Question

11)
find the area
of the triangle
(and the triangle image with side 37, angles 68° and 70°)

Explanation:

Step1: Find the third angle

The sum of angles in a triangle is \(180^\circ\). Let the third angle be \(C\). So, \(C = 180^\circ - 68^\circ - 70^\circ = 42^\circ\).

Step2: Use the formula for area of a triangle \(\frac{1}{2}ab\sin C\)

We know one side \(a = 37\). Let's denote the sides opposite to angles \(A = 68^\circ\), \(B = 70^\circ\), \(C = 42^\circ\) as \(a\), \(b\), \(c\) respectively. Using the Law of Sines \(\frac{a}{\sin A}=\frac{b}{\sin B}\), we can find another side. Let's find side \(b\) (opposite \(70^\circ\)): \(\frac{37}{\sin 68^\circ}=\frac{b}{\sin 70^\circ}\), so \(b=\frac{37\sin 70^\circ}{\sin 68^\circ}\approx\frac{37\times0.9397}{0.9272}\approx37.7\). Now use the area formula with sides \(a = 37\), \(b\approx37.7\) and included angle \(C = 42^\circ\). Area \(=\frac{1}{2}\times37\times37.7\times\sin 42^\circ\). First, \(\sin 42^\circ\approx0.6691\). Then, \(\frac{1}{2}\times37\times37.7\times0.6691\approx\frac{1}{2}\times37\times25.23\approx\frac{1}{2}\times933.51\approx466.76\). (Alternatively, using the formula \(\frac{1}{2}a^2\frac{\sin B\sin C}{\sin A}\): \(\frac{1}{2}\times37^2\times\frac{\sin 70^\circ\sin 42^\circ}{\sin 68^\circ}\). Calculate \(\sin 70^\circ\approx0.9397\), \(\sin 42^\circ\approx0.6691\), \(\sin 68^\circ\approx0.9272\). So, \(\frac{1}{2}\times1369\times\frac{0.9397\times0.6691}{0.9272}\approx\frac{1369}{2}\times\frac{0.629}{0.9272}\approx684.5\times0.678\approx464.1\). The slight difference is due to rounding. A more accurate way: Let's use the formula \(\text{Area}=\frac{1}{2}ab\sin C\) where we can use the two angles and one side. Wait, actually, the formula for area when two angles and a side are known is \(\text{Area}=\frac{a^2\sin B\sin C}{2\sin A}\). Let's compute that precisely. \(a = 37\), \(A = 68^\circ\), \(B = 70^\circ\), \(C = 42^\circ\). So, \(\text{Area}=\frac{37^2\times\sin 70^\circ\times\sin 42^\circ}{2\times\sin 68^\circ}\). Calculate numerator: \(37^2 = 1369\), \(\sin 70^\circ\approx0.9396926\), \(\sin 42^\circ\approx0.6691306\), so numerator \(= 1369\times0.9396926\times0.6691306\approx1369\times0.62903\approx861.14\). Denominator: \(2\times\sin 68^\circ\approx2\times0.9271839\approx1.8543678\). Then, Area \(\approx\frac{861.14}{1.8543678}\approx464.4\). (Using more precise values for sines: \(\sin 68^\circ\approx0.9271838546\), \(\sin 70^\circ\approx0.9396926208\), \(\sin 42^\circ\approx0.6691306064\). So, \(\frac{37^2\times0.9396926208\times0.6691306064}{2\times0.9271838546}=\frac{1369\times0.62903027}{1.854367709}=\frac{861.1403}{1.854367709}\approx464.4\))

Answer:

Approximately \(464.4\) (or using a calculator with more precision, the value may vary slightly due to rounding during steps, but around \(464 - 467\))