Calculate the angle alpha | Learn 3 Strategies for Finding Missing Angles

In this lesson, we calculated the missing angle in the given triangle in three different ways. During our calculations, we utilized properties of isosceles triangles, similar triangles, and the law of sines. Please watch the video until the end to learn these methods.

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[jimlocke]

A fourth method: Drop a perpendicular from A to BC and label the intersection as point E. AE = AB sin (20°) = (1)(sin(20°) = sin(20°). DE = AE/tan(80°) = sin(20°)/tan(80°). CE = 2 - DE. So tan(α) = sin(20°)/(2 - (sin(20°/tan(80°)). Using my scientific calculator, I find that α = 10° to within the precision of my calculator. So, I have a hint! If I can rewrite the expression in a manner where all the trigonometric functions are functions of 10°, the resulting expression should simplify to tan(10°). Note that tan(10°) = 1/tan(80°), so I can replace a trigonometric function of 80° with one of 10°: tan(α) = sin(20°)/(2 - (sin(20°)tan(10°))). To replace the 20° with 10°, apply the sine double angle formula sin(2Θ) = 2sin(Θ)cos(Θ) twice: tan(α) = (2sin(10°)cos(10°))/(2 - (2sin(10°)cos(10°)tan(10°)). Note that tan(Θ) = (sin(Θ))/(cos(Θ)), so replace tan(10°) by sin(10°)/cos(10°) and simplify: tan(α) = (2sin(10°)cos(10°))/(2 - 2sin²(10°)). Note that sin²(Θ) + cos²(Θ) = 1, so 2 - 2sin²(10°) = 2cos²(10°) and tan(α) = Factor out 2cos(10°) from numerator and denominator: tan(α) = (sin(10°))/(cos(10°)). However (sin(Θ))/(cos(Θ)) = tan (Θ) so tan(α) = tan(10°) and α = 10°, as GeoMathry also found.

[solimana-soli]

ADB angle is 100 degree then 1/sin 100, why 80?