Exercise 6.6 — Class 9 Mathematics Solutions

Solution

1. Draw the right triangle.

   Let $A$ be the point on the ground, $B$ the foot of the post and $C$ its top. Then $\angle B={90}^{\circ }$, $m\overline{AB}=40$ m and the angle of elevation at $A$ is ${60}^{\circ }$. Let the height be $h=m\overline{BC}$.

   $m\overline{AB}=40\text{ m},\theta ={60}^{\circ },m\overline{BC}=h$

2. Use the tangent ratio.

   The height is opposite $\theta$ and the 40 m distance is adjacent, so tangent links them.

   $$\begin{array}{cc}\tan {60}^{\circ }& =\frac{h}{40}\\ h& =40\tan {60}^{\circ }\\ & =40\sqrt{3}\\ & \approx 69.28\text{ m}\end{array}$$

Solution

1. Split the triangle with the altitude.

   Let $△ABD$ have $m\overline{AB}=m\overline{AD}$, vertical angle $\angle A={120}^{\circ }$ and base $m\overline{BD}=10$ cm. In an isosceles triangle the altitude $\overline{AC}$ from the vertex bisects both the vertical angle and the base.

   $$\begin{array}{cc}m\angle BAC& =\frac{1}{2}({120}^{\circ })\\ & ={60}^{\circ }\\ m\overline{BC}& =\frac{1}{2}(10)\\ & =5\text{ cm}\end{array}$$

2. Use the tangent ratio in the right triangle ABC.

   In right $△ABC$ (right angle at $C$), $\overline{BC}$ is opposite the ${60}^{\circ }$ angle at $A$ and the altitude $\overline{AC}$ is adjacent to it.

   $$\begin{array}{cc}\tan {60}^{\circ }& =\frac{m\overline{BC}}{m\overline{AC}}\\ \sqrt{3}& =\frac{5}{m\overline{AC}}\\ m\overline{AC}& =\frac{5}{\sqrt{3}}\\ & =\frac{5\sqrt{3}}{3}\\ & \approx 2.89\text{ cm}\end{array}$$

Solution

1. Draw the right triangle.

   Let $A$ be the point on the ground, $B$ the foot of the tree and $C$ its top, with $\angle B={90}^{\circ }$.

   $m\overline{BC}=72\text{ m},m\overline{AB}=100\text{ m},m\angle A=\theta$

2. Use the tangent ratio and its inverse.

   Height is opposite $\theta$, the ground distance is adjacent.

   $$\begin{array}{cc}\tan \theta & =\frac{72}{100}\\ & =0.72\\ \theta & ={\tan }^{-1}(0.72)\\ & \approx {35.75}^{\circ }\end{array}$$

Solution

1. Draw the right triangle.

   Let $A$ be the foot of the ladder, $B$ the foot of the wall and $C$ the top of the ladder on the wall, with $\angle B={90}^{\circ }$. The ladder $\overline{AC}$ is the hypotenuse.

   $m\overline{BC}=10\text{ m},m\angle A={60}^{\circ },m\overline{AC}=\ell$

2. Use the sine ratio.

   The wall height is opposite the ${60}^{\circ }$ angle and the ladder is the hypotenuse.

   $$\begin{array}{cc}\sin {60}^{\circ }& =\frac{10}{\ell }\\ \ell & =\frac{10}{\sin {60}^{\circ }}\\ & =\frac{10}{\sqrt{3}/2}\\ & =\frac{20}{\sqrt{3}}\\ & \approx 11.55\text{ m}\end{array}$$

Solution

1. Turn the angle of depression into an angle of elevation.

   Let $C$ be the top of the tower, $B$ its foot at sea level and $A$ the ship, with $\angle B={90}^{\circ }$. The horizontal line through $C$ is parallel to $\overline{AB}$, so the angle of depression at $C$ equals the alternate angle at $A$.

   $m\angle A={60}^{\circ },m\overline{BC}=150\text{ m},m\overline{AB}=d$

2. Use the tangent ratio.

   The tower is opposite $\angle A$ and the required distance is adjacent.

   $$\begin{array}{cc}\tan {60}^{\circ }& =\frac{150}{d}\\ d& =\frac{150}{\sqrt{3}}\\ & =50\sqrt{3}\\ & \approx 86.60\text{ m}\end{array}$$

Solution

1. Set up the figure.

   Let $\overline{CD}$ be the pole of height $h$ with foot $C$. The first point is $A$, the second point $B$ lies between $A$ and $C$ with $m\overline{AB}=100$ m. Let $m\overline{BC}=x$.

   $m\angle A={15}^{\circ },m\angle DBC={30}^{\circ },m\overline{AC}=100+x$

2. Express x in terms of h from triangle BCD.

   In right $△BCD$, the height is opposite ${30}^{\circ }$ and $x$ is adjacent.

   $$\begin{array}{cc}\tan {30}^{\circ }& =\frac{h}{x}\\ \frac{1}{\sqrt{3}}& =\frac{h}{x}\\ x& =\sqrt{3}h\end{array}$$

3. Use triangle ACD.

   In right $△ACD$, the height is opposite ${15}^{\circ }$ and $100+x$ is adjacent. Take $\tan {15}^{\circ }\approx 0.2679$ and substitute $x=\sqrt{3}h$.

   $$\begin{array}{cc}\tan {15}^{\circ }& =\frac{h}{100+\sqrt{3}h}\\ h& =\tan {15}^{\circ }(100+\sqrt{3}h)\\ h-\sqrt{3}\tan {15}^{\circ }h& =100\tan {15}^{\circ }\\ h& =\frac{100\tan {15}^{\circ }}{1-\sqrt{3}\tan {15}^{\circ }}\\ & =\frac{26.79}{1-0.4641}\\ & =\frac{26.79}{0.5359}\\ & \approx 50\text{ m}\end{array}$$

4. Check with the exact value.

   Since $\tan {15}^{\circ }=2-\sqrt{3}$, the denominator is $1-\sqrt{3}(2-\sqrt{3})=4-2\sqrt{3}=2(2-\sqrt{3})$, so the height is exactly 50 m.

   $$\begin{array}{cc}h& =\frac{100(2-\sqrt{3})}{2(2-\sqrt{3})}\\ & =50\text{ m}\end{array}$$

Solution

1. Draw the right triangle.

   Let $\overline{BC}$ be the tower with foot $B$ and top $C$, and let $A$ be the tip of the shadow, so $m\overline{AB}=450$ m. The Sun's angle of elevation is $\theta =m\angle A$.

   $m\overline{BC}=300\text{ m},m\overline{AB}=450\text{ m}$

2. Use the tangent ratio and its inverse.

   The tower is opposite $\theta$ and the shadow is adjacent.

   $$\begin{array}{cc}\tan \theta & =\frac{300}{450}\\ & =\frac{2}{3}\\ \theta & ={\tan }^{-1}\left(\frac{2}{3}\right)\\ & \approx {33.69}^{\circ }\end{array}$$

Solution

1. Set up the figure.

   Let $\overline{CD}$ be the cliff of height $h$ with foot $C$. The first point is $A$; after walking 100 m we reach $B$, and $m\overline{BC}=x$.

   $m\angle A={25}^{\circ },m\angle DBC={45}^{\circ },m\overline{AC}=100+x$

2. Use triangle BCD.

   In right $△BCD$, $\tan {45}^{\circ }=1$, so the distance equals the height.

   $$\begin{array}{cc}\tan {45}^{\circ }& =\frac{h}{x}\\ 1& =\frac{h}{x}\\ x& =h\end{array}$$

3. Use triangle ACD.

   In right $△ACD$, substitute $x=h$ and take $\tan {25}^{\circ }\approx 0.4663$ (4 d.p.).

   $$\begin{array}{cc}\tan {25}^{\circ }& =\frac{h}{100+h}\\ 0.4663(100+h)& =h\\ 46.63+0.4663h& =h\\ 46.63& =0.5337h\\ h& =\frac{46.63}{0.5337}\\ & \approx 87.37\text{ m}\end{array}$$

Solution

1. Set up the figure.

   Let $\overline{CD}$ be the hill with top $D$ and foot $C$. Let $B$ be the point on the nearer shore and $A$ the point on the far shore, all on one line with $C$. By alternate angles with the horizontal through $D$, the angles of depression equal the angles of elevation at $B$ and $A$. Let the river width be $m\overline{AB}=x$ and the distance from the foot of the hill be $m\overline{BC}=y$.

   $m\angle DBC={70}^{\circ },m\angle A={50}^{\circ },m\overline{CD}=300\text{ m}$

2. Find y from triangle BCD.

   In right $△BCD$, the hill is opposite ${70}^{\circ }$ and $y$ is adjacent. Take $\tan {70}^{\circ }\approx 2.7475$.

   $$\begin{array}{cc}\tan {70}^{\circ }& =\frac{300}{y}\\ y& =\frac{300}{\tan {70}^{\circ }}\\ & =\frac{300}{2.7475}\\ & \approx 109.19\text{ m}\end{array}$$

3. Find x + y from triangle ACD.

   In right $△ACD$, the hill is opposite ${50}^{\circ }$ and $x+y$ is adjacent. Take $\tan {50}^{\circ }\approx 1.1918$.

   $$\begin{array}{cc}\tan {50}^{\circ }& =\frac{300}{x+y}\\ x+y& =\frac{300}{1.1918}\\ & \approx 251.73\text{ m}\end{array}$$

4. Subtract to get the width.

   The width is the whole distance minus the distance to the nearer shore.

   $$\begin{array}{cc}x& =251.73-109.19\\ & =142.54\text{ m}\end{array}$$

Solution

1. Draw the right triangle.

   Let $A$ be the hand, $C$ the kite and $B$ the point directly below the kite at the level of the hand, so $\angle B={90}^{\circ }$. The tight string $\overline{AC}$ is the hypotenuse.

   $m\overline{AC}=120\text{ m},m\angle A={50}^{\circ },m\overline{BC}=h$

2. Use the sine ratio.

   The height is opposite the ${50}^{\circ }$ angle and the string is the hypotenuse. Take $\sin {50}^{\circ }\approx 0.7660$.

   $$\begin{array}{cc}\sin {50}^{\circ }& =\frac{h}{120}\\ h& =120\sin {50}^{\circ }\\ & =120(0.7660)\\ & \approx 91.93\text{ m}\end{array}$$