Tangent and solving right-angled triangles

Each side of a roof slopes at ${28}^{\circ }28^\backslash circ$, and the horizontal distance from the wall top to the middle is 4.5 m. How high is the ridge above the wall top?

Solution In triangle $WMRWMR$, $hh$ is opposite ${28}^{\circ }28^\backslash circ$ and 4.5 m is adjacent, so use tangent:

$$\tan {28}^{\circ }=\frac{h}{4.5}⟹h=4.5\tan {28}^{\circ }\approx 4.5\times 0.5317\approx 2.39\text{ m}\mathrm{.}\backslash tan\; 28^\backslash circ\; =\; \backslash frac\{h\}\{4.5\}\; \backslash quad\backslash Longrightarrow\backslash quad\; h\; =\; 4.5\backslash tan\; 28^\backslash circ\; \backslash approx\; 4.5\; \backslash times\; 0.5317\; \backslash approx\; 2.39\backslash \; \backslash text\{m\}.$$

(a) What angle does a road with a 10% gradient make with the horizontal? (b) What angle does the line $y=2x+1y\; =\; 2x\; +\; 1$ make with the $xx$-axis?

Solution A gradient is rise over run, which is opposite over adjacent, so gradient $=\tan \theta =\; \backslash tan\backslash theta$.

(a) $\tan \theta =0.1\backslash tan\backslash theta\; =\; 0.1$, so $\theta ={\tan }^{-1}(0.1)\approx {5.7}^{\circ }\backslash theta\; =\; \backslash tan^\{-1\}(0.1)\; \backslash approx\; 5.7^\backslash circ$. A "steep" 10% road is less than $6^{\circ }6^\backslash circ$.

(b) $\tan \theta =2\backslash tan\backslash theta\; =\; 2$, so $\theta ={\tan }^{-1}(2)\approx {63.4}^{\circ }\backslash theta\; =\; \backslash tan^\{-1\}(2)\; \backslash approx\; 63.4^\backslash circ$.

Solve the triangle: find $cc$, $\mathrm{\angle }A\backslash angle\; A$ and $\mathrm{\angle }B\backslash angle\; B$.

Solution Pythagoras: $c=\sqrt{7^2+{24}^2}=\sqrt{625}=25c\; =\; \backslash sqrt\{7^2\; +\; 24^2\}\; =\; \backslash sqrt\{625\}\; =\; 25$.

For $\mathrm{\angle }A\backslash angle\; A$, 7 is opposite and 24 is adjacent: $\tan A=\frac{7}{24}\backslash tan\; A\; =\; \backslash frac\{7\}\{24\}$, so $A\approx {16.3}^{\circ }A\; \backslash approx\; 16.3^\backslash circ$. Then $B={90}^{\circ }-{16.3}^{\circ }={73.7}^{\circ }B\; =\; 90^\backslash circ\; -\; 16.3^\backslash circ\; =\; 73.7^\backslash circ$.

Check: $\sin A=\frac{7}{25}=0.28\backslash sin\; A\; =\; \backslash frac\{7\}\{25\}\; =\; 0.28$ and $\sin {16.26}^{\circ }\approx 0.28\backslash sin\; 16.26^\backslash circ\; \backslash approx\; 0.28$ ✓.

Triangle $ABCABC$ has $\mathrm{\angle }C={90}^{\circ }\backslash angle\; C\; =\; 90^\backslash circ$, hypotenuse $c=10c\; =\; 10$ cm and $\mathrm{\angle }A={40}^{\circ }\backslash angle\; A\; =\; 40^\backslash circ$. Solve it.

Solution $\mathrm{\angle }B={90}^{\circ }-{40}^{\circ }={50}^{\circ }\backslash angle\; B\; =\; 90^\backslash circ\; -\; 40^\backslash circ\; =\; 50^\backslash circ$. $a=10\sin {40}^{\circ }\approx 6.43a\; =\; 10\backslash sin\; 40^\backslash circ\; \backslash approx\; 6.43$ cm (opposite $AA$), and $b=10\cos {40}^{\circ }\approx 7.66b\; =\; 10\backslash cos\; 40^\backslash circ\; \backslash approx\; 7.66$ cm (adjacent to $AA$).

Check: ${6.428}^2+{7.660}^2\approx 41.3+58.7=1006.428^2\; +\; 7.660^2\; \backslash approx\; 41.3\; +\; 58.7\; =\; 100$ ✓.

$\theta \backslash theta$ is acute and $\sin \theta =\frac{3}{5}\backslash sin\backslash theta\; =\; \backslash frac\{3\}\{5\}$. Find $\cos \theta \backslash cos\backslash theta$ and $\tan \theta \backslash tan\backslash theta$ without a calculator.

Solution ${\cos }^2\theta =1-{\sin }^2\theta =1-\frac{9}{25}=\frac{16}{25}\backslash cos^2\backslash theta\; =\; 1\; -\; \backslash sin^2\backslash theta\; =\; 1\; -\; \backslash frac\{9\}\{25\}\; =\; \backslash frac\{16\}\{25\}$, so $\cos \theta =\frac{4}{5}\backslash cos\backslash theta\; =\; \backslash frac\{4\}\{5\}$ (positive, as $\theta \backslash theta$ is acute). Then $\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{3}{5}÷\frac{4}{5}=\frac{3}{4}\backslash tan\backslash theta\; =\; \backslash frac\{\backslash sin\backslash theta\}\{\backslash cos\backslash theta\}\; =\; \backslash frac\{3\}\{5\}\; \backslash div\; \backslash frac\{4\}\{5\}\; =\; \backslash frac\{3\}\{4\}$.