Question

Find the area of the triangle with the given measurements. Round the solution to the nearest hundredth if necessary. B = 71°, a = 11 cm, c = 18 cm

Answer 1

$\boxed{\text{Area of the triangle:} \dfrac{a \times c \times sin(\measuredangle B)}{2}} \\ \\ A = \dfrac{a \times c \times sin(\measuredangle B)}{2} \\ \\ A= \dfrac{11 \times 18 \times sin71 ^o }{2} \\ \\ \\ Note : sin 71^o \approx 0,94 \\ \\ \\ A = \dfrac{198 \times 0,94}{2} \\ \\ A= \dfrac{186.12}{2} \\ \\ A=93.6 \ cm^2$

Answer 2

Answer:

93.606 cm^2

Step-by-step explanation:

Given that in a triangle two sides are 11 and 18 cm and included angle is 71 degrees

$a=11\\c=18\\angleB =71$

We know that area of a triangle when two sides and included angle is known is given by

Area of a triangle = $\frac{1}{2} ac sin B\\=\frac{1}{2} (11)(18)sin 71\\= 99 (0.9455)\\\\=-93.603$ cm^2