Question

calculate the value of all unknown sides.

Explanation:

Triangle 1: $\triangle MKL$ (right-angled at $K$)

Step1: Find $\angle L$

Sum of angles in triangle is $180^\circ$.
$\angle L = 180^\circ - 90^\circ - 40^\circ = 50^\circ$

Step2: Calculate side $KL$

Use tangent of $\angle M$: $\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$
$\tan(40^\circ)=\frac{KL}{10}$
$KL = 10 \times \tan(40^\circ) \approx 10 \times 0.8391 = 8.39$

Step3: Calculate side $ML$

Use cosine of $\angle M$: $\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$
$\cos(40^\circ)=\frac{10}{ML}$
$ML = \frac{10}{\cos(40^\circ)} \approx \frac{10}{0.7660} = 13.07$

Triangle 2: $\triangle GJH$ (right-angled at $J$)

Step4: Find $\angle H$

Sum of angles in triangle is $180^\circ$.
$\angle H = 180^\circ - 90^\circ - 55^\circ = 35^\circ$

Step5: Calculate side $GJ$

Use cosine of $\angle G$: $\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$
$\cos(55^\circ)=\frac{GJ}{20}$
$GJ = 20 \times \cos(55^\circ) \approx 20 \times 0.5736 = 11.47$

Step6: Calculate side $JH$

Use sine of $\angle G$: $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$
$\sin(55^\circ)=\frac{JH}{20}$
$JH = 20 \times \sin(55^\circ) \approx 20 \times 0.8192 = 16.38$

Answer:

For $\triangle MKL$:

• $\angle L = 50^\circ$
• $KL \approx 8.39$
• $ML \approx 13.07$

For $\triangle GJH$:

• $\angle H = 35^\circ$
• $GJ \approx 11.47$
• $JH \approx 16.38$