{"id":2124,"date":"2025-08-04T18:14:20","date_gmt":"2025-08-04T18:14:20","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2124"},"modified":"2025-08-13T16:53:47","modified_gmt":"2025-08-13T16:53:47","slug":"non-right-triangles-with-law-of-sines-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/non-right-triangles-with-law-of-sines-learn-it-3\/","title":{"raw":"Non-right Triangles with Law of Sines: Learn It 3","rendered":"Non-right Triangles with Law of Sines: Learn It 3"},"content":{"raw":"

Finding the Area of an Oblique Triangle Using the Sine Function<\/h2>\r\nNow that we can solve a triangle for missing values, we can use some of those values and the sine function to find the area of an oblique triangle. Recall that the area formula for a triangle is given as [latex]\\text{Area}=\\frac{1}{2}bh[\/latex], where [latex]b[\/latex] is base and [latex]h[\/latex] is height. For oblique triangles, we must find [latex]h[\/latex] before we can use the area formula. Observing the two triangles, one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric property [latex]\\sin \\alpha =\\frac{\\text{opposite}}{\\text{hypotenuse}}[\/latex] to write an equation for area in oblique triangles. In the acute triangle, we have [latex]\\sin \\alpha =\\frac{h}{c}[\/latex] or [latex]c\\sin \\alpha =h[\/latex]. However, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the base [latex]b[\/latex] to form a right triangle. The angle used in calculation is [latex]{\\alpha }^{\\prime }[\/latex], or [latex]180-\\alpha [\/latex].\r\n\r\n\"Two\r\n\r\nThus,\r\n
[latex]\\text{Area}=\\frac{1}{2}\\left(\\text{base}\\right)\\left(\\text{height}\\right)=\\frac{1}{2}b\\left(c\\sin \\alpha \\right)[\/latex]<\/div>\r\nSimilarly,\r\n
[latex]\\text{Area}=\\frac{1}{2}a\\left(b\\sin \\gamma \\right)=\\frac{1}{2}a\\left(c\\sin \\beta \\right)[\/latex]<\/div>\r\n
\r\n

area of an oblique triangle<\/h3>\r\nThe formula for the area of an oblique triangle is given by\r\n

[latex]\\begin{align}\\text{Area}&=\\frac{1}{2}bc\\sin \\alpha \\\\ &=\\frac{1}{2}ac\\sin \\beta \\\\ &=\\frac{1}{2}ab\\sin \\gamma \\end{align}[\/latex]<\/p>\r\nThis is equivalent to one-half of the product of two sides and the sine of their included angle.\r\n\r\n<\/section><\/div>\r\n

Find the area of a triangle with sides [latex]a=90,b=52[\/latex], and angle [latex]\\gamma =102^\\circ [\/latex]. Round the area to the nearest integer.[reveal-answer q=\"99495\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"99495\"]Using the formula, we have\r\n

[latex]\\begin{align}\\text{Area}&=\\frac{1}{2}ab\\sin \\gamma \\\\ \\text{Area}&=\\frac{1}{2}\\left(90\\right)\\left(52\\right)\\sin \\left(102^\\circ \\right) \\\\ \\text{Area}&\\approx 2289\\text{square units} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

Find the area of the triangle given [latex]\\beta =42^\\circ ,a=7.2\\text{ft},c=3.4\\text{ft}[\/latex]. Round the area to the nearest tenth.[reveal-answer q=\"426440\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"426440\"]about\u00a08.2 square feet[\/hidden-answer]<\/section>
[ohm_question hide_question_numbers=1]97465[\/ohm_question]<\/section>","rendered":"

Finding the Area of an Oblique Triangle Using the Sine Function<\/h2>\n

Now that we can solve a triangle for missing values, we can use some of those values and the sine function to find the area of an oblique triangle. Recall that the area formula for a triangle is given as [latex]\\text{Area}=\\frac{1}{2}bh[\/latex], where [latex]b[\/latex] is base and [latex]h[\/latex] is height. For oblique triangles, we must find [latex]h[\/latex] before we can use the area formula. Observing the two triangles, one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric property [latex]\\sin \\alpha =\\frac{\\text{opposite}}{\\text{hypotenuse}}[\/latex] to write an equation for area in oblique triangles. In the acute triangle, we have [latex]\\sin \\alpha =\\frac{h}{c}[\/latex] or [latex]c\\sin \\alpha =h[\/latex]. However, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the base [latex]b[\/latex] to form a right triangle. The angle used in calculation is [latex]{\\alpha }^{\\prime }[\/latex], or [latex]180-\\alpha[\/latex].<\/p>\n

\"Two<\/p>\n

Thus,<\/p>\n

[latex]\\text{Area}=\\frac{1}{2}\\left(\\text{base}\\right)\\left(\\text{height}\\right)=\\frac{1}{2}b\\left(c\\sin \\alpha \\right)[\/latex]<\/div>\n

Similarly,<\/p>\n

[latex]\\text{Area}=\\frac{1}{2}a\\left(b\\sin \\gamma \\right)=\\frac{1}{2}a\\left(c\\sin \\beta \\right)[\/latex]<\/div>\n
\n
\n

area of an oblique triangle<\/h3>\n

The formula for the area of an oblique triangle is given by<\/p>\n

[latex]\\begin{align}\\text{Area}&=\\frac{1}{2}bc\\sin \\alpha \\\\ &=\\frac{1}{2}ac\\sin \\beta \\\\ &=\\frac{1}{2}ab\\sin \\gamma \\end{align}[\/latex]<\/p>\n

This is equivalent to one-half of the product of two sides and the sine of their included angle.<\/p>\n<\/section>\n<\/div>\n

Find the area of a triangle with sides [latex]a=90,b=52[\/latex], and angle [latex]\\gamma =102^\\circ[\/latex]. Round the area to the nearest integer.<\/p>\n