SOLUTION: Given: △ABC with m∠A = 45°, m∠B = 30°, and BC = 14 Find: AB Draw altitude CD from C to AB. Determine the measures of ∠DCA and ∠DCB in degrees. m∠DCA =

Question 1193627: Given:
△ABC with m∠A = 45°, m∠B = 30°, and
BC = 14
Find: AB
Draw altitude
CD
from C to
AB.
Determine the measures of ∠DCA and ∠DCB in degrees.
m∠DCA =
°
m∠DCB =
°
Find the exact lengths of
CD, AD, and DB.
CD =

AD =

DB =

Find AB. Give both an exact solution and an approximate solution to two decimal places.
exact AB =

approximate AB =

Found 3 solutions by Boreal, Solver92311, greenestamps:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
a/sin A=c/sin C and angle C is 105deg. Side c is AB.
14/sin45=x/sin 105
x=14*sin 105/sin45
=19.12
-
Altitudes are perpendicular so triangle ADC is a right triangle and angle DCA is 45 degrees and angle DCB is 60 degrees.
-
Get AC or side b as well: 14/sin 45=b/sin 30
sin 30 is 0.5
b=14 sin30/sin45=7/sqrt(2)/2 or 7 sqrt(2)=9.90, which is the hypotenuse.
The legs are each hyp/sqrt(2) or 7 The altitude is 7
That is also AD
So DB is 12.12
the right triangles are 7/7/9.9 and 12.12/7/14
-
AD=x and DB is 19.12-x
ACD is a right triangle, so x^2+alt^2=7^2=49
(19-x)^2+49=196
(19-x)^2=147
19-x=12.12 AB approx
the exact value is sqrt(147)AB exact

Answer by Solver92311(821)   (Show Source): You can put this solution on YOUR website!

Illustration for the above solution.

.

John

My calculator said it, I believe it, that settles it

From
I > Ø

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

Use the figure shown by one of the tutors to solve the problem without the complicated trigonometry and decimal approximations shown in the response from the other tutor.

When altitude CD is drawn, triangle BCD is a 30-60-90 right triangle with hypotenuse BC=14; that makes CD=7 and BD=7*sqrt(3).

And triangle ACD is a 45-45-90 right triangle with CD=7, so AD=7 also.

And then AB = AD+DB = 7+7*sqrt(3).

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