Question

ASAP! triangle ACD is an isosceles triangle. the height is 12. the base is 10. Round your angle measures to the nearest tenth. Find the length of DC? (sorry if im asking too much. I usually dont do this but i've been doin math all day I just just get it done lol)

Answer 1

Answer:

Part 1) The length of DC is $13\ units$

Part 2) The measures of the angles in the isosceles triangle are

The base angles are 67.4° and the vertex angle is 45.2°

Step-by-step explanation:

step 1

In the right triangle BDC Find the length of DC

Applying the Pythagoras Theorem

$DC^{2}=BC^{2} +BD^{2}$

we have

$BC=AB=10/2=5\ units$

$BD=12\ units$

substitute

$DC^{2}=5^{2} +12^{2}$

$DC^{2}=169$

$DC=13\ units$

step 2

Find the measures of internal angles in the isosceles triangle ABC

we know that

∠DAC=∠DCA ------> base angles

∠ADC ------> vertex angle

Find the measure of angle DCA

In the right triangle BDC

sin(∠DCA)=BD/DC

substitute the values

sin(∠DCA)=12/13

∠DCA=arcsin(12/13)=67.4°

so

∠DAC=∠DCA=67.4°

Find the measure of angle ∠ADC

Remember that the sum of the internal angles of a triangle must be equal to 180 degrees

so

∠DAC+∠DCA+∠ADC=180°

substitute

67.4°+67.4°+∠ADC=180°

∠ADC=180°-134.8°=45.2°