This is Case 3. To obtain the angle at $B$ we use $tanB=\frac{4}{5}$ so $B={tan}^{-1}\left(0.8\right)=38.66^{\circ }$ .

Then the angle at $A$ is $180^{\circ }-\left(90^{\circ }-38.66^{\circ }\right)=51.34^{\circ }$ .

By Pythagoras’ theorem $c=\sqrt{4^2+5^2}=\sqrt{41}\approx 6.40$ .

Task!

This is Case 1. Since $31^{\circ }40^{\prime }=31.67^{\circ }$ then $cos31.67^{\circ }=\frac{a}{15}$ so $a=15cos31.67^{\circ }=12.77.$

The angle at $A$ is $180^{\circ }-\left(90+31.67^{\circ }\right)=58.33^{\circ }$ .

Finally $sin31.67^{\circ }=\frac{b}{15}$ $\therefore b=15sin31.67^{\circ }=7.85$ .

(Alternatively, of course, Pythagoras’ theorem could be used to calculate the length $b$ .)

Task!

This is Case 2.

Here $tan34.33^{\circ }=\frac{8}{a}$ so $a=\frac{8}{tan34.33^{\circ }}=11.7$

Also $c=\sqrt{8^2+11.7^2}=14.18$ and the angle at $A$ is $180^{\circ }-\left(90^{\circ }+34.33^{\circ }\right)=55.67^{\circ }$ .

1. $h=\sqrt{15^2+8^2}=17,\mathrm{\text{cosec}}\theta =\frac{1}{sin\theta }=\frac{17}{8}sec\theta =\frac{1}{cos\theta }=\frac{17}{15}cot\theta =\frac{1}{tan\theta }=\frac{15}{8}$

   $\theta ={sin}^{-1}\frac{8}{17}\left(\text{for example}\right)\therefore \theta =28.07^{\circ }$

2. 1. $sin\theta =\frac{2}{5}cos\theta =\frac{\sqrt{21}}{5}tan\theta =\frac{2\sqrt{21}}{21}$ $\mathrm{\text{cosec}}\theta =\frac{5}{2}$
   2. $sin\theta =\frac{y}{\sqrt{x^2+y^2}}cos\theta =\frac{x}{\sqrt{x^2+y^2}}tan\theta =\frac{y}{x}$ $\mathrm{\text{cosec}}\theta =\frac{\sqrt{x^2+y^2}}{y}$
3. Referring to the following diagram

   

   Hence $cos\theta =\frac{3\sqrt{5}}{7}tan\theta =\frac{2}{3\sqrt{5}}=\frac{2\sqrt{5}}{15}$

4. 1. $\theta ={sin}^{-1}0.5260=31.73^{\circ }$
   2. $\theta ={tan}^{-1}2.4=67.38^{\circ }$
   3. $\theta ={cos}^{-1}0.2=78.46^{\circ }$
5. $\beta =90-\alpha =32.5^{\circ }$ , $b=\frac{10}{tan57.5^{\circ }}\simeq 6.37$ $c=\frac{10}{sin57.5^{\circ }}\simeq 11.86$
6. $tan37^{\circ }=\frac{h}{z+0.5}tan41^{\circ }=\frac{h}{z}$ from which

   $h=\left(z+0.5\right)tan37^{\circ }=ztan41^{\circ }$ , so $ztan37^{\circ }-ztan41^{\circ }=-0.5tan37^{\circ }$

   $\therefore z=\frac{-0.5tan37^{\circ }}{tan37^{\circ }-tan41^{\circ }}\simeq 3.2556\text{km}$ , so $h=ztan41^{\circ }=3.2556tan41^{\circ }\simeq 2.83$ km

7. Since the required answer is in terms of $cot\alpha$ and $cot\beta$ we proceed as follows:

   Using $x$ to denote the distance $S_{1}P$ $cot\alpha =\frac{1}{tan\alpha }=\frac{x}{h}cot\beta =\frac{1}{tan\beta }=\frac{c-x}{h}$

   Adding: $cot\alpha +cot\beta =\frac{x}{h}+\frac{c-x}{h}=\frac{c}{h}$ $\therefore h=\frac{c}{cot\alpha +cot\beta }$ as required.

8. From the smaller right-angled triangle $d_1=\frac{200}{sin28^{\circ }}=426.0\text{m}$ . The base of this triangle then has length $\ell =426cos28^{\circ }=376.1\text{m}$

   From the larger right-angled triangle the straight-line distance from START to FINISH is $\frac{200}{sin15^{\circ }}=772.7$ m. Then, using Pythagoras’ theorem $\left(d_2+\ell \right)=\sqrt{772.7^2-200^2}=746.4\text{m}$ from which $d_2=370.3$ m $\therefore d_1+d_2=796.3$ m