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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.

 Oct 3, 2019
 #1
avatar+8879 
+3

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.)

 

\(tan (\alpha_C)=\frac{3}{h} \ (Part\ of\ \alpha \ on\ page\ C)\\ tan (\alpha_B)=\frac{2}{h} \ (Part\ of\ \alpha \ on\ page\ B)\\ arc\ tan \frac{3}{h}+arc\ tan \frac{3}{h}-45°=0 \)

     \( \alpha_ {\small B}\)      +         \(\alpha_ {\small C}\)         \(-45°=0\) 

\(h= 6\)

 

\(A=\frac{1}{2}\cdot a\cdot h\\ A=\frac{1}{2}\cdot (3+2)\cdot6\)

\(A=15\)

laugh  !

 Oct 3, 2019
edited by asinus  Oct 3, 2019
 #2
avatar+106539 
+3

 

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

 

cool cool cool

 Oct 3, 2019
edited by CPhill  Oct 3, 2019

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