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Construct a triangle ABC in which $BC\text{ }=\text{ }7cm$ $\angle B\text{ }=\text{ }{{75}^{o}}$ and $AB\text{ }+\text{ }AC\text{ }=\text{ }13cm$.

Answer
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Hint:
Here we will use compass to draw angles as well as to cut arc of required length and then we will start construction step by step where we will draw angle ${75}^{\circ}$ on the base side and the then cut an arc of length equal to the addition of other two sides on the ray making angle ${75}^{\circ}$ and we will join that point to the base point and then we will draw perpendicular bisector on the joined line to get the required triangle.

Complete step by step solution:
1) Step 1: Draw line BC which is 7cm in length 7cm.
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2) Step 2 Now, draw $\angle B\text{ }=\text{ }{{75}^{o}}$ using compass.
 Let the ray be BX.
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3) Step 3: Cut an arc of length 13 cm on ray BX using compass and name the point as D.
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4) Step 4: Join CD.
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5) Step 5: Now draw the perpendicular bisector of CD.
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6) Step 6: Mark point A where the perpendicular bisector intersects the line BD.
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7) Step 7: join the points CA.
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$\therefore \vartriangle \text{ABC}$ is the required triangle.

Note:
Always use a compass to draw the required angle. Here in this question, we have to draw ${75}^{\circ}$ . So for that we will first draw ${90}^{\circ}$ using compass and then we will bisect ${60}^{\circ}$ and ${90}^{\circ}$ to get ${75}^{\circ}$. Always use a compass wherever possible to draw angle with it in construction questions.