# If in a \[\Delta ABC\],\[\angle A = 45^\circ \],\[\angle C = 60^\circ \], then \[a + c\sqrt 2 \]

A. \[b\]

B. \[2b\]

C. \[\sqrt {2b} \]

D. \[\sqrt {3b} \]

Answer

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**Hint**

In this case we are given the values of some angles of like \[\angle A = {45^\circ },\angle C = {60^\circ }\] and asked to determine the value of \[a + c\sqrt 2 \] and we will use the extended sine rule to determine the relationship between the length of the triangle's sides and its circumradius to obtain the desired result.

**Formula used:**

Sine rule formula:

\[\frac{a}{{sinA}} = \frac{b}{{sinB}} = \frac{c}{{sinC}} = 2R\]

**Complete step-by-step solution:**

The given angle is \[A = 45^\circ \], \[C = 60^\circ \]

\[A + B + C = \pi \]

By substituting the values on the equation, it becomes

\[ = > B = 75^\circ \]

\[a + c\sqrt 2 = k\sin A + k\sin C(\sqrt 2 )\]

\[ = 2k(\frac{{\sqrt 3 + 1}}{{2\sqrt 2 }})\]

The values on the equation becomes,

\[ = 2k\sin 75^\circ \]

Then, the equation becomes

\[ = 2k\sin B\]

\[a + c\sqrt 2 = 2b\]

**So, option B is correct.**

**Note**

You need to first determine the lengths of ABC in order to solve this problem. The lengths of AB and BC are \[2\] and \[3\], respectively. As a result, dragging a downward will result in the intersection of AD and BC.

Two lines are said to intersect when they have exactly one point in common. There is a point at which the intersecting lines meet. The point of intersection is the same location that appears on all intersecting lines. There will be a place where the two coplanar, non-parallel straight lines intersect. Here, point O, the intersection point, is where lines A and B meet.

Last updated date: 30th Sep 2023

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