1. Problem 4: Find the lengths marked $a$ and $b$ in triangle $ABC$ where $\angle C=90^\circ$, $AB=2$ cm and $\angle B=45^\circ$. 1. In a right triangle the non-right angles sum to $90^\circ$, so $\angle A=90^\circ-\angle B=45^\circ$ and the triangle is isosceles with the two legs equal. 1. Use the relationship between hypotenuse and leg in a $45^\circ$-$45^\circ$-$90^\circ$ triangle: $\text{leg}=\dfrac{\text{hypotenuse}}{\sqrt{2}}$. 1. Compute $a=\dfrac{2}{\sqrt{2}}=\sqrt{2}\approx1.414$ cm. 1. Final: $a=\sqrt{2}\text{ cm}$ and $b=\sqrt{2}\text{ cm}$. 2. Problem 5: In triangle $PQR$ with right angle at $Q$, $PQ=4$ and $\angle R=45^\circ$, find $q=PR$. 2. If $\angle R=45^\circ$ and the triangle is right at $Q$, then $\angle P=45^\circ$ and the two legs are equal, so $PQ=QR=4$. 2. The hypotenuse $PR$ is $\text{leg}\times\sqrt{2}$, so $q=4\sqrt{2}\approx5.657$. 2. Final: $q=4\sqrt{2}\approx5.657$. 3. Problem 6: In isosceles triangle $XYZ$ with $XY=XZ=6$ cm and $\angle YXZ=120^\circ$, find $YZ$. 3. Use the cosine rule: $YZ^2=XY^2+XZ^2-2\,XY\,XZ\cos\angle YXZ$. 3. Substitute values: $YZ^2=6^2+6^2-2\cdot6\cdot6\cos120^\circ=36+36-72\cos120^\circ$. 3. Since $\cos120^\circ=-\tfrac{1}{2}$, we get $YZ^2=72-72(-\tfrac{1}{2})=72+36=108$. 3. Thus $YZ=\sqrt{108}=6\sqrt{3}\approx10.392$. 3. Final: $YZ=6\sqrt{3}\text{ cm}\approx10.392\text{ cm}$. 4. Problem 7: A cone has perpendicular height $42$ cm and semi-vertical angle $30^\circ$. Find the slant height $l$. 4. In the axial cross-section the slant height $l$ and height $h$ satisfy $\cos\alpha=\dfrac{h}{l}$ where $\alpha$ is the semi-vertical angle. 4. Thus $l=\dfrac{h}{\cos30^\circ}=\dfrac{42}{\cos30^\circ}$. 4. Using $\cos30^\circ=\tfrac{\sqrt{3}}{2}$ gives $l=\dfrac{42}{\tfrac{\sqrt{3}}{2}}=\dfrac{84}{\sqrt{3}}=28\sqrt{3}\approx48.497$. 4. Final: $l=28\sqrt{3}\text{ cm}\approx48.497\text{ cm}$. 5. Problem 8: From the top of a tower $60$ m high two ships to the south have angles of depression $45^\circ$ and $30^\circ$. Find the distance between the ships. 5. Let the horizontal distances from the tower base to the nearer and farther ships be $d_1$ and $d_2$ respectively. 5. For angle of depression $45^\circ$ we have $\tan45^\circ=\dfrac{60}{d_1}$ so $d_1=60$ m. 5. For angle $30^\circ$ we have $\tan30^\circ=\dfrac{60}{d_2}$ so $d_2=\dfrac{60}{\tan30^\circ}=60\sqrt{3}$ m. 5. The distance between the ships is $d_2-d_1=60(\sqrt{3}-1)\approx43.923$ m. 5. Final: distance $=60(\sqrt{3}-1)\text{ m}\approx43.923\text{ m}$. 6. Problem 9: Without using tables, find $\dfrac{1}{\sin60^\circ}$ given $\sin60^\circ=2\sin30^\circ\cos30^\circ$. 6. Compute $\sin30^\circ=\tfrac{1}{2}$ and $\cos30^\circ=\tfrac{\sqrt{3}}{2}$, so $\sin60^\circ=2\cdot\tfrac{1}{2}\cdot\tfrac{\sqrt{3}}{2}=\tfrac{\sqrt{3}}{2}$. 6. Therefore $\dfrac{1}{\sin60^\circ}=\dfrac{1}{\tfrac{\sqrt{3}}{2}}=\dfrac{2}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3}\approx1.155$. 6. Final: $\dfrac{1}{\sin60^\circ}=\dfrac{2\sqrt{3}}{3}\approx1.155$. 7. Problem 10: If $\tan x=\dfrac{1}{\sqrt{3}}$ and $0^\circ\le x\le90^\circ$, find $\cos x-\sin x$. 7. Solve $\tan x=\dfrac{1}{\sqrt{3}}$ to get $x=30^\circ$ in the given interval. 7. Compute $\cos30^\circ-\sin30^\circ=\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}=\dfrac{\sqrt{3}-1}{2}\approx0.366$. 7. Final: $\cos x-\sin x=\dfrac{\sqrt{3}-1}{2}\approx0.366$. 8. Example 12(a): Evaluate $\dfrac{1}{\sqrt{3}}$ to 3 significant figures given $\sqrt{3}=1.732$. 8. Compute $\dfrac{1}{\sqrt{3}}=\dfrac{1}{1.732}\approx0.577350$ and to 3 s.f. this is $0.577$. 8. Final: $\dfrac{1}{\sqrt{3}}\approx0.577$ (3 s.f.). 9. Example 12(b): Evaluate $\dfrac{2}{\sqrt{2}}$ to 3 significant figures given $\sqrt{2}=1.414$. 9. Note $\dfrac{2}{\sqrt{2}}=\sqrt{2}\approx1.414$ and to 3 s.f. this is $1.41$. 9. Final: $\dfrac{2}{\sqrt{2}}=\sqrt{2}\approx1.41$ (3 s.f.). 10. Example 12(c): Evaluate $\dfrac{2}{\sqrt{3}+\sqrt{2}}$ to 3 significant figures. 10. Rationalize: $\dfrac{2}{\sqrt{3}+\sqrt{2}}=\dfrac{2(\sqrt{3}-\sqrt{2})}{(\sqrt{3})^2-(\sqrt{2})^2}=2(\sqrt{3}-\sqrt{2})$. 10. Substitute numbers: $2(1.732-1.414)=2(0.318)=0.636$ and to 3 s.f. this is $0.636$. 10. Final: $\dfrac{2}{\sqrt{3}+\sqrt{2}}\approx0.636$ (3 s.f.). 11. Example 12(d): Evaluate $\dfrac{1}{\sqrt{3}-\sqrt{2}}$ to 3 significant figures. 11. Rationalize: $\dfrac{1}{\sqrt{3}-\sqrt{2}}=\dfrac{\sqrt{3}+\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2}=\sqrt{3}+\sqrt{2}$. 11. Substitute values: $1.732+1.414=3.146$ and to 3 s.f. this rounds to $3.15$. 11. Final: $\dfrac{1}{\sqrt{3}-\sqrt{2}}\approx3.15$ (3 s.f.). 12. Summary of final answers: 12. Problem 4: $a=\sqrt{2}\text{ cm},\; b=\sqrt{2}\text{ cm}$. 12. Problem 5: $q=4\sqrt{2}\approx5.657$. 12. Problem 6: $YZ=6\sqrt{3}\approx10.392$ cm. 12. Problem 7: Slant height $=28\sqrt{3}\approx48.497$ cm. 12. Problem 8: Distance between ships $=60(\sqrt{3}-1)\approx43.923$ m. 12. Problem 9: $\dfrac{1}{\sin60^\circ}=\dfrac{2\sqrt{3}}{3}\approx1.155$. 12. Problem 10: $\cos x-\sin x=\dfrac{\sqrt{3}-1}{2}\approx0.366$. 12. Example 12(a)--(d): (a) $0.577$, (b) $1.41$, (c) $0.636$, (d) $3.15$.