Question

Complete the steps to determine how many triangles abc can exist with the given measures. b = 18.7, c = 16.1 and c = 35° what is the approximate value of the height, h, of the triangle? round to the nearest tenth. brainly

Answer 1

Answer with explanation:

it is given that ,In ΔABC, b=18.7 units,c=16.1 units and Angle C=35°.

Using law of Sines

$\frac{a}{SinA}=\frac{b}{SinB}=\frac{c}{SinC}\\\\\frac{18.7}{SinB^{\circ}}=\frac{16.1}{Sin35^{\circ}}\\\\ Sin B^{\circ}=\frac{18.7\times 0.57358}{16.1}\\\\Sin B^{\circ}=0.66620=0.67\\\\ B^{\circ}=60^{\circ}$

Sum of angles of triangle is equal to 180°.

∠A+∠B+∠C=180°

∠A+60°+35°=180°

∠A=180°-95°

∠A=85°

$\frac{a}{Sin85^{\circ}}=\frac{c}{SinC}\\\\\frac{a}{Sin85^{\circ}}=\frac{16.1}{Sin35^{\circ}}\\\\a=\frac{16.1 \times0.57358}{0.99619}\\\\ a=9.26$

  Also, Area of triangle is given by

           $=\frac{abSinC}{2}=\frac{cbSinA}{2}=\frac{acSinB}{2}$

         Area of triangle (ΔABC)

            $=\frac{cbSinA}{2}\\\\=\frac{18.7\times 16.1\times sin85^{\circ}}{2}$  

              =150.535 × 0.99619

              =149.9614=149.96 square units(Approx)

It is not given ,we have to find the length of altitude along which side.

Supposing that. Altitude is drawn from vertex B , on the side AC,and altitude cuts the side AC at point M.

BM= h units

Also, Area of Triangle

   $=\frac{\text{Base}\times{\text{Height}}}{2}}$

$149.96=\frac{18.7 \times \text{Height}}{2}\\\\ \frac{299.9228}{18.7}=\text{height}$

So, Height =16.038 units

BM=Length of altitude on side CA=16.038 units(approx)

$149.96=\frac{16.7 \times \text{Height(CN)}}{2}\\\\ \frac{299.9228}{16.7}=\text{height}$

So, Height =17.95 units

C N=Length of altitude on side BA=17.95 units(approx)

$149.96=\frac{9.26 \times \text{Height(CN)}}{2}\\\\ \frac{299.9228}{9.26}=\text{height}$

So, Height =32.39 units

AO=Length of altitude on side BC=32.39 units(approx)

Answer 2

Answer:

The answer is 10.7