Question

Triangle ABC has sides AB= 17cm, AC= 13cm and BC= 23cm, as shown

below.
13 cm
23cm
17 cm
Diagram not drawn to scale
Calculate the size of angle CAB to the nearest integer.
(3 marks)

Answer 1

The measure of angle CAB is approximately 108°.

Step-by-step explanation:

To find the measure of angle CAB, we can use the Law of Cosines. The Law of Cosines states that in a triangle with sides a, b, and c, and opposite angles A, B, and C respectively, the following equation holds: c^2 = a^2 + b^2 - 2ab * cos(C).

In this case, we know that side AB is 17cm, side AC is 13cm, and side BC is 23cm. Let's substitute these values into the equation:

23^2 = 17^2 + 13^2 - 2 * 17 * 13 * cos(C)

529 = 289 + 169 - 442 * cos(C)

442 * cos(C) = -71

cos(C) = -71/442

C = arccos(-71/442)

C ≈ 108° (to the nearest integer).

Answer 2

`\boxed{m \angle CAB =99^(\circ)}`

Step-by-step explanation:

We can use the law of cosines to determine measure of ∠CAB

If a, b and c are the three sides of a triangle and C is the angle opposite side c then the law of cosines says
c² = a² + b² - 2ab cos(C)

Here we are asked to find m∠CAB

The side opposite ∠CAB is BC which is 23 cm long. This is c in the formula

The other two sides, a and b are 13 cm and 17 cm

Applying the formula, using C to represent m∠CAB

`23^2=\:13^2+\:17^2-\:2\:\cdot 13\cdot 17\cdot \cos \left(C\right)`

First switch sides:

`13^2+17^2-2\cdot \:13\cdot \:17\cos \left(C\right)=23^2`

`\rightarrow 69+289-442\cos \left(C\right)=529\\\\\rightarrow 458 -4 42\cos \left(C\right) = 529\\`

`\rightarrow -442\cos \left(C\right) = 529-458\\\\\rightarrow -442\cos \left(C\right) = 71\\\\`

`\rightarrow \cos \left(C\right)=-(71)/(442)\\\\\rightarrow \cos \left(C\right) = - 0.16063\\\\`

`\rightarrow C = \cos^(-1) (-0.16063)`

`\rightarrow C = 99.2437^(\circ)`

Rounded to the nearest integer, this would become

`\rightarrow C = \boxed{99^(\circ)}`