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In △ ABC, a and B are acute angles. The opposite sides of angles a, B and C are a, B, C, cos2a = 3 / 5, SINB = √ 10 / 10, A-B = √ 2-1 respectively. Find the value of a, B and C
cos2A=1-2(sinA)^2=3/5sinA=√5/5cosA=2√5/5sinB=√10/10cosB=3√10/10cosC=-cos(A+B)=-cosAcosB+sinAsinB=-√2/2sinC=√2/2c/sinC=(a-b)/(sinA-sinB)c=√55=c^2=a^2+b^2-2abcosC=a^2+b^2+√2ab=(a-b)^2+2ab+√2aba...
In △ ABC, a and B are acute angles, and cos2a = 3 / 5, SINB = √ 10 / 10?
A. B is an acute angle, cos2a = 2cos & # 178; A-1 = 3 / 5, so cosa = 2 √ 5 / 5sina = √ (1-cos & # 178; a) = √ 5 / 5cosb = √ (1-sin & # 178; b) = 3 √ 10 / 101. Sin (a + B) = sinacosb + sinbcosa = (√ 5 / 5) (3 √ 10 / 10) + (√ 10 / 10) (2 √ 5 / 5) = √ 2 / 2, so a + B = 135 ° 2
The length of three sides of a triangle is ABC. If a and B are equal to 10, AB is equal to 18, and C is equal to 8, try to find the area of the triangle
If you are a junior high school student, you can do this: a + B = 10, ab = 18, a = (28 under 10 + radical) / 2, a = (28 under 10 - radical) / 2, so the square of a + B = 64 = 8 square = C square, so the triangle is a right triangle with a and B as right angles, so the area is 1 / 2 × (28 under 10 + radical) / 2 × (28 under 10 - radical) / 2
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