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Help with this please

Help with this please-example-1

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AB = 6 cm, AC = 12 cm, CD = ?

In triangle ABC, ∠CBA = 90°, therefore in triangle BCD ∠CBD = 90° also.

Since ∠BDC = 55°, ∠CBD = 90°, and there are 180 degrees in a triangle, we know ∠DCB = 180 - 55 - 90 = 35°

In order to find ∠BCA, use the law of sines:

sin(∠BCA)/BA = sin(∠CBA)/CA
sin(∠BCA)/6 cm = sin(90)/12 cm
sin(∠BCA) = 6*(1)/12 = 0.5
∠BCA = arcsin(0.5) = 30° or 150°
We know the sum of all angles in a triangle must be 180°, so we choose the value 30° for ∠BCA

Now add ∠BCA (30°) to ∠DCB = 35° to find ∠DCA.
∠DCA = 30 + 35 = 65°

Since triangle DCA has 180°, we know ∠CAD = 180 - ∠DCA - ∠ADC = 180 - 65 - 55 = 60°

In triangle DCA we now have all three angles and one side, so we can use the law of sines to find the length of DC.

12cm/sin(∠ADC) = DC/sin(∠DCA)
12cm/sin(55°) = DC/sin(60°)
DC = 12cm*sin(60°)/sin(55°)
DC = 12.686 cm
User Lee Mighdoll
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