The angles: B = 35^{0}, A_{1} = 23^{0} and h = 3.5cm ;

(The angle A_{1} is located between *b *and *h *)

Find the area of a triangle ABC.

civonamzuk
May 14, 2015

#1**+18 **

BD = 3.5/tan(35°)

CD = 3.5*tan(23°)

Area of triangle ABC = 1/2(3.5^{2}/tan(35°) - 3.5^{2}*tan(23°))

$${\mathtt{Area}} = \left({\frac{{{\mathtt{3.5}}}^{{\mathtt{2}}}}{{\mathtt{2}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{35}}^\circ\right)}}}{\mathtt{\,-\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{23}}^\circ\right)}\right) \Rightarrow {\mathtt{Area}} = {\mathtt{6.147\: \!498\: \!292\: \!005\: \!575\: \!7}}$$

Area ≈ 6.147cm^{2}.

Alan
May 14, 2015

#1**+18 **

Best Answer

BD = 3.5/tan(35°)

CD = 3.5*tan(23°)

Area of triangle ABC = 1/2(3.5^{2}/tan(35°) - 3.5^{2}*tan(23°))

$${\mathtt{Area}} = \left({\frac{{{\mathtt{3.5}}}^{{\mathtt{2}}}}{{\mathtt{2}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{35}}^\circ\right)}}}{\mathtt{\,-\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{23}}^\circ\right)}\right) \Rightarrow {\mathtt{Area}} = {\mathtt{6.147\: \!498\: \!292\: \!005\: \!575\: \!7}}$$

Area ≈ 6.147cm^{2}.

Alan
May 14, 2015

#2**+8 **

Area of a triangle ABC = area ABD - area ACD

Area ABC = 8.75cm^{2} - 2.6cm^{2}

Area ABC = ≈6.15cm^{2}

civonamzuk
May 14, 2015

#3**+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**+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)]

BD=AD/tan(7pi/36)

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 CD=AD/tan(67pi/180)

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 angle ADE=90 degrees ,and angle DAE=angle CAD= 23 degree,plus segment AD=segment AD

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