Formula: tan∠T=√tan∠A²+tan∠B²
∠A=38.00°
∠T=41.30°
tan 41.3°=√tan(38°)²+tan(x°)²
Solve for x
+118654 tan 41.3°=√tan(38°)²+tan(x°)²
I suspect your brackets are in the wrong place. Maybe this is what you mean.
\(tan 41.3°=\sqrt{(tan38°)²+tan(x°)² }\\ (tan 41.3°)^2=(tan38°)²+(tanx°)² \\ (tan 41.3°)^2-(tan38°)²=(tanx°)² \\ (tanx°)²=(tan 41.3°)^2-(tan38°)² \\ tanx°=\pm \sqrt{(tan 41.3°)^2-(tan38°)²} \\ x°=180n\pm tan^{-1} \sqrt{(tan 41.3°)^2-(tan38°)²} \qquad n\in Z\\ x°=180n\pm 21.887^\circ \qquad n\in Z\\\)
