Given: ∆abc, ab = 12, ac = 17 area ∆abc = 65 find: bc, m∠a, m∠b, m∠c

Step-by-step explanation:

There are at least a couple of ways you could go at this. Here, we'll use an area formula to find m∠A, then use the law of cosines to find BC. Using BC, we can use the law of sines to find another angle.

  Area = (1/2)·AB·AC·sin(∠A)

  65·2/(12·17) = sin(∠A) ≈ 65/102

  ∠A = arccos(65/102) ≈ 39.587°

From the law of cosines, ...

  BC² = AB² +AC² -2·AB·AC·cos(∠A)

  BC² = 12² +17² -2·12·17·cos(39.587°) ≈ 118.5735

  BC ≈ √118.5735 ≈ 10.889

Then ∠C can be found from the law of sines:

  sin(∠C)/AB = sin(∠A)/BC

  ∠C = arcsin(AB/BC·sin(∠A)) ≈ 44.609°

The measure of ∠B will be the angle that makes the total be 180°:

  39.587° +∠B +44.609° = 180°

  ∠B = 95.804°

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There is actually another solution, in which ∠A is obtuse. We thought the diagram showed an acute triangle, so we didn't investigate the other alternative. The above calculations show the triangle is obtuse in any event.

See the second attachment for the other solution.