Question

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

Answer 1

In this problem, we have to solve the $\Delta HJK$ from the given h = 18, j = 10, k = 23. We can use cosine rule from the Laws of cosine, to find H, J, K by substituting the values of h, j, k in the formula for the cosine rule. We can substitute the values for the area of the triangle to find its area.

Complete step by step solution:
We have to solve the, $\Delta HJK$.
We are given h = 18, j = 10, k = 23.
We know that the cosine rule is,

& \cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc} \\\ & \cos B=\dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac} \\\ & \cos C=\dfrac{{{b}^{2}}+{{a}^{2}}-{{c}^{2}}}{2ab} \\\ \end{aligned}$$ We can now apply the cosine rule for H, J, K, we get  $$\Rightarrow \cos K=\dfrac{{{h}^{2}}+{{j}^{2}}-{{k}^{2}}}{2hj}$$ We can now substitute the value h = 18, j = 10, k = 23 in the above formula, we get $$\Rightarrow \cos K=\dfrac{{{18}^{2}}+{{10}^{2}}-{{23}^{2}}}{2\times 18\times 10}$$ We can now simplify the above step, we get $$\Rightarrow \cos K=\dfrac{{{18}^{2}}+{{10}^{2}}-{{23}^{2}}}{2\times 18\times 10}=-0.2917$$ We can now take cos inverse, we get $$\begin{aligned} & \Rightarrow K={{\cos }^{-1}}\left( -0.2917 \right) \\\ & \Rightarrow K={{106.96}^{\circ }} \\\ \end{aligned}$$ Similarly, we can find for H, we get $$\begin{aligned} & \Rightarrow \cos H=\dfrac{{{23}^{2}}+{{10}^{2}}-{{18}^{2}}}{2\times 23\times 10}=0.663 \\\ & \Rightarrow H={{\cos }^{-1}}\left( 0.663 \right) \\\ & \Rightarrow H={{48.47}^{\circ }} \\\ \end{aligned}$$ $$$$ We can find J, $$\begin{aligned} & \Rightarrow \cos J=\dfrac{{{23}^{2}}+{{18}^{2}}-{{10}^{2}}}{2\times 23\times 18}=0.9094 \\\ & \Rightarrow J={{\cos }^{-1}}\left( 0.9094 \right) \\\ & \Rightarrow J={{24.57}^{\circ }} \\\ \end{aligned}$$ Now we have to solve $$\Delta HJK$$. Area of $$\Delta HJK$$$$=\dfrac{1}{2}hj\sin K$$. We can substitute the required values, we get Area of $$\Delta HJK$$$$=\dfrac{1}{2}\times 18\times 10\times \sin \left( {{106.96}^{\circ }} \right)$$. We can now simplify the above step, we get Area of $$\Delta HJK$$$$=86.086$$square units. Therefore, the Area of $$\Delta HJK$$$$=86.086$$ square units. **Note:** Students make mistakes while finding the value of H, J, K, by using the cosine rule. We can use a calculator to find the cos inverse values to get the value of angles. We should know that we can solve a triangle by finding the area of the triangle. We can use a calculator to find the trigonometric degree values.