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# The angles: B = 350, A1 = 230 and h = 3.5cm ;

+3
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+1068

The angles: B = 350,  A1 = 230  and   h = 3.5cm ;

(The angle  A1 is located between  and  )

Find the area of a triangle  ABC.

civonamzuk  May 14, 2015

#1
+27062
+18

BD = 3.5/tan(35°)

CD = 3.5*tan(23°)

Area of triangle ABC = 1/2(3.52/tan(35°) - 3.52*tan(23°))

$${\mathtt{Area}} = \left({\frac{{{\mathtt{3.5}}}^{{\mathtt{2}}}}{{\mathtt{2}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{\underset{\,\,\,\,^{{360^\circ}}}{{tan}}{\left({\mathtt{35}}^\circ\right)}}}{\mathtt{\,-\,}}\underset{\,\,\,\,^{{360^\circ}}}{{tan}}{\left({\mathtt{23}}^\circ\right)}\right) \Rightarrow {\mathtt{Area}} = {\mathtt{6.147\: \!498\: \!292\: \!005\: \!575\: \!7}}$$

Area ≈ 6.147cm2.

Alan  May 14, 2015
#1
+27062
+18

BD = 3.5/tan(35°)

CD = 3.5*tan(23°)

Area of triangle ABC = 1/2(3.52/tan(35°) - 3.52*tan(23°))

$${\mathtt{Area}} = \left({\frac{{{\mathtt{3.5}}}^{{\mathtt{2}}}}{{\mathtt{2}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{\underset{\,\,\,\,^{{360^\circ}}}{{tan}}{\left({\mathtt{35}}^\circ\right)}}}{\mathtt{\,-\,}}\underset{\,\,\,\,^{{360^\circ}}}{{tan}}{\left({\mathtt{23}}^\circ\right)}\right) \Rightarrow {\mathtt{Area}} = {\mathtt{6.147\: \!498\: \!292\: \!005\: \!575\: \!7}}$$

Area ≈ 6.147cm2.

Alan  May 14, 2015
#2
+1068
+8

Area of a triangle ABC = area ABD - area ACD

Area ABC = 8.75cm2 - 2.6cm2

Area ABC = ≈6.15cm2

civonamzuk  May 14, 2015
#3
+90180
+13

Area of ABC =  (1/2)(BC)(AD) =

(1/2)(3.5 )(3.5)  ≈ 6.125 sq cm

Here's a pic.......

CPhill  May 14, 2015
#4
+537
+8

Hello everybody.I am fiora.And I going to do this problem with you all.

First,I think you guys only get part of the answers right.

Now,let me repeat the question, The angles: B = 35o,  A1 = 23o  and   h = 3.5cm ; (The angle  A1 is located between  b and  h ) Find the area of a triangle  ABC.

At first, what is b? b in this question can be the base of the triangle or the opposite side of the angle B.

However, if b is the base of the triangle ,then b is perpendicular (90 degrees) to the h (height),but according to given,the angle between b and h is 23 degrees,so b can not be the base.

Now,b can only be the opposite side of angle B.

Here is my picture of this situations.

In triangle ABC, the opposite side of angle B is AC

and in triangle ABE,the opposite side of angle B is AE

Look at triangle ABC,area of triangle ABC=1/2(AD*BC)=1/2[AD*(BD-CD)]

AD=h is the height of the triangle,so angle ADC=90 degrees

Threfore angle ACD=180 degrees -angle CAD- angle ADC=180 degrees -23 degrees -90 degrees=67 degree

so area fo triangle ABC=1/2[AD*(BD-CD)]=

1/2[3.5cm*（3.5cm/tan(7pi/36)-3.5cm/tan67(pi/180)=1/2[3.5cm*3.5128561668618683cm]

=6.14749829200826952 cm^2

In triangle ABE,h (AD) is the height of the triangle

so triangle CAD is congruent to triangle EAD.

so CD=ED=3.5/tan(7pi/36) (the corresopding sides of congruent triangles are equal)

area of triangle ABE=1/2*[AD*BE]=1/2[AD*(BD+DE)]=1/2[3.5 cm*(3.5 cm/tan7(pi/36)+3.5 cm/tan(67pi/180)]=11.347314790575381975 cm^2

fiora  May 15, 2015
#5
+537
+8

Here is my graph from desmos.

https://www.desmos.com/calculator/mf39mbkv1i

Is it cool?and Thank You CPhill for correted my mistook.

fiora  May 15, 2015
#6
+537
0

LOL! You changed your question!.....

fiora  May 15, 2015
#7
+93691
+10

Hi Fiora,

Your Desmos graph has spun me out!

Your graph is brilliant but have you ever tried to use GeoGebra?

It is heaps easier for this and it is a free download.

Your GeoGebra diagram is really impressive too CPhill - You are becoming an expert

Melody  May 15, 2015