In the shape below, the length of BC is 3.1 cm and the length of CD is 5.9cm
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.
We can solve this problem using the Law of Cosines.
a)
To find the size of angle BDC, we will use the Law of Cosines for triangle BDC.
Let's denote angle BDC as θ.
Using the Law of Cosines:
CD^2 = BC^2 + BD^2 - 2(BC)(BD)cos(θ)
5.9^2 = 3.1^2 + BD^2 - 2(3.1)(BD)cos(θ)
34.81 = 9.61 + BD^2 - 6.2BDcos(θ)
25.2 = BD^2 - 6.2BDcos(θ)
Also, BDcos(θ) = BC which is equal to 3.1
25.2 = BD^2 - 6.2(3.1)
25.2 = BD^2 - 18.62
BD^2 = 25.2 + 18.62
BD^2 = 43.82
BD = √43.82
BD ≈ 6.62 cm
Now we can use the Law of Cosines again to find angle BDC:
cos(θ) = (BC^2 + CD^2 - BD^2) / (2*BC*CD)
cos(θ) = (3.1^2 + 5.9^2 - 6.62^2) / (2*3.1*5.9)
cos(θ) = (9.61 + 34.81 - 43.82) / 36.58
cos(θ) = 0.478
θ ≈ cos^(-1)(0.478)
θ ≈ 61.6 degrees
b)
Given that AC is twice the length of BC, we know that AC = 2(3.1) = 6.2 cm.
To find the size of angle ADC, we can use a similar approach as above:
cos(θ) = (CD^2 + AC^2 - AD^2) / (2*CD*AC)
cos(θ) = (5.9^2 + 6.2^2 - 6.62^2) / (2*5.9*6.2)
cos(θ) = (34.81 + 38.44 - 43.82) / 72.98
cos(θ) = 0.252
θ ≈ cos^(-1)(0.252)
θ ≈ 75.5 degrees
Therefore, the size of angle ADC is approximately 75.5 degrees.