A triangle has sides A, B and C. The side BC has length 20cm, the angle ABC is 50 deg and angle BAC is 68 deg. a) Show that the length of AC is 16.5cm, correct to three significant figures. b) The midpoint of BC is M, hence find the length of AM

a) The answer to this question is quite simple, it is all dependent on the pupil realising that they need to use the sine rule. This rule for this triangle is: BC/sin(BAC) = AC/sin(ABC) Once they get this equation then it is a simple case of rearranging the equation for AC: AC = BC * sin(ABC) / sin(BAC) Then they just need to plug in the numbers that are given in the question: AC = 20 * sin(50) / sin(68) AC = 16.5cm b) From the previous part we know that AC = 16.5 cm and we can figure out from the information in the question that angle ACB = 62 deg. We then have a new triangle of sides A, M and C. As M is the midpoint of BC, the length of MC is 10cm. We now have two sides and the angle in between them, hence we need to use the cosine rule to find AM. The cosine rule here is: AM^2 = MC^2 + AC^2 - (2 * MC * AC * cos(ACB)) AM^2 = 10^2 + 16.5^2 - (2 * 10 * 16.5 * cos(62)) This gives a length of AM = 14.7 cm