780 CHAPTER 8 Applications of Trigonometry Area of a Triangle A familiar formula for the area of a triangle is = 1 2 bh, where represents area, b base, and h height. This formula cannot always be used easily because in practice, h is often unknown. To find another formula, refer to acute triangle ABC in Figure 9(a) or obtuse triangle ABC in Figure 9(b). A perpendicular has been drawn from B to the base of the triangle (or the extension of the base). Consider right triangle ADB in either figure. sin A = h c , or h = c sin A Substitute into the formula for the area of a triangle. = 1 2 bh = 1 2 bc sin A Any other pair of sides and the angle between them could have been used. A C B D h c a Acute triangle ABC C B h c a A b D Obtuse triangle ABC (b) Figure 9 (a) Area of aTriangle (SAS) In any triangle ABC, the area is given by the following formulas. = 1 2 bc sin A, = 1 2 ab sin C, and = 1 2 ac sin B That is, the area is half the product of the lengths of two sides and the sine of the angle between them. NOTE If the included angle measures 90°, its sine is 1 and the formula becomes the familiar = 1 2 bh. EXAMPLE 8 Finding the Area of aTriangle (SAS) Find the area of triangle ABC in Figure 10. SOLUTION Substitute B = 55° 10′, a = 34.0 ft, and c = 42.0 ft into the area formula. = 1 2 ac sin B = 1 2 134.02142.02 sin 55° 10′ ≈586 ft2 S Now Try Exercise 75. 42.0 ft 34.0 ft B C A 558 109 Figure 10 b = 27.3 cm c a A B C 248 409 528 409 Figure 11 EXAMPLE 9 Finding the Area of aTriangle (ASA) Find the area of triangle ABC in Figure 11. SOLUTION Before the area formula can be used, we must find side a or c. 180° = A + B + C Angle sum formula B = 180° - 24° 40′ - 52° 40′ Substitute and solve for B. B = 102° 40′ Subtract. First find remaining angle B.
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