Question

In ΔABC, m∠ACB = 90°, CD ⊥ AB and m∠ACD = 45°. Find AC, if CD = 6 sqrt 3

Answer 1

Answer:

$6\sqrt{6}\ un.$

Step-by-step explanation:

Consider right triangle ABC with right angle ACB. If m∠ACD = 45°, then

m∠BCD=m∠ACB-m∠ACD=90°-45°=45°.

Since  CD ⊥ AB, then

m∠CDA=m∠CDB=90°.

Thus, triangles ACD and BCD are right triangles with right angles ADC and BDC. In these triangles:

m∠CAD=90°-m∠ACD=90°-45°=45°;

m∠CBD=90°-m∠BCD=90°-45°=45°.

Thus, triangles ACD and BCD are isosceles right triangles with

$CD=AD=BD=6\sqrt{3}\ un.$

By the Pythagorean theorem,

$AC^2=CD^2+AD^2,\\ \\AC^2=(6\sqrt{3})^2+(6\sqrt{3})^2,\\ \\AC^2=216,\\ \\AC=6\sqrt{6}\ un.$

Answer 2

Answer:

$AC =6\sqrt{6}$

Step-by-step explanation:

In ΔACD,

AC is the hypotenuse.

CD is the adjacent side to the angle 45°.

So, $cos 45=\frac{CD}{AC}$

$\frac{1}{\sqrt{2} } =\frac{6\sqrt{3} }{AC}$

$AC=6\sqrt{3} \sqrt{2}$

$=6\sqrt{6}$