (perimeter $72$ ✓).

Height from $C$ to the midpoint of $AB$: $\sqrt{26^2 - 10^2} = \sqrt{576} = 24$.

Area $= \frac{1}{2} \times 20 \times 24 = 240\text{ cm}^2$

AB is opposite the $62°$ angle and BC (7.6 cm) is adjacent.

$\tan 62° = \frac{AB}{7.6}$

$AB = 7.6 \times \tan 62° = 14.29... \approx 14.3$ cm

This is one of the exact trig values to learn.

$\cos 60^\circ = \frac{1}{2}$

AB is opposite the $49°$ angle and BC (28 cm) is adjacent, so use tan.

$AB = 28 \times \tan 49° = 32.21\ldots$

So $AB = 32.2$ cm (3 s.f.)

The diagonal of the square is a diameter of the circle.

$d = \sqrt{3.5^2 + 3.5^2} = \sqrt{24.5} = 4.949...$ cm

Circumference $= \pi d = \pi \times 4.949... = 15.55... \approx 15.6$ cm

This is one of the exact trig values to learn: $\sin 30^\circ = \frac{1}{2} = 0.5$

$x$ is adjacent to $53°$ and 14.5 cm is the hypotenuse, so use cosine.

$x = 14.5 \cos 53° = 8.726...$

$x = 8.73$ (3 s.f.)

Hypotenuse: $\sqrt{8^2 + 10^2} = \sqrt{164} = 12.806\ldots$ mm.

Outer edges: 8, 10, part of the first hypotenuse ($12.806 - 10 = 2.806$), the second hypotenuse $12.806$, and 8.

Perimeter $= 8 + 10 + 2.806 + 12.806 + 8 = 41.6$ mm (3 s.f.)

(a) $BC$ is opposite $56^\circ$ and 12 is adjacent, so use tan.

$BC = 12 \tan 56^\circ = 17.79\ldots = 17.8$ cm

(b) 15 is adjacent and 18 is the hypotenuse, so use cos.

$x = \cos^{-1}\left(\frac{15}{18}\right) = 33.557\ldots = 33.6^\circ$

$x$ is opposite $34^\circ$ and 178 is the hypotenuse, so use sine.

$x = 178 \times \sin 34^\circ = 99.536\ldots$

$x = 99.5$ mm (1 d.p.)

AC is the hypotenuse (16 cm) and AB is opposite the 38° angle.

$\sin 38^\circ = \frac{AB}{16}$

$AB = 16 \sin 38^\circ = 9.850... = 9.85\text{ cm}$

(a) $\tan 45° = 1$

(b) The 4 cm side is next to the 60° angle and $x$ is the hypotenuse.

$\cos 60° = \frac{4}{x}$, so $0.5 = \frac{4}{x}$

$x = 4 \div 0.5 = 8$

Angles A and B share $180 - 90 = 90°$ in the ratio $3 : 2$ (B : A)

One part $= 90 \div 5 = 18°$, so $A = 36°$ and $B = 54°$

AC is next to angle A and AB is the hypotenuse: $\cos 36° = \frac{14}{AB}$

$AB = \frac{14}{\cos 36°} = 17.3$ cm

(a) $AB$ is the hypotenuse and $BC$ is adjacent to angle $ABC$.

$\cos ABC = \frac{7}{11}$, so $ABC = \cos^{-1}(\frac{7}{11}) = 50.5^\circ$.

(b) New $\cos ABC = \frac{7}{10} = 0.7$, which is more than $\frac{7}{11} \approx 0.636$, so it increases.

(a) $\cos 30° = \frac{\sqrt3}{2}$

(b) $x$ is opposite the $30°$ angle and 12 cm is the hypotenuse, so use sine.

$\sin 30° = \frac{x}{12}$

$x = 12 \times 0.5 = 6$

Draw a right-angled triangle: across $9 - 7 = 2$, down 5

$CD = \sqrt{5^2 + 2^2} = \sqrt{29} = 5.385$ cm

Perimeter $= 9 + 7 + 5 + 5.385 = 26.385$ cm

Side of square $= 26.385 \div 4 = 6.596$ cm

Area $= 6.596^2 = 43.5$ cm²

$\sin a = \dfrac{\text{opp}}{\text{hyp}} = \dfrac{7-2x}{8}$ and $\tan b = \dfrac{\text{opp}}{\text{adj}} = \dfrac{1+x}{4}$

$\dfrac{7-2x}{8} = \dfrac{1+x}{4} \Rightarrow 7-2x = 2+2x \Rightarrow 4x = 5 \Rightarrow x = 1.25$

$M = \left(\frac{7 + (-3)}{2}, \frac{13 + 21}{2}\right) = (2, 17)$

$N = \left(\frac{-3 + 15}{2}, \frac{21 + 23}{2}\right) = (6, 22)$

$MN = \sqrt{4^2 + 5^2} = \sqrt{41} = 6.403... \approx 6.4$

In triangle $ABC$: $AC = 6.8 \sin 41° = 4.461...$ cm

In triangle $ACD$: $\tan 55° = \dfrac{AC}{DC}$

$DC = \dfrac{4.461...}{\tan 55°} = 3.12...$, so $DC = 3.1$ cm

Horizontal difference: $8 - (-7) = 15$

Vertical difference: $6 - (-5) = 11$

$AB = \sqrt{15^2 + 11^2} = \sqrt{346} = 18.60...$

$AB = 18.6$ cm

In triangle ACD, $\tan 45^\circ = \frac{AC}{8}$, so $AC = 8$ cm.

In triangle ABC, $AC$ is opposite the $20^\circ$ angle and $AB$ is the hypotenuse:

$AB = \frac{8}{\sin 20^\circ} = 23.39\ldots \approx 23.4$ cm

$\tan 30^\circ = \frac{1}{\sqrt{3}}$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$

$\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{2} = \frac{1}{2}$

In triangle BDC: $BD = 8.6 \sin 40° = 5.528$ cm

E is the midpoint, so $ED = 5.528 \div 2 = 2.764$ cm

In triangle EAD: $\tan(EAD) = \frac{2.764}{4.4}$

$EAD = \tan^{-1}(0.6282) = 32.1°$

(a) $\cos 30° = \frac{\sqrt{3}}{2}$

(b) $x$ is opposite the $30°$ angle and 12 is the hypotenuse.

$\sin 30° = \frac{x}{12}$, so $x = 12 \times 0.5 = 6$.

$CD = 1.5 \times 1.7 = 2.55$ m, so $C$ is $2.55 - 1.7 = 0.85$ m above $A$.

In the right-angled triangle from $A$ across to below $C$: $\tan 52^\circ = \frac{0.85}{BD}$

$BD = \frac{0.85}{\tan 52^\circ} = 0.6640... = 0.664$ m (3 s.f.)

(a) Start at 0, rise to a maximum of 1 at $90^\circ$, back to 0 at $180^\circ$, down to $-1$ at $270^\circ$ and back to 0 at $360^\circ$.

(b) $\sin x = -\frac{1}{2}$. Since $\sin 30^\circ = \frac{1}{2}$, the negative values are below the axis:

$x = 180 + 30 = 210$ and $x = 360 - 30 = 330$

The angle is $FCA$ in the right-angled triangle $FAC$ (right angle at $A$).

$\sin x = \frac{\text{opp}}{\text{hyp}} = \frac{6.8}{13.6} = \frac{1}{2}$

So $x = 30^\circ$

Height $= 6 \sin 30^\circ = 6 \times \frac{1}{2} = 3$ cm

Area: $\frac{1}{2}(AB + CD) \times 3 = 66$, so $AB + CD = 44$

$AB : CD = 2 : 3$, so $AB = \frac{2}{5} \times 44 = 17.6$ cm

Diagonal $AC = \sqrt{5^2 + 5^2} = \sqrt{50} = 7.071$ m, so half of it is $3.536$ m.

The triangle from A to the centre to T is right-angled:

$\tan TAC = \frac{12}{3.536}$

$TAC = \tan^{-1}(3.394) = 73.6°$

Let the unknown edge be $d$. The space diagonal satisfies $7^2 + 5^2 + d^2 = 15^2$

$d^2 = 225 - 49 - 25 = 151$, so $d = \sqrt{151} = 12.288...$ cm

Volume $= 7 \times 5 \times 12.288... = 430.09... = 430\text{ cm}^3$ (3 s.f.)

In triangle $VFC$: $FC = \frac{8}{\tan 52°} = 6.250…$

$MF : FC = 1 : 2$, so $MC = 1.5 \times 6.250… = 9.375…$

$MC$ is the height of the equilateral triangle: side $= \frac{9.375…}{\sin 60°} = 10.8$ cm

$AP = \frac{5}{7} \times 14.7 = 10.5$ cm and $AM = 3.8 \div 2 = 1.9$ cm.

Angle $MAP = 90°$, so $MP = \sqrt{10.5^2 + 1.9^2} = 10.67...$ cm

T is above M, so the angle is $TPM$:

$\tan TPM = \dfrac{2.3}{10.67...}$, so angle $= 12.16...° \approx 12.2°$

Front face: $AF = \sqrt{13^2 - 9^2} = \sqrt{88} = 9.380...$

Base: $GH = 9$, so $FH = \frac{9}{\cos 49°} = 13.71...$

Triangle $FAH$ is right-angled at $F$: $\tan FAH = \frac{13.71...}{9.380...}$

Angle $FAH = 55.6... \approx 56°$

$TC = 14 \times \frac{3}{7} = 6$ cm

$TD = \sqrt{14^2 + 6^2} = \sqrt{232} = 15.23\ldots$ cm

Trapezium: $\frac{1}{2}(SD + 12)\times14 = 147 \Rightarrow SD = 9$ cm

Angle $= \tan^{-1}\left(\frac{9}{15.23\ldots}\right) = 30.6^\circ$

$BE = 15 \tan 35^\circ = 10.50$ cm

$MA = \frac{3}{5} \times 15 = 9$ cm, so $MB = \sqrt{9^2 + 15^2} = \sqrt{306} = 17.49$ cm

The angle is at M in triangle EMB, right-angled at B:

$\tan \theta = \frac{10.50}{17.49}$, so $\theta = 30.98... = 31.0^\circ$

In right-angled triangle $ABC$: $AC = \frac{7.3}{\sin 48°} = 9.823$ cm

In right-angled triangle $ACH$: $\tan(CAH) = \frac{8.1}{9.823}$

Angle $= \tan^{-1}(0.8246) = 39.5°$