1. **Problem:** Calculate the height of a tree given two angles of elevation from points P and Q, where Q is 5 m closer to the tree than P. 2. **Formula:** Use the tangent function for right triangles: $$\tan \theta = \frac{\text{height}}{\text{distance}}$$ 3. Let the height of the tree be $h$ meters and the distance from point Q to the tree base be $x$ meters. 4. From point P, distance is $x + 5$ meters. Using tangent: $$\tan 24.5^\circ = \frac{h}{x+5}$$ $$\tan 32.5^\circ = \frac{h}{x}$$ 5. From the two equations: $$h = (x+5) \tan 24.5^\circ = x \tan 32.5^\circ$$ 6. Equate and solve for $x$: $$x \tan 32.5^\circ = (x+5) \tan 24.5^\circ$$ $$x (\tan 32.5^\circ - \tan 24.5^\circ) = 5 \tan 24.5^\circ$$ $$x = \frac{5 \tan 24.5^\circ}{\tan 32.5^\circ - \tan 24.5^\circ}$$ 7. Calculate values: $$\tan 24.5^\circ \approx 0.4551, \quad \tan 32.5^\circ \approx 0.6360$$ $$x = \frac{5 \times 0.4551}{0.6360 - 0.4551} = \frac{2.2755}{0.1809} \approx 12.58$$ 8. Find height $h$: $$h = x \tan 32.5^\circ = 12.58 \times 0.6360 \approx 8.0$$ **Answer:** Height of the tree is approximately $8.0$ meters. --- 1. **Problem:** Given $\sin(3a) = \frac{1}{2}$, find $\cos a$ where $a$ is acute. 2. **Formula:** Use the triple angle identity for sine: $$\sin 3a = 3 \sin a - 4 \sin^3 a$$ 3. Since $\sin 3a = \frac{1}{2}$, and $3a$ is acute, $3a = 30^\circ$ or $\frac{\pi}{6}$ radians. 4. Then: $$a = \frac{30^\circ}{3} = 10^\circ$$ 5. Calculate $\cos a$: $$\cos 10^\circ \approx 0.9848$$ **Answer:** $\cos a \approx 0.9848$. --- 1. **Problem:** A man walks 150 m towards a building 240 m away. At that point, angle of elevation is $45^\circ$. Find angle of elevation from starting point A. 2. **Formula:** Use tangent: $$\tan \theta = \frac{\text{height}}{\text{distance}}$$ 3. Let height of building be $h$, distance from man at 150 m is $240 - 150 = 90$ m. 4. From point after walking 150 m: $$\tan 45^\circ = \frac{h}{90} \Rightarrow h = 90$$ 5. From point A (distance 240 m): $$\tan \theta = \frac{h}{240} = \frac{90}{240} = 0.375$$ 6. Calculate $\theta$: $$\theta = \tan^{-1}(0.375) \approx 20.6^\circ$$ **Answer:** Angle of elevation from A is approximately $20.6^\circ$. --- 1. **Problem:** Given $\tan \theta = \frac{4}{3}$, find $\sin \theta + \cos \theta$. 2. **Formula:** Use Pythagorean identity: $$\sin^2 \theta + \cos^2 \theta = 1$$ 3. From $\tan \theta = \frac{4}{3}$, construct a right triangle with opposite = 4, adjacent = 3. 4. Hypotenuse: $$\sqrt{3^2 + 4^2} = 5$$ 5. Calculate $\sin \theta$ and $\cos \theta$: $$\sin \theta = \frac{4}{5}, \quad \cos \theta = \frac{3}{5}$$ 6. Sum: $$\sin \theta + \cos \theta = \frac{4}{5} + \frac{3}{5} = \frac{7}{5} = 1.4$$ **Answer:** $\sin \theta + \cos \theta = 1.4$. --- 1. **Problem:** Given $\sin(x + 60^\circ) = \cos(2x)$, find $\tan(2x + 60^\circ)$. 2. **Formula:** Use identity: $$\cos \alpha = \sin(90^\circ - \alpha)$$ 3. Rewrite: $$\sin(x + 60^\circ) = \sin(90^\circ - 2x)$$ 4. Equate angles: $$x + 60^\circ = 90^\circ - 2x \Rightarrow 3x = 30^\circ \Rightarrow x = 10^\circ$$ 5. Calculate: $$\tan(2x + 60^\circ) = \tan(20^\circ + 60^\circ) = \tan 80^\circ$$ 6. Approximate: $$\tan 80^\circ \approx 5.67$$ **Answer:** $\tan(2x + 60^\circ) \approx 5.67$. --- 1. **Problem:** Given $\sin 3x - \cos 2x = 0$ and $3x$ acute, find $x$. 2. **Rewrite:** $$\sin 3x = \cos 2x$$ 3. Use identity: $$\cos 2x = \sin(90^\circ - 2x)$$ 4. Equate: $$3x = 90^\circ - 2x \Rightarrow 5x = 90^\circ \Rightarrow x = 18^\circ$$ **Answer:** $x = 18^\circ$. --- 1. **Problem:** Calculate area of triangular stool top with sides 27 cm, 35 cm, 42 cm. 2. **Formula:** Use Heron's formula: $$s = \frac{a+b+c}{2}$$ $$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$$ 3. Calculate semi-perimeter: $$s = \frac{27 + 35 + 42}{2} = 52$$ 4. Calculate area: $$\sqrt{52(52-27)(52-35)(52-42)} = \sqrt{52 \times 25 \times 17 \times 10}$$ 5. Simplify: $$\sqrt{221000} \approx 470.1$$ **Answer:** Area is approximately $470.1$ cm². --- 1. **Problem:** Triangle XYZ with sides $ZY=13.4$ cm, $XY=5$ cm, angle $XYZ=57.7^\circ$. Find area. 2. **Formula:** Area = $\frac{1}{2}ab \sin C$ 3. Sides adjacent to angle $XYZ$ are $ZY=13.4$ cm and $XY=5$ cm. 4. Calculate area: $$\frac{1}{2} \times 13.4 \times 5 \times \sin 57.7^\circ$$ 5. Calculate $\sin 57.7^\circ \approx 0.844$ 6. Area: $$0.5 \times 13.4 \times 5 \times 0.844 = 28.3$$ **Answer:** Area is approximately $28.3$ cm². --- 1. **Problem:** Parallelogram ABCD with sides 5 cm, 10 cm, angle $70^\circ$. Find area. 2. **Formula:** Area = base $\times$ height = $ab \sin \theta$ 3. Calculate area: $$5 \times 10 \times \sin 70^\circ$$ 4. $\sin 70^\circ \approx 0.9397$ 5. Area: $$50 \times 0.9397 = 47.0$$ **Answer:** Area is approximately $47.0$ cm². --- 1. **Problem:** Rhombus with side 15 cm and one angle $120^\circ$. Find area. 2. **Formula:** Area = $a^2 \sin \theta$ 3. Calculate area: $$15^2 \times \sin 120^\circ = 225 \times 0.8660 = 194.9$$ **Answer:** Area is approximately $194.9$ cm². --- 1. **Problem:** Trapezium with bases 10 cm, 16 cm, height 8 cm, area 128 cm². Find side $a$. 2. **Formula:** Area = $\frac{1}{2} (a+b) h$ 3. Calculate area with given bases: $$128 = \frac{1}{2} (10 + 16) \times 8 = 104$$ 4. Since area given is 128, $a$ must be a non-parallel side, likely height or slant side. More info needed for $a$. **Answer:** Insufficient data to find $a$ precisely. --- 1. **Problem:** Calculate area of regular octagon with side 6 cm. 2. **Formula:** Area = $2(1+\sqrt{2})a^2$ 3. Calculate: $$2(1+1.414) \times 6^2 = 2(2.414) \times 36 = 173.8$$ **Answer:** Area is approximately $173.8$ cm². --- 1. **Problem:** Regular pentagon inscribed in circle radius 10 cm. Find side and shaded area. 2. **Formula:** Side $s = 2r \sin(\frac{\pi}{5})$ 3. Calculate side: $$s = 2 \times 10 \times \sin 36^\circ = 20 \times 0.5878 = 11.76$$ 4. Area of pentagon: $$\frac{5}{2} s r \cos 36^\circ = \frac{5}{2} \times 11.76 \times 10 \times 0.8090 = 237.9$$ 5. Shaded area = Circle area - Pentagon area: $$\pi 10^2 - 237.9 = 314.16 - 237.9 = 76.3$$ **Answer:** Side $\approx 11.76$ cm, shaded area $\approx 76.3$ cm². --- 1. **Problem:** Regular octagon area 101.8 cm². Find side length. 2. **Formula:** Area = $2(1+\sqrt{2})a^2$ 3. Solve for $a$: $$a^2 = \frac{101.8}{2(1+1.414)} = \frac{101.8}{4.828} = 21.09$$ $$a = \sqrt{21.09} = 4.59$$ **Answer:** Side length is approximately $4.59$ cm. --- 1. **Problem:** Sector radius 6.3 cm, area 41.58 cm². Find angle at center. 2. **Formula:** Area = $\frac{\theta}{360} \pi r^2$ 3. Solve for $\theta$: $$41.58 = \frac{\theta}{360} \pi (6.3)^2$$ $$\theta = \frac{41.58 \times 360}{\pi \times 39.69} = 120^\circ$$ **Answer:** Angle is $120^\circ$. --- 1. **Problem:** Two intersecting circles with radii 8 cm and 10 cm, chord AB = 12 cm. Find shaded lens area. 2. **Formula:** Area of lens = sum of two circular segments. 3. Calculate segment areas using formula: $$A = r^2 \cos^{-1}\left(\frac{d}{r}\right) - d \sqrt{r^2 - d^2}$$ 4. For circle P (r=8, half chord $d=6$): $$A_P = 8^2 \cos^{-1}(\frac{6}{8}) - 6 \sqrt{64 - 36} = 64 \times 0.6435 - 6 \times 5 = 41.18 - 30 = 11.18$$ 5. For circle Q (r=10, $d=6$): $$A_Q = 100 \cos^{-1}(\frac{6}{10}) - 6 \sqrt{100 - 36} = 100 \times 0.9273 - 6 \times 8 = 92.73 - 48 = 44.73$$ 6. Total lens area: $$11.18 + 44.73 = 55.91$$ **Answer:** Shaded area is approximately $55.9$ cm². --- 1. **Problem:** Find surface area of sphere with diameter 21 cm. 2. **Formula:** Surface area = $4 \pi r^2$ 3. Radius: $$r = \frac{21}{2} = 10.5$$ 4. Calculate surface area: $$4 \pi (10.5)^2 = 4 \pi \times 110.25 = 1385.4$$ **Answer:** Surface area is approximately $1385.4$ cm².