10.1 Angle of Elevation and Angle of Depression (Part 2)

Posted on May 27, 2020 by Myhometuition

3. The angle of elevation or angle of depression is always measured from the horizontal line.

Example 3:

Diagram below shows two vertical poles JK and NL on a horizontal plane. M is a point on NL such that JK = ML.

The angle of depression of point J from point N is

Solution:

The angle of depression of point J from N is the angle between line JN and the horizontal line through N.

Angle of depression of J from N

= angle of elevation of N from J
=∠NJM

10.2 Solving Problems Involving Angle of Elevation and Angle of Depression

Posted on April 23, 2020 by user

Example:

The angle of depression of a cycling kid measured from a hill with 10.9 m high is 52^o. When the kid cycles along the hillside and stops, the angle of depression becomes 25.3^o. What is the distance cycled by the kid along the hillside?

Solution:

Step 1: Draw a diagram to represent the situation.

Step 2: Devise a plan.

Find the lengths of QS and QR. Then, QS – QR = distance cycled by the kid.

\begin{array}{l} \tan 52^{o} = \frac{10.9}{Q R} \\ Q R = \frac{10.9}{\tan 52^{o}} \\ Q R = 8.5 m \end{array}

\begin{array}{l} \tan {25.3}^{o} = \frac{10.9}{Q S} \\ Q S = \frac{10.9}{\tan {25.3}^{o}} \\ Q S = 23.1 m \end{array}

QS – QR = (23.1 – 8.5) m = 14.6 m

Therefore the kid has cycled 14.6 metres.

SPM Practice (Short Questions)

Posted on April 23, 2020 by user

Question 1:

Diagram below shows two vertical poles on a horizontal plane. J, K, L, M and N are five points on the poles such that KL = MN.

Name the angle of elevation of point J from point M.

Solution:

Angle of elevation of point J from point M = ∠KMJ

Question 2:

Diagram below shows two vertical poles JM and KL, on a horizontal plane.

The angle of depression of vertex K from vertex J is 42^o.

Calculate the angle of elevation of vertex K from M.

Solution:

JN = 10 tan 42^o = 9.004 m

NM = 25 – 9.004 = 15.996 m

KL = NM = 15.996 m

\begin{array}{l} \tan ∠ K M L = \frac{K L}{M L} \\ = \frac{15.996}{10} \\ = 1.5996 \\ ∠ K M L = \tan^{- 1} 1.5996 \\ = 57^{o} 59^{'} \end{array}

Question 3:

Diagram below shows a vertical pole KMN on a horizontal plane. The angle of elevation of M from L is 20^o.

Calculate the height, in m, of the pole.

Solution:

\begin{array}{l} \tan 20^{o} = \frac{K M}{K L} \\ 0.3640 = \frac{K M}{15} \\ K M = 5.4 m \\ Height of the pole = 9 + 5.4 = 14.4 m \end{array}

10.1 Angle of Elevation and Angle of Depression (Part 1)

Posted on April 23, 2020 by user

10.1 Angle of Elevation and Angle of Depression (Part 1)

1. Angle of elevation is the acute angle measured from a horizontal line up to the line of sight.

Example 1:

Angle of elevation of object O from P = ∠OPA

2. Angle of depression is the acute angle measured down from a horizontal line to the line of sight.

Example 2:

Angle of depression of object O from P = ∠OPA

SPM Practice (Short Questions)

Posted on April 23, 2020 by user

Question 4:

Diagram below shows three vertical poles JP, KN and LM, on a horizontal plane.

The angle of elevation of N from P is 15^o.

The angle of depression of M from N is 35^o.

Calculate the distance, in m, from K to L.

Solution:

\begin{array}{l} \tan ∠ A P N = \frac{A N}{P A} \\ \tan 15^{o} = \frac{A N}{4} \\ A N = 4 \times 0.268 \\ A N = 1.072 m \\ Length of B N = 2 + 1.072 = 3.072 c m \\ \tan ∠ B M N = \frac{B N}{B M} \\ \tan 35^{o} = \frac{3.072}{B M} \\ B M = \frac{3.072}{0.700} = 4.389 \\ Distance of K L = 4.389 m \end{array}

Question 5:

Diagram below shows two vertical poles JM and LN, on a horizontal plane.

The angle of elevation of M from K is 70^oand the angle of depression of K from N is 40^o.

Find the difference in distance, in m, between JK and KL.

Solution:

\begin{array}{l} \tan ∠ J K M = \frac{14}{J K} \\ J K = \frac{14}{\tan 70^{o}} \\ J K = 5.096 m \\ \tan ∠ L K N = \frac{8}{K L} \\ K L = \frac{8}{\tan 40^{o}} \\ K L = 9.534 m \\ Difference in distance of J K and K L \\ = 9.534 - 5.096 \\ = 4.438 m \end{array}