In the shape below, the length of BC is 4.1cm and the length of CD is 5.3cm.

Question

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

a) To find the size of angle BDC, we can use the cosine rule.
Let's denote angle BDC as x. Then we have:

cos(x) = (BC^2 + CD^2 - BD^2) / (2 * BC * CD)
cos(x) = (4.1^2 + 5.3^2 - 4.1^2) / (2 * 4.1 * 5.3)
cos(x) = (16.81 + 28.09 - 16.81) / (8.63)
cos(x) = 28.09 / 44.73
cos(x) = 0.628

x = cos^(-1)(0.628)
x ≈ 51.3 degrees

Therefore, the size of angle BDC is approximately 51.3 degrees.

b) Since the length of AC is twice the length of BC, we have:
AC = 2 * BC
AC = 2 * 4.1
AC = 8.2 cm

Now, to find the size of angle ADC, we can use the cosine rule again. Let's denote angle ADC as y. Then we have:

cos(y) = (AC^2 + CD^2 - AD^2) / (2 * AC * CD)
cos(y) = (8.2^2 + 5.3^2 - 4.1^2) / (2 * 8.2 * 5.3)
cos(y) = (67.24 + 28.09 - 16.81) / (17.26)
cos(y) = 78.52 / 87.68
cos(y) = 0.895

y = cos^(-1)(0.895)
y ≈ 26.8 degrees
Therefore, the size of angle ADC is approximately 26.8 degrees.