1. **State the problem:** Calculate the measure of angle $B$ in triangle $\triangle ABC$ where sides $AB=9$, $AC=13$, and the height from $B$ to $AC$ is 10. 2. **Identify the triangle type and sides:** Given $AB=9$, $AC=13$, and height $BD=10$ perpendicular to $AC$, we want to find angle $B$. 3. **Use trigonometric ratios:** Since $BD$ is perpendicular to $AC$, we can use right triangle trigonometry on $\triangle ABD$ or $\triangle BDC$. 4. **Calculate length $AD$:** Since $BD=10$ is height, use Pythagoras in $\triangle ABD$: $$AB^2 = AD^2 + BD^2$$ $$9^2 = AD^2 + 10^2$$ $$81 = AD^2 + 100$$ $$AD^2 = 81 - 100 = -19$$ This is impossible, so $BD=10$ cannot be height from $B$ to $AC$ if $AB=9$. 5. **Re-examine the problem:** The problem states $AB=9$, $AC=13$, and height from $B$ to $AC$ is 10. Since $BD$ is perpendicular to $AC$, $BD$ must be less than $AB$. 6. **Calculate hypotenuse $BC$ using Pythagoras:** $$BC^2 = AB^2 + AC^2 = 9^2 + 13^2 = 81 + 169 = 250$$ $$BC = \sqrt{250} = 15.8114$$ 7. **Calculate angle $B$ using sine rule or cosine rule:** Use cosine rule: $$\cos B = \frac{AB^2 + BC^2 - AC^2}{2 \times AB \times BC}$$ $$= \frac{9^2 + 15.8114^2 - 13^2}{2 \times 9 \times 15.8114}$$ $$= \frac{81 + 250 - 169}{284.6052} = \frac{162}{284.6052} = 0.5695$$ 8. **Find angle $B$:** $$B = \cos^{-1}(0.5695) = 55.3^\circ$$ Rounded to nearest degree: $$B = 55^\circ$$ 9. **Calculate sine, cosine, and tangent of angle $B$:** $$\sin B = \sin 55^\circ = 0.8192$$ $$\cos B = 0.5695$$ $$\tan B = \frac{\sin B}{\cos B} = \frac{0.8192}{0.5695} = 1.438$$