A triangle has sides A, B, and C. Sides A and B have lengths of 5 and 9, respectively. The angle between A and C is `(17pi)/24` and the angle between B and C is `(pi)/8`. What is the area of the triangle?

`color(blue)("Method")`
Find the height h then use: `"area=1/2xxAxxh`
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`color(blue)("Consider the option to find C and hence h")`

Cosine Rule:
`C^2=A^2+B^2-2ABcos(/_bca)`

Sine Rul:
`C/(sin(/_bca)) = B/(sin(/_abc))=A/(sin(/_bac))`

Both require that the angle `/_bca` is determined.
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Using: Sum of internal angles of a triangle`=180^o-> (pi " radians")`

`color(blue)(=> /_bca=pi-(17pi)/24-pi/8 =pi/6 -> (30^o))`

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`color(blue)("Determining length of C")`

Using the Sine Rule (simpler: no square roots!)

`C/(sin(pi/6))=A/(sin(pi/8))`

Known: `color(white)(...)sin(pi/6)=sin(30^o)= 1/2`

`color(brown)(=>C=1/2xx5/(sin(pi/8)) ~~6.499)` to 3 decimal places
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`color(blue)("Determining height h")`

Project the line cb to d such that a vertical line from 'a' forms 'ad' and `/_bda=pi/2 ->(90^o)`

Then `/_dba= pi-(17pi)/24=(7pi)/24 ->52 1/2 " degrees"`

Using basic trig
`sin((7pi)/24) = h/C`

`=>h=Csin((7pi)/24)`

`color(brown)(h~~ 5.156)` to 3 decimal places
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`color(blue)("To determine area of triangle")`

using: area `= 1/2 xx Axx h`

`color(brown)("area "= 1/2xx 5xx5.156... ~~ 12.889" square unites"` to 3 decimal places