**Question 15 (12 marks)**:

You are

**not**allowed

**to use graph paper to answer this question.**

**(a)**Diagram 11.1 shows a solid prism with a rectangular base

*JKLM*on a horizontal plane. The surface

*MQRSTL*is the uniform cross section of the prism. Triangle

*STU*and

*PQR*are horizontal planes. Edges

*PJ*,

*RS*and

*UK*are vertical.

*PQ*=

*TU*= 3 cm.

**Diagram 11.1**

Draw to full scale, the elevation of the solid on a vertical plane parallel to

*KL*as viewed from

*X*.

**Another solid cuboid with rectangle base**

(b)

(b)

*JMDE*is combined to the prism in Diagram 11.1 at the vertical plane

*JMQP*. The composite solid is as shown in Diagram 11.2. The base

*EJKLMD*lies on a horizontal plane.

**Diagram 11.2**

Draw to full scale,

**(i)**the elevation of the composite solid on a vertical plane parallel to

*EJK*as viewed from

*Y*.

**(ii)**the plan of the composite solid.

*Solution:***(a)**

**(b)(i)**

**(b)(ii)**

**Question 16 (12 marks)**:

Diagram 12 in the answer space shows the locations of points

*J*,

*L*and

*M*, which lie on the surface of the earth.

*O*is the centre of the earth. The longitude of

*M*is 30

^{o}W.

*K*is another point on the surface of the earth such that

*KJ*is the diameter of the common parallel of latitude 45

^{o}S.

**Mark and label point**

(a)(i)

(a)(i)

*K*on Diagram 12 in the answer space.

**Hence, state the longitude of point**

(ii)

(ii)

*K*.

(b)

(b)

*L*lies due north of

*M*and the shortest distance from

*M*to

*L*measured along the surface of the earth is 7500 nautical miles.

Calculate the latitude of

*L*.

**Calculate the distance, in nautical mile, from**

(c)

(c)

*K*due east to

*M*measured along the common parallel of latitude.

**An aeroplane took off from**

(d)

(d)

*K*and flew due east to

*M*along the common parallel of latitude. The average speed of the aeroplane for the flight was 750 knots.

Calculate the total time, in hour, taken for the whole flight.

Answer:

*Solution:***(a)(i)**

**(a)(ii)**

Longitude of point K = 130

^{o}W

**(b)**

$\begin{array}{l}\angle \text{}LOM\times 60=7500\\ \angle \text{}LOM=\frac{7500}{60}\\ \angle \text{}LOM={125}^{o}\\ \\ \text{Latitudeof}L={125}^{o}-{45}^{o}\\ ={80}^{o}N\end{array}$

**(c)**

*KM*= (130

^{o}– 30

^{o}) × 60 × cos 45

^{o}= 4242.64 nautical miles

**(d)**

$\begin{array}{l}\text{Time}=\frac{\text{Distance}}{\text{speed}}\\ \text{}=\frac{4242.64}{750}\\ \text{}=5.66\text{hours}\end{array}$