In triangle mathrm{FGH}, g=910 mathrm{~cm}, m angle mathrm{G}=98^{circ} and m angle mathrm{H}=51^{circ}. Find the length of h, to the nearest 1oth of a centimeter.

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Step:1 ：Given a triangle FGH with side g=910 cm, angle G=98°, and angle H=51°. We need to find the length of side h using the Law of Sines.Step:2 ：Using the Law of Sines, we have: (frac{h}{sin{H}} = frac{g}{sin{G}})Step:3 ：Solving for h, we get: (h = g cdot frac{sin{H}}{sin{G}})Step:4 ：Substituting the given values, we have: (h = 910 cdot frac{sin{51°}}{sin{98°}})Step:5 ：Calculating the value of h, we get: (h approx 714.2,text{cm})Step:6 ：(boxed{h approx 714.2,text{cm}})

Answer

Step: 1 ：Given a triangle FGH with side g=910 cm, angle G=98°, and angle H=51°. We need to find the length of side h using the Law of Sines.Step: 2 ：Using the Law of Sines, we have: (frac{h}{sin{H}} = frac{g}{sin{G}})Step: 3 ：Solving for h, we get: (h = g cdot frac{sin{H}}{sin{G}})Step: 4 ：Substituting the given values, we have: (h = 910 cdot frac{sin{51°}}{sin{98°}})Step: 5 ：Calculating the value of h, we get: (h approx 714.2,text{cm})Step: 6 ：(boxed{h approx 714.2,text{cm}})

In $\triangle \mathrm{FGH}, g=910 \mathrm{~cm}, m \angle \mathrm{G}=98^{\circ}$ and $m \angle \mathrm{H}=51^{\circ}$. Find the length of $h$, to the nearest 1oth of a centimeter.

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