Question

In the shape below, the length of BC is 4.4cm and the length of CD is 5.2cm
a) Calculate the size of angle BDC.
b) The length of AC is twice the length of BC. Calculate the size of angle ADC.
Give each of your answers in degrees to 1 d.p.

Answer 1

To solve this problem, we can use the cosine rule.

a) To find the size of angle BDC, we can use the cosine rule on triangle BDC:
cos(BDC) = (BC^2 + CD^2 - BD^2) / (2*BC*CD)
cos(BDC) = (4.4^2 + 5.2^2 - 4.4^2) / (2*4.4*5.2)
cos(BDC) = (19.36 + 27.04 - 19.36) / (45.76)
cos(BDC) = 26.72 / 45.76
cos(BDC) = 0.5835

Taking the inverse cosine of 0.5835, we get:
BDC = cos^(-1)(0.5835)
BDC = 54.56 degrees

b) To find the size of angle ADC, we can first find the length of AC:
AC = 2*BC = 2*4.4 = 8.8cm

Now, we can use the cosine rule on triangle ADC:
cos(ADC) = (AC^2 + CD^2 - AD^2) / (2*AC*CD)
cos(ADC) = (8.8^2 + 5.2^2 - 4.4^2) / (2*8.8*5.2)
cos(ADC) = (77.44 + 27.04 - 19.36) / 91.52
cos(ADC) = 85.12 / 91.52
cos(ADC) = 0.9297

Taking the inverse cosine of 0.9297, we get:
ADC = cos^(-1)(0.9297)
ADC = 21.30 degrees

Therefore, angle BDC is 54.56 degrees and angle ADC is 21.30 degrees.