In acute triangle ABC, \(\angle A=45^\circ\) Let D be the foot of the altitude from A to BC if BD=2 and CD=3 then find the area of triangle ABC.
In acute triangle ABC, \(\angle \alpha=45^\circ \) Let D be the foot of the altitude from A to BC if BD=2 and CD=3 then find the area of triangle ABC.
(Im spitzen Dreieck ABC,\(\angle \alpha=45^\circ \) sei D der Fuß der Höhe von A nach BC, wenn BD = 2 und CD = 3, dann finde die Fläche des Dreiecks ABC.)
\(tan (\alpha_C)=\frac{3}{h} \ (Part\ of\ \alpha \ on\ page\ C)\\ tan (\alpha_B)=\frac{2}{h} \ (Part\ of\ \alpha \ on\ page\ B)\\ arc\ tan \frac{3}{h}+arc\ tan \frac{3}{h}-45°=0 \)
\( \alpha_ {\small B}\) + \(\alpha_ {\small C}\) \(-45°=0\)
\(h= 6\)
\(A=\frac{1}{2}\cdot a\cdot h\\ A=\frac{1}{2}\cdot (3+2)\cdot6\)
\(A=15\)
!
Let A be the altitude, AD, of ABC
Let one base angle = x
Then the other base angle is (180 - 45 - x) = 135 - x
So.....
tan x = A/2 (1)
tan (135 -x) = A/3 (2)
Using a trig identity to simplify (2) and subbing in (1)
tan 135 - tan x A
_____________ = ____
1 + tan135 tan x 3
-1 - A/2 A
__________ = ____
1 + (-1) A/2 3
-1 - A/2 A
_______ = _____ cross-multiply
1 - A/2 3
-3 ( 1 + A/2) = A (1 - A/2)
- 3 -(3/2)A = A - (1/2)A^2
(1/2)A^2 - (5/2)A - 3 = 0
A^2 - 5A - 6 = 0 factor
(A - 6) ( A + 1) = 0
Assuming that A is positive.....then A = 6
So....the area of ABC =
(1/2)(BD + DC) (A) =
(1/2)(2 + 3) (6) =
(1/2) (5) (6) =
15