Question

1. find the length of the indicated side, to the nearest centimetre.
2. find the length of the indicated side, to the nearest tenth of a metre.
3. find the measure of the indicated angle, to the nearest degree.
4. find the measure of the indicated angle, to the nearest degree.
5. two wires are supporting a tent pole, as shown.

a) how far apart are the wires fixed in the ground, to the nearest tenth of a metre?
b) find the angle each wire makes with the ground, to the nearest degree. state any assumptions you make.

1. cheryl is trying to hit her golf ball between two trees. she estimates the distances shown. within what angle must cheryl make her shot, in order to pass between the trees? round to the nearest tenth of a degree.
2. solve $\triangle xyz$. round the side length to the nearest tenth of a metre and the angle measures to the nearest degree.
3. in acute $\triangle ntw$, $\angle n = 54^\circ$, $n = 2.3$ km, and $w = 1.8$ km.

a) sketch this triangle and label the given information.
b) solve this triangle. round the side length to the nearest tenth of a metre and the angle measures to the nearest degree.

1. students of fowler’s aeronautical school have this crest on their pilot’s cap. find the total length of gold trim needed to make a crest, to the nearest centimetre. state any assumptions you make.
2. use the measurements given to find the height of the peace tower in ottawa, to the nearest metre.

$ab = 50$ m
$\angle xay = 43^\circ$
$\angle xab = 60^\circ$
$\angle abx = 82^\circ$

1. tess was flying from toronto to hamilton at night. she noticed that her heading indicator was malfunctioning and decided to check her position. she called hamilton tower and st. catharines radio, knowing that the stations were 50 km apart. hamilton reported that the position of her plane formed an angle of $65^\circ$ with the line joining the two stations. st. catharines reported that her position made an angle of $48^\circ$. how far was the plane from hamilton, to the nearest kilometre?
2. scuba divers count fin strokes to estimate the distance they have travelled under water. felipe uses this method and a compass to swim 300 m north from the dive boat. he then turns right $120^\circ$ and swims 400 m. how far from the boat is he...
3. omitted as per image content
4. a fire station serves an area bordered by three highways that join the towns of west port, sackville, and jonestown, as shown.

the fire station serves a total area of $126$ km². find the number of minutes it would take for a fire truck to drive the perimeter of this region, assuming an average speed of 80 km/h.

1. from his cockpit, the pilot of a flying plane observes a jet flying...

Explanation:

Since you haven't specified a particular sub - question from the given set of problems, here's a general approach to solve a triangle - related problem (using the Law of Sines or Law of Cosines which falls under the sub - field of Geometry in Mathematics). Let's take problem 1 as an example:

Problem 1: Find the length of the indicated side, to the nearest centimetre.

We have triangle \(ABC\) with \(BC = 25\space cm\), \(\angle B=45^{\circ}\), \(\angle A = 74^{\circ}\). First, we find \(\angle C\) using the angle - sum property of a triangle (\(\angle A+\angle B+\angle C = 180^{\circ}\)).

Step 1: Find \(\angle C\)

The sum of the interior angles of a triangle is \(180^{\circ}\). So, \(\angle C=180^{\circ}-\angle A - \angle B\)
\(\angle C = 180^{\circ}-74^{\circ}-45^{\circ}=61^{\circ}\)

Step 2: Use the Law of Sines to find side \(b\) (opposite \(\angle B\))

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\), where \(a, b, c\) are the sides opposite to angles \(A, B, C\) respectively.
We know \(BC = a = 25\space cm\) (opposite \(\angle A\)), \(\angle B = 45^{\circ}\), \(\angle A=74^{\circ}\)
From \(\frac{a}{\sin A}=\frac{b}{\sin B}\), we can solve for \(b\):
\(b=\frac{a\sin B}{\sin A}\)
\(a = 25\), \(\sin B=\sin45^{\circ}=\frac{\sqrt{2}}{2}\approx0.7071\), \(\sin A=\sin74^{\circ}\approx0.9613\)
\(b=\frac{25\times\sin45^{\circ}}{\sin74^{\circ}}=\frac{25\times0.7071}{0.9613}\)
\(b=\frac{17.6775}{0.9613}\approx18.4\space cm\approx18\space cm\) (to the nearest centimetre)

If you want a solution for a different problem (such as problem 2, 3, etc.), please specify the problem number, and I will provide a detailed step - by - step solution.

Answer:

Since you haven't specified a particular sub - question from the given set of problems, here's a general approach to solve a triangle - related problem (using the Law of Sines or Law of Cosines which falls under the sub - field of Geometry in Mathematics). Let's take problem 1 as an example:

Problem 1: Find the length of the indicated side, to the nearest centimetre.

We have triangle \(ABC\) with \(BC = 25\space cm\), \(\angle B=45^{\circ}\), \(\angle A = 74^{\circ}\). First, we find \(\angle C\) using the angle - sum property of a triangle (\(\angle A+\angle B+\angle C = 180^{\circ}\)).

Step 1: Find \(\angle C\)

The sum of the interior angles of a triangle is \(180^{\circ}\). So, \(\angle C=180^{\circ}-\angle A - \angle B\)
\(\angle C = 180^{\circ}-74^{\circ}-45^{\circ}=61^{\circ}\)

Step 2: Use the Law of Sines to find side \(b\) (opposite \(\angle B\))

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\), where \(a, b, c\) are the sides opposite to angles \(A, B, C\) respectively.
We know \(BC = a = 25\space cm\) (opposite \(\angle A\)), \(\angle B = 45^{\circ}\), \(\angle A=74^{\circ}\)
From \(\frac{a}{\sin A}=\frac{b}{\sin B}\), we can solve for \(b\):
\(b=\frac{a\sin B}{\sin A}\)
\(a = 25\), \(\sin B=\sin45^{\circ}=\frac{\sqrt{2}}{2}\approx0.7071\), \(\sin A=\sin74^{\circ}\approx0.9613\)
\(b=\frac{25\times\sin45^{\circ}}{\sin74^{\circ}}=\frac{25\times0.7071}{0.9613}\)
\(b=\frac{17.6775}{0.9613}\approx18.4\space cm\approx18\space cm\) (to the nearest centimetre)

If you want a solution for a different problem (such as problem 2, 3, etc.), please specify the problem number, and I will provide a detailed step - by - step solution.