Chapter 13: Q. 12 (page 870)

Solve the following triangle and determine its area.

Short Answer

Expert verified

The value of side: $a = 6$

$m ∠ B = 33^{o} m ∠ C = 90^{o}$

Step by step solution

Step 1. Given information.

Triangle is :

Step 2. Apply cosine formula to the triangle and find a.

$\cos (40^{o}) = \frac{5^{2} + 9^{2} - a^{2}}{2 \times 5 \times 9} \cos (40^{o}) = \frac{25 + 81 - a^{2}}{90} 106 - a^{2} = 90 \cos (40^{o}) 106 - a^{2} \approx 69 a^{2} \approx 106 - 69 a \approx 37 a \approx 6$

Step 3. Apply sine formula to the triangle.

$\frac{\sin (40^{o})}{6} = \frac{\sin (B)}{5} = \frac{\sin (C)}{9}$

By taking first two fractions,

$\frac{\sin (40^{o})}{6} = \frac{\sin B}{5} \sin B = \frac{5 \sin 40^{o}}{6} \sin B \approx \frac{3.21}{6} \sin B \approx 0.53 B \approx 33^{o}$

By taking first and last fraction,

$\frac{\sin (40^{o})}{6} = \frac{\sin (C)}{9} \sin C = \frac{9 \sin 40^{o}}{6} \sin C \approx 1 C = 90^{o}$

Step 4. To find the area of the triangle.

$A = \sqrt{s (s - a) (s - b) (s - c)}$

$s = \frac{a + b + c}{2} s = \frac{6 + 5 + 9}{2} s = 10$

$A = \sqrt{10 (10 - 6) (10 - 5) (10 - 9)} A = \sqrt{10 (4) (5) (1)} A = \sqrt{200} A = 10 \sqrt{2} A \approx 14.14$

So the area of the triangle is approximately 14.14 square units.

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Solve the following triangle and determine its area.

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Apply cosine formula to the triangle and find a.

Step 3. Apply sine formula to the triangle.

Step 4. To find the area of the triangle.

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