How do you solve the triangle HJK given h=18, j=10, k=23?

Then teach the underlying concepts
Don't copy without citing sources
preview
?

Write a one sentence answer...

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

1
Mar 25, 2018

color(brown)(hat H = 48.47^@, hat J = 24.57^@, hat K = 106.96^@

color(indigo)("Area of triangle " A_t = )color(indigo)(86.086 " sq units"

Explanation:

$\text{Given : " h = 18, j = 10, k = 23, " To solve the triangle}$

Applying Cosine Rule,

$\cos K = \frac{{h}^{2} + {j}^{2} - {k}^{2}}{2 \cdot h \cdot j}$

$\cos K = \frac{{18}^{2} + {10}^{2} - {23}^{2}}{2 \cdot 18 \cdot 10} = - 0.2917$

$\hat{K} = {\cos}^{- 1} - 0.2917 = {106.96}^{\circ}$

Similarly, $\cos H = \frac{{23}^{2} + {10}^{2} - {18}^{2}}{2 \cdot 23 \cdot 10} = 0.663$

$\hat{H} = {\cos}^{- 1} 0.663 = {48.47}^{\circ}$

Similarly, $\cos J = \frac{{23}^{2} + {18}^{2} - {10}^{2}}{2 \cdot 23 \cdot 18} = 0.9094$

$\hat{J} = {\cos}^{- 1} 0.9094 = {24.57}^{\circ}$

color(indigo)("Area of triangle " A_t = )(1/2) h j sin K = (1/2) * 18 * 10 * sin (106.96^@) = color(indigo)(86.086 " sq units"

Just asked! See more
• 12 minutes ago
• 13 minutes ago
• 16 minutes ago
• 21 minutes ago
• A minute ago
• A minute ago
• 4 minutes ago
• 12 minutes ago
• 12 minutes ago
• 12 minutes ago
• 12 minutes ago
• 13 minutes ago
• 16 minutes ago
• 21 minutes ago