Chapter 10: Distance and Elevation

A straight line drawn on the horizontal plane is called a—

A straight line parallel to the horizon is also called a—

A line perpendicular to the horizontal plane is called a—

A vertical line is also known as a—

When a horizontal line and a vertical line intersect at right angles on a plane, the plane they define is called the—

The vertical plane is perpendicular to the—

In the standard tree figure with $AB$ vertical and $BC$ on the ground, the segment $BC$ represents the—

In the same figure, $AB$ represents the—

The plane formed by points $A$, $B$, $C$ where $AB$ is the vertical tree and $BC$ is on the ground is a—

Two horizontal lines drawn in the same plane are always—

"Normal line" with respect to the horizontal plane means a line that is—

A flagpole standing erect on level ground best represents a—

The angle formed at the observer's eye by the line of sight to a point above the horizontal line is called the—

The angle of elevation is always measured from the—

In the textbook figure where $O$ lies on the horizontal line $AB$ and $P$ lies above $AB$, the angle of elevation of $P$ at $O$ is—

Looking up at the top of a tower from a point on the ground produces an angle of—

The theoretical maximum value of an angle of elevation is—

If the angle of elevation of an object is $0^\circ$, the object lies on the—

As an observer moves towards the foot of a tower (on level ground), the angle of elevation to its top—

The angle of elevation of an object is $90^\circ$ when the object is—

If the sun is directly overhead, the angle of elevation of the sun is—

The angle of elevation of the top of a hill is greater for an observer who is—

The angle formed at the observer's eye by the line of sight to a point below the horizontal line is called the—

From the top of a tower, the angle to a point on the ground is an angle of—

An aeroplane flying high looks down at a building on the ground; the angle from the plane is one of—

The angle of depression from a higher point and the angle of elevation back to that point are equal because they form—

If the angle of depression of a point from the top of a tower is $30^\circ$, the angle of elevation of the tower top from that point is—

As an observed point on the ground moves further from the foot of a tall building, the angle of depression from the top—

The angle of depression equals $0^\circ$ when the line of sight is—

The angle of depression to the foot of a vertical pole, observed from directly above its top, is—

While drawing a $30^\circ$ angle for a problem, the figure should approximately satisfy—

While drawing a $45^\circ$ angle, the figure should approximately satisfy—

While drawing a $60^\circ$ angle, the figure should approximately satisfy—

In a right triangle with one acute angle $30^\circ$, the side opposite to that angle is—

In a right triangle with one acute angle $60^\circ$, the side opposite to that angle is—

In a $45^\circ$–$45^\circ$–$90^\circ$ triangle, the two legs are—

$\tan 30^\circ = ?$

$\tan 60^\circ = ?$

$\tan 45^\circ = ?$

$\sin 30^\circ = ?$

$\cos 60^\circ = ?$

$\sin 45^\circ = ?$

$\sin 60^\circ = ?$

$\cos 30^\circ = ?$

$\cos 45^\circ = ?$

$\sin 90^\circ = ?$

With $AB$ the height of a tower, $BC=75$ m the horizontal distance, and $\angle ACB = 30^\circ$, the equation $\tan 30^\circ$ equals—

Therefore the height $h$ satisfies—

Numerically $75 \cdot \tan 30^\circ$ equals—

After rationalizing, $\dfrac{75}{\sqrt{3}}$ becomes—

The approximate height of the tower in Example 1 is—

To rationalize $\dfrac{75}{\sqrt{3}}$, the numerator and denominator are multiplied by—

In Example 2, with tree $AB=105$ m, foot distance $BC=x$, and $\angle ACB=60^\circ$, $\tan 60^\circ$ equals—

The distance $x$ satisfies—

$\dfrac{105}{\sqrt{3}}$ simplifies to—

The approximate ground distance of the point in Example 2 is—

The relation $\tan 60^\circ = \sqrt{3}$ tells us that the side opposite the $60^\circ$ angle is $\sqrt{3}$ times the—

With the same height $105$ m, if the elevation were $30^\circ$ instead of $60^\circ$, the foot distance would be—

A ladder of length $18$ m leans against a wall making $45^\circ$ with the ground; with $AC=18$ m and wall height $AB=h$, $\sin 45^\circ$ equals—

Therefore the wall height $h$ equals—

$18 \cdot \sin 45^\circ$ in surd form equals—

The approximate wall height in Example 3 is—

The horizontal distance from the foot of the wall to the foot of the ladder is—

In a $45^\circ$–$45^\circ$–$90^\circ$ right triangle, the two legs are—

The stick contacts the tree at $AB=7$ m above the ground; the depression angle at the contact is $30^\circ$. By alternate angles, $\angle ACB$ equals—

From right $\triangle ABC$ with stick $BC$ as hypotenuse, $\sin \angle ACB$ equals—

Therefore $BC = \dfrac{AB}{\sin 30^\circ} = \dfrac{7}{1/2}$ equals—

The length of the stick in Example 4 is—

The reasoning that the depression angle at $B$ equals $\angle ACB$ at $C$ uses the fact that the two horizontal lines at $B$ and at $C$ are—

Angle of depression from a higher point and angle of elevation from a lower point form—

Building height $AB=h$, $BC=x$, with $\angle ACB=60^\circ$. Then $\tan 60^\circ$ equals—

From the equation $\tan 60^\circ = \sqrt{3}$, we get $h = ?$

Moving back $CD=42$ m to point $D$ gives elevation $\angle ADB = 45^\circ$, so $\tan 45^\circ$ equals—

Therefore $h$ also equals—

Combining $\sqrt{3}\,x = x + 42$ gives $x = ?$

Hence $h = \dfrac{42\sqrt{3}}{\sqrt{3}-1}$ approximately equals—

Rationalizing, $\dfrac{42\sqrt{3}}{\sqrt{3}-1}$ is equivalent to—

As the observer moves backwards on level ground, the angle of elevation of a fixed building top—

The pole stands $BC=x$, breaks without separation, and the broken piece $CD$ slopes down to $D$ at distance $BD=10$ m, with $\angle BCD=30^\circ$. From $\triangle BCD$, $\tan 30^\circ$ equals—

Solving for the standing piece, $x$ equals—

Numerically $x$ equals—

From the same triangle, $\sin 30^\circ$ equals—

So the broken piece $CD = \dfrac{BD}{\sin 30^\circ} = \dfrac{10}{1/2}$ equals—

The total height of the pole is $x + CD = 10\sqrt{3} + 20$ approximately—

The unbroken (still standing) piece measures approximately—

The slanted (broken) piece measures exactly—

A balloon flies above a point between two posts $1$ km apart on level ground. From the balloon, the depression of the nearer post is $60^\circ$ and the farther post $30^\circ$. With balloon height $h$, the distance to the foot of the $60^\circ$ post is—

The horizontal distance to the foot of the $30^\circ$ post equals—

The two horizontal distances together equal the spacing between the posts—

Solving $\dfrac{h}{\sqrt{3}} + h\sqrt{3} = 1000$ gives $h = ?$

Approximate height of the balloon above the ground in metres—

Given $(\text{stick})^2 = \dfrac{1}{3}(\text{shadow})^2$, the ratio stick : shadow equals—

Therefore $\tan(\text{elevation})$ equals—

Hence the angle of elevation of the sun is—

If the length of a stick equals the length of its shadow, then $\tan(\text{angle})$ equals—

The corresponding angle of depression of the sun is—

This situation describes a stick whose shadow is—

With height $26$ m and elevation $30^\circ$ from a point, the ground distance $d$ satisfies—

Numerically $\dfrac{26}{1/\sqrt{3}}$ equals—

The approximate distance from the tower is—

The exact (rationalized) form of this distance is—

With base distance $20$ m and elevation $60^\circ$, the height $h$ equals—

$20\sqrt{3}$ m is approximately—

With the same $20$ m base distance, if the elevation were $30^\circ$, the height would be—

With ladder $18$ m and ground angle $45^\circ$, the height of the wall equals—

$18 \cdot \sin 45^\circ$ in simplest surd form equals—

The approximate wall height is—

From the top of a house, the angle of depression to a point on the ground $20$ m from the foot is $30^\circ$. The angle of elevation from that point to the top equals—

The equation $\tan 30^\circ = \dfrac{h}{20}$ gives $h = ?$

$\dfrac{20}{\sqrt{3}}$ is approximately—

After rationalization, $\dfrac{20}{\sqrt{3}}$ becomes—